Using ANSYS for Finite Element Analysis, Volume I: A Tutorial for Engineers [1] 1947083201, 9781947083202

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Using ANSYS for Finite Element Analysis, Volume I: A Tutorial for Engineers [1]
 1947083201, 9781947083202

Table of contents :
Cover
Contents
List of Figures
Preface
Chapter 1: Introduction to Finite Element Analysis
Chapter 2: Static Analysis Using ANSYS
Chapter 3: Geometric Modeling
Chapter 4: Static Analysis Using Line Elements
Chapter 5: Static Analysis Using Area Elements
Chapter 6: Static Analysis Using Volume Elements
Chapter 7: Thermal Stress Analysis
Summary
Bibliography
Index
Adpage
Backcover

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A Tutorial for Engineers, Volume I Wael A. Altabey • Mohammad Noori • Libin Wang Over the past two decades, the use of finite element method as a design tool has grown rapidly. Easy to use commercial software, such as ANSYS, have become common tools in the hands of students as well as practicing engineers. The objective of this book is to demonstrate the use of one of the most commonly used Finite Element Analysis software, ANSYS, for linear static, dynamic, and thermal analysis through a series of tutorials and examples. Some of the topics covered in these tutorials include development of beam, frames, and Grid Equations; 2-D elasticity problems; dynamic analysis; composites, and heat transfer problems. These simple, yet, fundamental tutorials are expected to assist the users with the better understanding of finite element modeling, how to control modeling errors, and the use of the FEM in designing complex load bearing components and structures. These tutorials would supplement a course in basic finite element or can be used by practicing engineers who may not have the advanced training in finite element analysis. Wael A. Altabey is an assistant professor in the department of mechanical engineering, faculty of engineering, Alexandria University, Alexandria, Egypt and has been a postdoctoral researcher at the International Institute for Urban Systems Engineering, Southeast University, Nanjing, China. Mohammad Noori is a professor of mechanical engineering at California Polytechnic State University in San Luis Obispo, California, USA, and a fellow of the American Society of Mechanical Engineers. Dr Noori has over 34 years of experience as a scholar and educator. He has also been a distinguished visiting professor at the International Institute for Urban Systems Engineering, Southeast University, Nanjing, China. Libin Wang is a professor and the dean of the school of civil engineer-

For further information, a free trial, or to order, contact:  [email protected]

ing at Nanjing Forestry University, in Nanjing, China. He has been an educator and scholar, for over 20 years, and has taught the subject of finite element analysis both at the undergraduate and graduate level.

ISBN: 978-1-94708-320-2

Using ANSYS for Finite Element Analysis, Volume I

• Manufacturing Engineering • Mechanical & Chemical Engineering • Materials Science & Engineering • Civil & Environmental Engineering • Advanced Energy Technologies

Using ANSYS for Finite Element Analysis

ALTABEY • NOORI • WANG

EBOOKS FOR THE ENGINEERING LIBRARY

SUSTAINABLE STRUCTURAL SYSTEMS COLLECTION Mohammad Noori, Editor

Using ANSYS for Finite Element Analysis A Tutorial for Engineers Volume I

Wael A. Altabey Mohammad Noori Libin Wang

Using ANSYS for Finite Element Analysis

Using ANSYS for Finite Element Analysis A Tutorial for Engineers Volume I

Wael A. Altabey, Mohammad Noori, and Libin Wang

MOMENTUM PRESS, LLC, NEW YORK

Using ANSYS for Finite Element Analysis: A Tutorial for Engineers, Volume I Copyright © Momentum Press®, LLC, 2018. 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, photocopy, recording, or any other—except for brief quotations, not to exceed 400 words, without the prior permission of the publisher. First published by Momentum Press®, LLC 222 East 46th Street, New York, NY 10017 www.momentumpress.net ISBN-13: 978-1-94708-320-2 (print) ISBN-13: 978-1-94708-321-9 (e-book) Momentum Press Sustainable Structural Systems Collection Collection ISSN: 2376-5119 (print) Collection ISSN: 2376-5127 (electronic) Cover and interior design by Exeter Premedia Services Private Ltd., Chennai, India 10 9 8 7 6 5 4 3 2 1 Printed in the United States of America

Abstract Finite Element Method (FEM) is a well-established technique for analyzing the behavior and the response of structures or mechanical components under static, dynamic, or thermal loads. Over the past two decades the use of finite element analysis as a design tool has grown rapidly. Easy to use commercial software have become common tools in the hands of students as well as practicing engineers. The objective of this two volume book is to demonstrate the use of one of the most commonly used Finite ­Element Analysis software, ANSYS, for linear static, dynamic, and ­thermal analysis through a series of tutorials and examples. Some of the topics and ­concepts covered in these tutorials include development of beam, frames, and grid equations; 2-D elasticity problems; dynamic analysis; and heat transfer problems. We are hoping these simple, yet, fundamental ­tutorials will assist the users with the better understanding of finite element ­modeling, how to control modeling errors, the safe use of the FEM in support of designing complex load bearing components and structures. There are many good textbooks currently used for teaching the fundamentals of finite element methods. There are also detailed users manuals available for commercial software (ANSYS). However, those sources are useful for advanced students and users. Therefore, there was a need to develop a tutorial that would supplement a course in basic finite element or can be used by practicing engineers who may not have the advanced training in finite element analysis. That is the gap addressed by this book.

Keywords ANSYS, composite materials, Dynamics, Failure analysis, Fatigue loads, FEM, optimization, statistics

Contents List of Figures

ix

Preface

xi

1  Introduction to Finite Element Analysis 1.1  Finite Element Method

1 1

1.2  Review Topics

15

1.3  General Steps of FEM

26

1.4  The Most Common Finite Element Types

27

1.5 Types of Element Formulation Methods

27

1.6 Derivation of Spring Element Equations Using Direct Method

28

1.7 Bar Element Formulation Using Direct Method

31

1.8  Examples of Linear FEM

34

1.9  FEA: Modeling, Errors, and Accuracy

48

2  Static Analysis Using ANSYS

53

2.1  Overview of Structural Analysis

53

2.2  Static Analysis Procedure

54

3  Geometric Modeling

65

3.1 Typical Steps Involved in Model Generation Within ANSYS

65

3.2 Importing Solid Models Created in CAD Systems

66

3.3  Solid Modeling

68

3.4 Tutorial 1: Solid Modeling Using 2D Primitives

69

3.5 Tutorial 2: Solid Modeling Using 3D Primitives

79

viii  •  Contents

4  Static Analysis Using Line Elements 4.1 Tutorial 3: Static Analysis Using Truss Elements

91 91

4.2 Tutorial 4 (a): Static Analysis Using BEAM Elements

112

4.3 Tutorial 4 (b): Static Analysis Using Beam Elements with Distributed Load

115

5  Static Analysis Using Area Elements

121

5.1 Tutorial 5: Static Analysis Using Area Elements: Plane Problem (Bracket)

121

5.2 Tutorial 6: Static Analysis Using Area Elements: Plane Problem (Wrench)

141

6  Static Analysis Using Volume Elements

147

6.1 Tutorial 7: Static Analysis Using Volume Elements: Component Design

147

6.2 Tutorial 8: Static Analysis Using Volume Elements: Assembly Design

153

7  Thermal Stress Analysis

161

7.1 Tutorial 9: Thermal Analysis of Mechanical Structure

161

7.2 Tutorial 10 (a): Thermal-Stress Analysis-Sequential Coupled Field

168

7.3 Tutorial 10 (b): Thermal-Stress Analysis: Direct-Coupled Field177 Summary

185

Bibliography

187

Index

189

List of Figures Figure 1.1. A diagram of the two common branches of the general modeling solution.

2

Figure 1.2. The thermomechanical stresses in an air-cooled turbine blade depicted.

3

Figure 1.3. The complex three-dimensional geometry of the blade along with the combined thermal and mechanical loadings.

3

Figure 1.4.  The mathematical FEM.

5

Figure 1.5.  The physical FEM.

6

Figure 1.6.  A two dimensional FE model for a gear tooth.

7

Figure 1.7.  Linear and nonlinear behavior of the structure.

9

Figure 1.8.  Three-dimensional body.

17

Figure 1.9.  Equilibrium of elemental volume.

18

Figure 1.10. (a) Plane stress (b) Plane strain.

21

Figure 1.11.  Outline of an FE analysis project.

49

Figure 1.12. (a) A tapered bar loaded by axial force P, (b) Discretization of the bar into four uniform ­ two-node elements of equal length.

50

Figure 1.13. Lateral midpoint displacement versus time for a beam loaded by a pressure pulse. The material is ­elastic-perfectly plastic. Plots were generated by various users and various codes.

51

Preface The finite element method (FEM) is a well-established technique for analyzing the structural behavior of mechanical components and systems. In recent years, the use of finite element analysis as a design tool has grown rapidly. Easy-to-use commercial software have become common tools in the hands of students, as well as practicing engineers. The objectives of this work include: • To teach students the basic concepts in the linear FEM as related to solving engineering problems in solids and heat transfer. • To provide students with a working knowledge of finite element analysis tools and their use in mechanical design. • The topics covered in this course include: introduction to finite ­element; finite element formulation; introduction to a general FE software (ANSYS); development of beam, frames, and grid equations; 2D elasticity problems; dynamic analysis; and heat ­ transfer problems. Gain insight into appropriate use of finite element modeling, understand how to control modeling errors, benefit from hands-on exercise at the computer workstation, and understand the safe use of FEM in support of designing complex load-bearing components and structures. There are many good textbooks already in existence that cover the theory of FEMs. Similarly, there are detailed user manuals available for commercial software (ANSYS). But, these are useful for advanced ­students and users. Therefore, there was a need to develop a computer ­session manual in line with the flow of the course and utilizing the ­software platform available in the department. Students will be able to acquire the required level of understanding and skill in modeling, analysis, validation, and report generation for various design problems.

xii  •  Preface

This work could also be very helpful for the students of senior design (mechanical system design) and (FEA for large deformation problems). In addition, it could be used for computer sessions of short courses on stress analysis techniques and finite element analysis offered by the Mechanical Engineering department. After giving a brief introduction to the finite element analysis and modeling, various guided tutorials have been included in this manual. Several new tutorials have been developed and others adapted from ­different sources including ANSYS manuals in two volumes, ANSYS workshops and Internet resources. Tutorials have been arranged in each volume according to the flow of the course and cover topics such as solid modeling using 2D and 3D primitives available in ANSYS, static structural analysis (truss, beam, 2D and 3D structures), dynamic analysis ­(harmonic and modal analysis), and thermal analysis. Wael A. Altabey, Assistant Professor, Department of Mechanical Engineering, Faculty of Engineering, Alexandria University, Alexandria (21544), Egypt. Postdoctoral follow, International Institute for Urban Systems Engineering, Southeast University, Nanjing, Jiangsu (210096), China. Mohammad Noori, Professor, Department of Mechanical Engineering, California Polytechnic State University, San Luis Obispo, CA 93405, USA. E-mail: [email protected] [email protected]

Chapter 1

Introduction to Finite Element Analysis 1.1 Finite Element Method The field of mechanics can be subdivided into three major areas: theoretical, applied, and computational. Theoretical mechanics deals with fundamental laws and principles of mechanics studied for their intrinsic scientific value. Applied mechanics transfers this theoretical knowledge to scientific and engineering applications, especially through the construction of mathematical models of physical phenomena. Computational mechanics solves specific problems by simulation through numerical methods implemented on digital computers. One of the most important advances in applied mathematics in the 20th century has been the development of the finite element method as a general mathematical tool for obtaining approximate solutions to boundary-value problems. The theory of finite elements draws on almost every branch of mathematics and can be considered as one of the richest and most diverse bodies of the current mathematical knowledge. 1.1.1 Mathematical Modeling of Physical Systems In general, engineering problems are mathematical models of physical situations. Two main goals of engineering analysis are to be able to identify the basic physical principle(s) and fundamental laws that govern the behavior of a system or a control volume and to translate those principles into a mathematical model involving an equation or equations that can be solved accurately to predict qualitative and quantitative behavior of the system. The resulting mathematical model is frequently a single

2  •   Using ANSYS for Finite Element Analysis

d­ ifferential equation or a set of differential equations with a set of corresponding boundary and initial conditions whose solution should be consistent with and accurately represent the physics of the system. These governing equations represent balance of mass, force, or energy. When possible, the exact solution of these equations renders detailed behavior of a system under a given set of conditions. In situations where the system is relatively simple, it may be possible to analyze the problem by using some of the classical methods learned in elementary courses in ordinary and partial differential equations. Far more frequently, however, there are many practical engineering problems for which we cannot obtain exact solutions. This inability to obtain an exact solution may be attributed to either the complex nature of governing differential equations or the difficulties that arise from dealing with the boundary and initial conditions. To deal with such problems, we resort to numerical approximations. In contrast to analytical solutions, which show the exact behavior of a system at any point within the system, numerical solutions approximate exact solutions only at discrete points, called nodes. Due to the complexity of physical systems, some approximation must be made in the process of turning physical reality into a mathematical model. It is important to decide at what points in the modeling process these approximations are made. This, in turn, determines what type of analytical or computational scheme is required in the solution process. Let us

Physical problem

Simplified model

Exact solution for approximate model

Complicated model

Approximate solution for exact model

FEM approach

Figure 1.1.  A diagram of the two common branches of the general modeling solution.

Introduction to Finite Element Analysis   •  3

consider a diagram of the two common branches of the general modeling solution process as shown in Figure 1.1. For many real-world problems, the second approach is in fact the only possibility. For instance, suppose that the aim is to find the thermomechanical stresses in an air-cooled turbine blade depicted in Figure 1.2. Figure 1.3 shows the complex three-dimensional geometry of the blade along with the combined thermal and mechanical loadings, which makes the analysis of the blade a formidable task. Nevertheless, many powerful commercial finite element packages are available that can be implemented to perform this task with relative ease.

Figure 1.2.  The thermomechanical stresses in an air-cooled turbine blade depicted.

Temperature max

min

Equivalent stress min

max

Figure 1.3.  The complex three-dimensional geometry of the blade along with the combined thermal and mechanical loadings.

4  •   Using ANSYS for Finite Element Analysis

1.1.2  Basic Concept of Numerical Methods The basic concept of these methods is based on the idea of building a complicated object with simple blocks, or, dividing a complicated object into small and manageable pieces. Application of this simple idea can be found everywhere in everyday life, as well as in engineering. ­Examples include Lego (kids’ play), buildings, and approximation of the area of a circle: Element Si

Area of one triangle = Si = N

Area of one circle = ∑Si = i =1

1 2 R sin qi 2 1 2  2p  R N sin   N 2

Where N = total number of triangles (elements). The first step of any numerical procedure is discretization. This process divides the medium of interest into a number of small subregions and nodes. There are two common classes of numerical methods: finite difference methods and finite element methods (FEMs). With finite difference methods, the differential equation is written for each node, and the derivatives are replaced by difference equations. This approach results in a set of simultaneous linear equations. Although finite difference methods are easy to understand and employ in simple problems, they become ­difficult to apply to problems with complex geometries or complex boundary ­conditions. This situation is also true for problems with nonisotropic ­properties. By contrast, FEM uses integral formulations, rather than difference equations to create a system of algebraic equations. Moreover, an approximate continuous function is assumed to represent the solution for each element. The complete solution is then generated by connecting or assembling the individual solutions, allowing for continuity at the ­interelemental boundaries. Thus, FEM is a numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.

Introduction to Finite Element Analysis   •  5

1.1.3 A Brief History of FEM Finite element analysis (FEA) was first developed in 1943 by R. ­Courant, who utilized the Ritz method of numerical analysis and minimization of variational calculus to obtain approximate solutions to vibration systems. Shortly thereafter, a paper published in 1956 by M. J. Turner, R. W. Clough, H. C. Martin, and L. J. Topp established a broader definition of numerical analysis. The paper centered on the “stiffness and deflection of complex structures.” By the early 1970s, FEA was limited to expensive mainframe computers generally owned by the aeronautics, automotive, defense, and nuclear industries. Since the rapid decline in the cost of computers and the phenomenal increase in computing power, FEA has been developed to an incredible precision. Present day super computers are now able to produce accurate results for all kinds of parameters. 1.1.4 The FEM Analysis Process A model-based simulation process using FEM consists of a sequence of steps. This sequence takes two basic configurations depending on the environment in which FEM is used. These are referred to as the mathematical FEM and the physical FEM. The mathematical FEM as depicted in Figure 1.4, the centerpiece in the process steps of the mathematical FEM is the mathematical mode, which is often an ordinary or partial differential equation in space and time. Using the methods provided by the variational calculus, a discrete finite element model is generated from the mathematical model. The resulting FEM equations are processed by an equation solver, which provides a discrete solution. In this process, we may also think of an ideal physical system, which may be regarded as a realization of the mathematical model. For example, if the mathematical model is the ­ Mathematical model FEM Physical problem

Complicated model

Verification discretization + solution error

Solution

Verification solution error

Figure 1.4.  The mathematical FEM.

Discrete solution

6  •   Using ANSYS for Finite Element Analysis Mathematical model Physical problem

Complicated model

Solution

Discrete solution

Verification solution error Simulation error = Modeling + Solution error

Figure 1.5.  The physical FEM.

Poisson’s equation, realizations may be a heat conduction problem. In mathematical FEM, this step is unnecessary and indeed FEM discretizations may be constructed without any reference to physics. The concept of error arises when the discrete solution is substituted in the mathematical and discrete models. This replacement is generically called verification. The solution error is the amount by which the discrete solution fails to satisfy the discrete equations. This error is relatively unimportant when using computers. More relevant is the discretization error, which is the amount by which the discrete solution fails to satisfy the mathematical model. The physical FEM is depicted in Figure 1.5; in the physical FEM process, the centerpiece is the physical system to be modeled. The processes of idealization and discretization are carried out concurrently to produce the discrete model. Indeed, FEM discretizations may be constructed and adjusted without reference to mathematical models, simply from experimental measurements. The concept of error arises in the physical FEM in two ways, known as verification and validation. The verification is the same as in the mathematical FEM: the discrete solution is replaced into the discrete model to get the solution error. As noted earlier, this error is not generally important. Validation tries to compare the discrete solution against observation by computing the simulation error, which combines modeling and solution errors. As the latter is typically insignificant, the simulation error in practice can be identified with the modeling error. Comparing the discrete solution with the ideal physical system would in principle quantify the modeling errors. 1.1.5 How Does FEA Work? FEA uses a complex system of points called nodes, which make a grid called a mesh. Figure 1.6 shows a two-dimensional FE model for a gear

Introduction to Finite Element Analysis   •  7

Typical element

Typical node

Figure 1.6.  A two dimensional FE model for a gear tooth.

tooth. All nodes and elements lie in the plane of the paper. This mesh is programmed to contain the material and structural properties, which define how the structure will react to certain loading conditions. Nodes are assigned at a certain density throughout the material depending on the anticipated stress levels of a particular area. Regions that will receive large amounts of stress usually have a higher node density than those that experience little or no stress. Points of interest may consist of: fracture point of previously tested material, fillets, corners, complex detail, and high-stress areas. The mesh acts like a spider web, in that from each node, there extends a mesh element to each of the adjacent nodes. This web of vectors is what carries the material properties to the object creating many elements (theory). A wide range of objective functions (variables within the system) is available for minimization or maximization: mass, volume, temperature, strain energy, stress strain, force, displacement, velocity, acceleration, synthetic (user-defined). There are multiple loading conditions that may be applied to a system. Point, pressure, thermal, gravity, and centrifugal static loads, thermal loads from solution of heat transfer analysis, enforced displacements, heat flux and convection, point, pressure, and gravity dynamic loads. Each FEA program may come with an element library or one is constructed over time. Some sample elements are: rod elements, beam elements, plate or shell or composite elements, shear panel, solid elements, spring elements, mass elements, rigid elements, and viscous damping elements. Many FEA programs also are equipped with the capability to use multiple materials within the structure such as:

8  •   Using ANSYS for Finite Element Analysis

i­ sotropic: identical throughout, orthotropic: identical at 90 degrees, general anisotropic: different throughout. The basic ingredients of any FEA of structural problems are: • • • •

Geometry Material model (constitutive model) Loading Support or boundary conditions

1.1.6 Types of Structural Analysis Structural analysis is probably the most common application of the FEM. Structure is defined as something that is designed to bear load. Therefore, the term structural (or structure) implies not only civil engineering structures such as bridges and buildings, but also naval, aeronautical, and mechanical structures such as ship hulls, aircraft bodies, and machine housings, as well as mechanical components such as pistons, machine parts, and tools. FEA has become a solution to the task of predicting failure due to unknown stresses by showing problem areas in a material and allowing designers to see all of the theoretical stresses within. This method of product design and testing is far superior to the manufacturing costs, which would accrue if each sample was actually built and tested. FEA consists of a computer model of a material or design that is stressed and analyzed for specific results. It is used in new product design and existing product refinement. A company is able to verify whether a proposed design will be able to perform to the client’s specifications prior to manufacturing or construction. Modifying an existing product or structure is utilized to qualify the product or structure for a new service condition. In case of structural failure, FEA may be used to help determine the design modifications to meet the new condition. There are generally two types of analysis that are used in industry: 2D modeling and 3D modeling. While 2D modeling conserves simplicity and allows the analysis to be run on a relatively normal computer, it tends to yield less accurate results. 3D modeling, however, produces more accurate results while sacrificing the ability to run on all, but the fastest computers effectively. Within each of these modeling schemes, the programmer can insert numerous algorithms (functions), which may make the system behave linearly or nonlinearly. Linear systems are far less complex and generally do not take into account plastic deformation. Nonlinear systems do account for plastic deformation, and many also are capable of testing a material all the way to fracture.

Introduction to Finite Element Analysis   •  9 F

Linear

Nonlinear

Δ

Figure 1.7.  Linear and nonlinear behavior of the structure.

Structural analysis consists of linear and nonlinear behavior of the structural material, as shown in Figure 1.7. Linear behavior assumes that the material is not plastically deformed. Nonlinear behavior consists of stressing the material past its elastic capabilities (into the plastic range). The stresses in the material then vary with the amount of deformation. Vibrational analysis is used to test a material against random vibrations, shock, and impact. Each of these incidences may act on the natural vibrational frequency of the material, which, in turn, may cause resonance and subsequent failure. Fatigue analysis helps designers to predict the life of a material or structure by showing the effects of cyclic loading on the specimen. Such analysis can show the areas where crack propagation is most likely to occur. Failure due to fatigue may also show the damage tolerance of the material. Heat transfer analysis models the conductivity or thermal fluid dynamics of the material or structure. This may consist of a steadystate or transient transfer. Steady-state transfer refers to constant thermal properties in the material that yield linear heat diffusion. Thus, some of the common types of structural analysis performed using FEA are as follows. 1.1.6.1 Static Analysis (Linear and Nonlinear) A static analysis calculates the effects of steady loading conditions on a structure, while ignoring inertia and damping effects, such as those caused by time-varying loads. A static analysis can, however, include steady inertia loads (such as gravity and rotational velocity) and time-varying loads that can be approximated as static equivalent loads (such as the static equivalent wind and seismic loads commonly defined in many building codes). Static analysis is used to determine the displacements, stresses, strains, and forces in structures or components caused by loads that do

10  •   Using ANSYS for Finite Element Analysis

not induce significant inertia and damping effects. Steady loading and response conditions are assumed; that is, the loads and the structure’s response are assumed to vary slowly with respect to time. The kinds of loading that can be applied in a static analysis include: • • • • •

Externally applied forces and pressures Steady-state inertial forces (such as gravity or rotational velocity) Imposed (nonzero) displacements Temperatures (for thermal strain) Fluences (for nuclear swelling)

1.1.6.2 Transient Dynamic Analysis (Linear and Nonlinear) Transient dynamic analysis (sometimes called time history analysis) is a technique used to determine the dynamic response of a structure under the action of any general time-dependent loads. You can use this type of analysis to determine the time-varying displacements, strains, stresses, and forces in a structure, as it responds to any combination of static, transient, and harmonic loads. The time scale of the loading is such that the inertia or damping effects are considered to be important. If the inertia and damping effects are not important, you might be able to use a static analysis instead. 1.1.6.3  Modal Analysis You use modal analysis to determine the vibration characteristics ­(natural frequencies and mode shapes) of a structure or a machine component while it is being designed. It also can be a starting point for another, more detailed, dynamic analysis, such as a transient dynamic analysis, a ­harmonic response analysis, or a spectrum analysis. 1.1.6.4 Harmonic Analysis Any sustained cyclic load will produce a sustained cyclic response (a harmonic response) in a structural system. Harmonic response analysis gives you the ability to predict the sustained dynamic behavior of your structures, thus enabling you to verify whether or not your designs will successfully overcome resonance, fatigue, and other harmful effects of forced vibrations.

Introduction to Finite Element Analysis   •  11

1.1.6.5 Spectrum Analysis Spectrum analysis is one in which the results of a modal analysis are used with a known spectrum to calculate displacements and stresses in the model. It is mainly used in place of a time history analysis to determine the response of structures to random or time-dependent loading conditions such as earthquakes, wind loads, ocean wave loads, jet engine thrust, rocket motor vibrations, and so on. 1.1.6.6  Buckling Analysis Buckling analysis is a technique used to determine buckling loads, critical loads at which a structure becomes unstable, and buckled ­ mode shapes—the characteristic shape associated with a structure’s buckled response. 1.1.6.7 Nonlinear Analysis If a structure experiences large deformations, its changing geometric configuration can cause the structure to respond nonlinearly. Nonlinear stress–strain relationships are a common cause of non­ linear structural behavior. Many factors can influence a material’s stress–strain properties, including load history (as in elastoplastic response), environmental conditions (such as temperature), and the amount of time that a load is applied (as in creep response). 1.1.6.8 Contact Problems Contact problems are highly nonlinear and require significant computer resources to solve. It is important that you understand the physics of the problem and take the time to set up your model to run as efficiently as possible. Contact problems present two significant difficulties. First, you generally do not know the regions of contact until you have run the problem. Depending on the loads, material, boundary conditions, and other factors, surfaces can come into and go out of contact with each other in a largely unpredictable and abrupt manner. Second, most contact problems need to account for friction. There are several friction laws and models to choose

12  •   Using ANSYS for Finite Element Analysis

from, and all are nonlinear. Frictional response can be chaotic, making solution convergence difficult. In addition to these two difficulties, many contact problems must also address multi-field effects, such as the conductance of heat and electrical currents in the areas of contact. 1.1.6.9 Fracture Mechanics Cracks and flaws occur in many structures and components, sometimes leading to disastrous results. The engineering field of fracture mechanics was established to develop a basic understanding of such crack propagation problems. Fracture mechanics deals with the study of how a crack or flaw in a structure propagates under applied loads. It involves correlating ­analytical predictions of crack propagation and failure with experimental results. The analytical predictions are made by calculating fracture parameters such as stress intensity factors in the crack region, which you can use to estimate crack growth rate. Typically, the crack length increases with each ­application of some cyclic load, such as cabin pressurization-­ depressurization in an airplane. Further, environmental conditions such as temperature or extensive exposure to irradiation can affect the fracture propensity of a given material. 1.1.6.10 Composites Composite materials have been used in structures for a long time. In recent times, composite parts have been used extensively in aircraft structures, automobiles, sporting goods, and many consumer products. Composite materials are those containing more than one bonded material, each with different structural properties. The main advantage of composite materials is the potential for a high ratio of stiffness to weight. Composites used for typical engineering applications are advanced fiber or laminated composites, such as fiberglass, glass epoxy, graphite epoxy, and boron epoxy. FEA allows you to model composite materials with specialized elements called layered elements. Once you build your model using these elements, you can do any structural analysis (including nonlinearities such as large deflection and stress stiffening).

Introduction to Finite Element Analysis   •  13

1.1.6.11 Fatigue Fatigue is the phenomenon in which a repetitively loaded structure fractures at a load level less than its ultimate static strength. For instance, a steel bar might successfully resist a single static application of a 300 kN tensile load, but might fail after 1,000,000 repetitions of a 200 kN load. The main factors that contribute to fatigue failures include: • • • •

Number of load cycles experienced Range of stress experienced in each load cycle Mean stress experienced in each load cycle Presence of local stress concentrations

A formal fatigue evaluation accounts for each of these factors as it calculates how used up a certain component will become during its anticipated lifecycle. 1.1.6.12 Other Types of Engineering Analysis Other types of engineering that can be solved using FEA are: • • • • • •

Heat transfer (linear and nonlinear) Fluid flow (laminar and turbulent) Crash dynamics Electromagnetics Optimization Kinematics

1.1.7 Advantages and Disadvantages Advantages: • Model irregularly shaped bodies quite easily • Handle general load conditions without difficulty • Model bodies composed of several different materials because the element equations are evaluated individually • Handle unlimited number and kinds of boundary conditions • Vary the size of the elements to make it possible to use small ­elements where necessary

14  •   Using ANSYS for Finite Element Analysis

• Alter the FE model relatively easily and cheaply • Corrective measures can be taken before any hardware is built (also, the number of prototypes can be reduced) • Physically meaningful to engineers and resemble actual structure Disadvantages • Experience and judgment needed • Requires good computing facility 1.1.8 Areas of Application 1.1.8.1 Structural •  A1 •  A2 •  A3 •  A4

Stress analysis Buckling Vibration analysis …..

1.1.8.2 Nonstructural •  B1 •  B2 •  B3 •  B4 •  B5

Heat transfer Fluid flow Distribution of electric or magnetic Lubrication …..

1.1.8.3 Examples of Multi-physics Applications •  A + B1 •  B1 + B2 •  A + B1 + B2

Thermal analysis Convective flow Manufacturing processes

1.1.8.4 Common Fields of Application • Mechanical engineering • Aerospace engineering • Civil engineering

Introduction to Finite Element Analysis   •  15

• • • • •

Automobile engineering Electrical engineering Chemical engineering Geomechanics Biomechanics

1.2 Review Topics 1.2.1  Matrix Operations Matrix Algebra • A matrix is an m × n array of numbers arranged in m rows and n columns. • m = n A square matrix. • m = 1 A row matrix. • n = 1 A column matrix. • aij Element of matrix a row i, column j Multiplication of a matrix by a scalar: [a] = k [c] aij = k cij Addition of matrices: Matrices must be of the same order (m × n). Add them term by term: [c] = [a] + [b]  cij = aij + bij Multiplication of two matrices: If [a] is m × n, then [b] must have n rows: [c] = [a] [b] n

cij = ∑aie � bie e =1

Transpose of a matrix: Interchange of rows and columns: [aij] = [aji]T • If [a] is m × n, then [a]T is n × m • If [a] = [a]T, then [a] is symmetric. [a] must be a square matrix

16  •   Using ANSYS for Finite Element Analysis

The identity matrix (or unit matrix) is denoted by the symbol [I]:

[a ][ I ] = [ I ][a ] = [a] 1 0 0  [ I ] = 0 1 0 0 0 1  Differentiating a matrix:  daij  d a] =  [  dx  dx  Example of differentiating a matrix: U=

1 [x 2

 a11 y]   a21

 ∂U   ∂x   a 11  ∂U  =  a  21    ∂y 

a12   x    a22   y 

a12   x    a22   y 

Integrating a matrix:

[a ]� dx = ∫a� ij dx The inverse of a matrix is such that:

[a ][a ]−1 = [ I ] To find the inverse of a matrix, first find the determinant of the matrix: |a| = determinant of the matrix [a] Next, calculate the cofactors of [a]. Cofactors of [aij] are given by: cij = (−1)i + j d

Introduction to Finite Element Analysis   •  17

Where matrix d is the first minor of [aij] and is matrix [a] with row i and column j deleted. Finally, determine the inverse  aij   

−1

=

[c ]T a

1.2.2 Elasticity Equations 1.2.2.1 Stress Equilibrium Equations A three-dimensional body occupying a volume V and having a surface S is shown in Figure 1.8. Points in the body are located by x, y, and z coordinates. The boundary is constrained on some region, where displacement is specified. On part of the boundary, distributed force per unit area T, also called traction, is applied. Under the force, the body deforms. The deformation of a point (x = [x y z]T) is given by the three components of its displacement: u = [u v w]T The distributed force per unit volume, for example, the weight per unit volume, is the body force vector f given by: f =  f x

fy

f z 

T

The body force acting on the elemental volume dV is shown in ­ igure 1.8. The surface traction T may be given by its component values F at points on the surface: y

σy x

z

τ yx τ yz

τx

τ zy τzx Loaded material body σz

Figure 1.8.  Three-dimensional body.

xy

τxz

18  •   Using ANSYS for Finite Element Analysis

T = Tx

Tz 

Ty

T

Examples of traction are distributed contact force and action of p­ ressure. A concentrated load P acting at a point i is represented by its three components: Pi =  Px

Pz 

Py

T i

The stresses acting on the elemental volume dV are shown in ­ igure 1.9. When the volume dV shrinks to a point, the stress tensor is repF resented by placing its components in a (3 × 3) symmetric matrix. However, we represent stress by the six independent components as follows:

s = s x

sy

sz

txy

txz 

t yz

Where s x , s y , and s z are normal stresses and txy , t yz , and tzx are shear stresses. Let us consider equilibrium of the elemental volume shown in Figure 1.9. First, we get forces on faces by multiplying the stresses by the corresponding areas. Writing ∑ Fzx = 0, ∑ Fy = 0, and ∑ Fz = 0 and recognizing dV = dx dy dz, we get the equilibrium equations: ∂s x ∂txy ∂txz + + + Fx = 0 ∂x ∂y ∂z ∂txy ∂x

+

∂s y ∂y

+

∂t yz ∂z

+ Fy = 0

∂txz ∂t yz s z + + + Fz = 0 ∂x ∂y ∂z y

σyy σyx

σyz σzy σzz

σzx

σxy σxz

z

Figure 1.9.  Equilibrium of elemental volume.

σxx x

Introduction to Finite Element Analysis   •  19

1.2.2.2 Strain–Displacement Relationship Strains can be calculated by differentiating displacement functions. ­Differentiation of a function is possible only if it is continuous. Therefore, the strain–displacement relations are also known as compatibility ­equations and are given as follows: ∂u ∂u ∂v + gzy = ∂x ∂y ∂x ∂v ∂u ∂w ey = gxz = + ∂y ∂z ∂x ∂w ∂w ∂v ez = g yz = + ∂z ∂y ∂z

ex =

In matrix form: ∂  ∂x    ex   0 e    y   e   0  z =  gxy   ∂    ∂y gxz   g   ∂  yz   ∂z  0 

0 ∂ ∂y 0 ∂ ∂x 0 ∂ ∂z

 0  0   ∂ u  ∂z     v  0   w   ∂ ∂x   ∂ ∂y 

1.2.2.3 Stress-Strain Relationships For linear elastic materials, the stress–strain relations come from the generalized Hooke’s law. For isotropic materials, the two material properties are Young’s modulus (or modulus of elasticity) E and Poisson’s ratio ν. For a three-dimensional case, the state of stress at any point in relation to the state of strain as follows:

20  •   Using ANSYS for Finite Element Analysis

v v 0 1 − v  v 0 s x   v 1 − v s   v v 1 − v 0  y  s   1 − 2v 0 0  z =  0 2 txy      0 0 0 t yz   0 t    zx  0 0 0  0 

0 0 0 0 1 − 2v 2 0

  e  x   e y      0   ez   g   xy    0  g yz     1 − 2v  gzx   2  0 0 0

In matrix notation:

{s} = [ D] {e} Where [D] is known as a stress–strain matrix or material properties matrix and is given by: v v 0 1 − v  v 1− v v 0   v v 1− v 0  − 1 2v  E 0 0 [ D ] = (1 + v)(1 − 2v)  0 2   0 0 0 0   0 0 0  0 

0 0 0 0 1 − 2v 2 0

     0    0   1 − 2v   2  0 0 0

1.2.2.4 Special Cases One dimension: In one dimension, we have normal stress along the x-axis and the corresponding normal strain. Stress–strain relations are simply to: {σx}=[E]{εx} Where [D] = [E] Plane stress: A thin planar body subjected to in-plane loading on its edge surface is said to be in plane stress. A ring press-fitted on a shaft as shown in Figure 1.10(a) is an example. Here, stresses s x , txz, and tzy are set as zero. The Hooke’s law relations then give us:

Introduction to Finite Element Analysis   •  21

P

P

z σx = 0 τxz = 0 τzy = 0

(a)

P

P

(b)

z

εz = 0 γzx = 0 γyz = 0

Figure 1.10.  (a) Plane stress (b) Plane strain.

   s  1 u 0   ex  x      0  e y  s y  = u 1      txy  0 0 1 − u  gxy  2   Plane strain: If a long body of uniform cross-section is subjected to transverse loading along its length, a small thickness in the loaded area, as shown in Figure 1.10(b), can be treated as subjected to plane strain. Here e z, gzx, and g yz are taken as zero. Stress s z may not be zero in this case. The stress–strain relations can be obtained directly as:   s  1 u 0   ex  x      E 0  e y  s y  = u 1   (1 + v ) (1 − 2v )  1 − u  gxy  txy    0 0 2  

22  •   Using ANSYS for Finite Element Analysis

1.2.3 Solution of Set of Linear Algebraic Equations 1.2.3.1 Cramer’s Rule Consider a set of linear algebraic equations written as follows: In matrix notation: [a]{x} = {c} Or, in index notation:

n

∑ j =1 aij x j = ci

Let matrix  d (i)  be matrix [a] with column i replaced by [c]. Then:   xi =

d (i ) a

Example: Consider a set of three linear algebraic equations given as follows: − x1 + 3 x2 − 2 x3 2 x1 − 4 x2 + 2 x3 4 x2 + x3 In matrix form:  −1 3 −2   x1  2   2 −4 2   x  = 1   2     0 4 1   x3  3  Solution using Cramer’s rule: −1 3 −2 a = 2 −4 2 = ( −1) ( −4 − 8) − (3) ( 2 − 0) + ( −2) (8 − 0) = 12 − 6 − 16 = 10 0 4 1

x1 =

d (1) a

=

2 3 −2 1 −41 1 −4 2 = = 4.1 (−10) −10 3 4 1

Introduction to Finite Element Analysis   •  23

x2 =

x3 =

d ( 2) a

d (3) a

−1 2 −2 1 −11 2 1 2 = = = 1.1 (−10) −10 0 3 1 −1 3 2 1 −14 2 −4 1 = = = 1.4 (−10) −10 0 4 3

1.2.3.2 Inversion

[a ]{ x} = {c} [a ]−1 [a ]{ x} = [a ]−1 {c} [ I ]{ x} = [a ]−1 {c} ∴ { x} = [ a ]

−1

{c}

Example:  −1 3 −2   x1  2   2 −4 2   x  = 1    2    using inversion,   0 4 1   x3  3  1.1 0.2  2   4.1   x1   1.2        ∴  x2 x  =  0.2 0.1 0.2  1  =  1.1   x   −0.8 −0.4 0.2  3  −1.4       3   1.2.3.3 Gaussian Elimination General system of n equations with n unknowns:  a11  a1n   x1   c1         =          an1  ann   xn  cn 

24  •   Using ANSYS for Finite Element Analysis

Steps in Gaussian elimination: 1. Eliminate the coefficient of x1 in every equation except the first one. Select a11 as the pivot element. • Add the multiple −a21 / a11 of the first row to the second row. • Add the multiple −a31 / a11 of the first row to the third row. • Continue this procedure through the nth row

After this step:  a11  0      0

a12 ´

a22  ´

an 2

 a1n   x1   c1  ´  ´  a2 n   x2  c    =  2        `  x     ann   n  cn 

2. Eliminate the coefficient of x2 in every equation below the second ′ one. Select a22 as the pivot element. ′ ′ / a22 • Add the multiple a32 of the second row to the third row. ′ ′ • Add the multiple a42 / a22 of the second row to the fourth row. • Continue this procedure through the nth row. After this step:  a11  0  0      0

a12

a13  a1n  c  x1   1  ´ ´   ´  a23  a2 n   x2  c      2  '' a33  a3'' n   x3  =  c ''  3             xn    '' cn  an'' 3  ann 

´

a22 0  ´

an 2

3. Repeat the process for the remaining rows until we have a triangularized system of equations:  a11  0  0  0      0

a12

a13

a14

´

´

´



a22

a23

a24 

0

'' a33

'' a34 

0 

0 

'''  a44 

an'' 3

an''' 4 

´

an 2

a1n   c1  ´   x1   ´  a2 n   x2   c     2  a3'' n   x3   c ''    =  3  a4'''n   x4   c '''  4            x  n −1   n  c n −1  ann n   

Introduction to Finite Element Analysis   •  25



Solve using back substitution: xn =

xi =

1 aii

cnn −1

n −1 ann

n   a − air xr  ∑ 1 , n + 1    r = i +1

Example:  2 2 1   x1  9   2 1 0   x  = 4    2   1 1 1   x3  6  1. Eliminate the coefficient of x1 in every equation except the first one. Select a11 as the pivot element. • Select a11 = 2 as the pivot element. • Add the multiple − a21 / a11 = − 2 / 2 = − 1 of the first row to the second row. 1 • Add the multiple − a31 / a11 = − = − 0.5 of the first row to the 2 third row. 1   x1   9  2 2  0 −1 −1   x  =  −5   2     0 0 0.5  x3  1.5 2. Eliminate the coefficient of x2 in every equation below the second ′ one. Select a22 as the pivot element (already done in this example). 1   x1   9  2 2  0 −1 −1   x  =  −5   2     0 0 0.5  x3  1.5 3. Solve using back substitution: x3 =

c3'' '' a33

x2 =

=

( 3 2) = 3 ( 12)

( −5 + 3) = 2 −1

26  •   Using ANSYS for Finite Element Analysis

x1 =

(9 − 2 ( 2) − 3) = 1 2

1.3  General Steps of FEM The objective of structural analysis is usually to determine the displacements and stresses throughout the structure, which is in equilibrium and is subjected to applied loads. At any point in the continuum body, there are 15 unknowns (three displacements, six stresses, and six strains). To determine these 15 unknowns, we have 15 equations in three-dimensional case shown as follows: Unknowns

Unknowns governing equations

Displacements (u,v,w) Stresses  ex e y  Strains  ex e y 

ez

gxy

g yz

gxz 

ez

gxy

g yz

gxz 

3

Stress equilibrium equations

3

6

Compatibility equations

6

6

Stress–strain equations

6

15

15

There are two general approaches associated with the finite element method to solve the governing equations: force (or flexibility) method and displacement (or stiffness) method. The force method uses internal forces as the unknown of the problem, whereas the displacements are the system variable in displacement method. The displacement method is more desirable because its formulation is simpler for most structural analysis problems. Furthermore, a vast majority of general-purpose finite element programs have incorporated the displacement formulation for solving structural problems. Consequently, only the displacement method will be used throughout this course. The basic steps involved in any FEA consist of the following: Preprocessing phase (build the FE model, loads, and constraints) 1. 2. 3. 4. 5.

Discretize and select element type. Select a displacement function. Define strain/displacement and stress/strain relationships. Derive element stiffness matrix and equations. Assemble equations and introduce Boundary Conditions (BCs).

Solution phase: (assemble and solve the system of equations) 6. Solve for the unknown degrees of freedom.

Introduction to Finite Element Analysis   •  27

Post-processing Phase: (sort and display the results) 7. Solve for element stresses and strains. 8. Interpret the results.

1.4 The Most Common Finite Element Types The basic idea of FEA is to make calculations at only limited (finite) ­number of points and then interpolate the results for the entire domain (line, surface, or volume). Any continuous object has infinite degrees of freedom, and it is just not possible to solve the problem in this format. FEM reduces the degrees of freedom from infinite to finite with the help of discretization or meshing (nodes and elements). 1D (line) element

(Spring, truss, beam, pipe, etc.)

2D (plane) element

(Membrane, plate, shell, etc.)

3D (solid) element

(3-D fields, temperature, displacement, stress, flow velocity)

1.5 Types of Element Formulation Methods The element characteristic matrix has different names in different problem areas. In structural mechanics, it is called a stiffness matrix; it relates nodal displacements to nodal forces. There are three important ways to derive an element characteristic matrix. 1.5.1 Direct (Equilibrium) Method The direct method is based on physical reasoning. It is limited to very simple elements (spring, bar, and beam), but is worth studying because it enhances our physical understanding of FEM. According to this method, the stiffness matrix and element equations relating nodal forces to nodal displacements are obtained using force equilibrium conditions for a basic element, along with force/deformation relationships.

28  •   Using ANSYS for Finite Element Analysis

1.5.2 Work or Energy Methods To develop the stiffness matrix and equations for two- and three-­ dimensional elements, it is much easier to apply a work or energy method. These are based on variational calculus. The variational method is applicable to problems that can be stated by certain integral expressions such as the expression for potential energy. The principle of virtual work (using virtual displacement), the principle of minimum potential energy, and Castigliano’s theorem are methods frequently used for the purpose of derivation of element equations. The principle of virtual work is applicable for any material behavior, whereas the principle of minimum potential energy and Castigliano’s theorem are applicable only to elastic materials. For the purpose of extending, FEM outside the structural stress analysis field, a functional (a scalar function of other functions) analogous to the one to be used with the principle of minimum potential energy is quite useful in deriving the element stiffness matrix and equations. 1.5.3  Methods of Weighted Residuals Weighted residual methods are particularly suited to problems for which differential equations are known, but no variational statement or functional is available. For stress analysis and some other problem areas, the variational method and the most popular weighted residual method (the Galerkin method) yield identical finite element formulations.

1.6 Derivation of Spring Element Equations using Direct Method To understand the FE formulation, we start with the concept “Everything important is simple.” Figure shows a spring element 1

k

2

ƒˆ1x ,dˆ1x

ƒˆ2x ,dˆ2x L

Two nodes: Local nodal displacements: Local nodal forces:

Node 1, node 2 dˆ1x ,dˆ2x (inch, m, mm) ƒˆ ,fˆ (lb, newton)

Spring constant (stiffness)

K (lb/in, N/m, N/mm)

1x

2x



Introduction to Finite Element Analysis   •  29

The stiffness equation for a single spring element in a local coordinate system can be written as: ∧   f 1x   k11  ∧ =  f   k21  2x 

∧  k12   d 1x   ∧   ∧   ∧   ⇒  f  = k d  k22   ∧        d  2x 

Step 1: Select the element type Consider linear spring element subjected to tensile forces k

2

1

ˆx

L 1

k

2

T dˆ1x

T

ˆx

dˆ2x

Step 2: Select a displacement function • • • •

Degree of freedom (DOF) per node = 1 Number of nodes per element = 2 Total (DOF) per element = 2 × 1 = 2 Number of coefficients = 2

x4

x3

x2 x3 y

x x2 y

1 xy x2 y2

y x y2

y2

y3

x y3

y4

Assume a linear displacement function: ∧



u = a1 + a2 x

Write in matrix form:  ∧   a1  ∧ u = 1 x       a2  



Express u^ as a function of nodal displacements (  d 1x , d 2 x  ). ∧



30  •   Using ANSYS for Finite Element Analysis

Apply boundary condition: ∧







d 2 x − d 1x L

At x∧ = 0 u∧ = d 1x ∴ u∧ (0) = a1 + a2 ( 0) = d 1x ∴ a1 = d 1x At



x = L u = d 2 x ∴ u ( L ) = a1 + a2 ( L ) = d 2 x ∴ a1 = ∧









Substituting values of coefficients: ∧   ∧    ∧ ∧ ∧ ∧ ∧ x d 2 x − d 1x  ∧  x ∧ x ∧ ∴ u = d 1x +  x = 1 −  d 1x +   d 2 x = 1 −  L L L     L       

∧  ∧ x   d 1x    L ∧   d 2 x 

 ∧ ∧ ∴ u = [ N ] d   

Where N1 = 1 −





x x and N 2 = L L

N1 and N2 are called shape functions or interpolation functions. They express the shape of the assumed displacements. The sum of all shape functions at any point within an element should be equal to 1. N1 = 1    N2 = 0

at node 1

N1 = 0    N2 = 1

at node 2

N1 + N2 = 1 N1

N1

N2

2

1

1

2

L

N2

1

2 L

L

Step 3: Define the strain/displacement and stress/strain relationships ∧



Deformation, d = u∧ ( L) − u∧ ( 0) = d 2 x − d 1x From the force/deformation relationship:

∧   ∧ T = k d = k  d 2 x − d 1x   

Where T is the tensile force and δ is the total elongation. Step 4: Derive the element stiffness matrix and equations Consider the equilibrium of forces for the spring. At node 1,

∧ ∧  ∧  f 1x = −T = k  d 1x − d 2 x   

At node 2,

f

∧ 2x

∧   ∧ = T = k  d 2 x − d 1x   

Introduction to Finite Element Analysis   •  31

In matrix form,  ∧   f 1x   k  ∧ =  f  −k  2x 

∧  e − k   d 1x     ∧ ( e)    ∧( )    ∧ ( e)    ⇒  f  =  k  d  k   ∧              d 2 x 

Note k is symmetric. Is k singular or nonsingular? That is, can we solve the equation? If not, why? Step 5: Assemble the element equations to obtain the global equations and introduce the boundary conditions (e)  N ∧ Global stiffness matrix: [ K ] = ∑ e =1  k    e ( ) N   ∧  Global load vector: { F } = ∑ e =1  f    ∴ { F } = [ K ] {d } This vector does not imply a simple summation of the element matrices, but rather denotes that these element matrices must be assembled properly satisfying compatibility conditions. Step 6: Solve for nodal displacements Displacements are then determined by imposing boundary conditions, such as support conditions, and solving a system of equations, {F} = [K]{d}, simultaneously. Step 7: Solve for element forces Once displacements at each node are known, then substitute back into element stiffness equations to obtain element nodal forces.

1.7 Bar Element Formulation using Direct Method Step 1: Select the element type y

Tˆ x

Cx ˆ (force/length)

ˆy

2 L

T

1

dˆ1x , fˆ1x

θ

T

ˆx, uˆ

dˆ2x , fˆ2x

x

32  •   Using ANSYS for Finite Element Analysis

Assumptions: • • • •





The bar cannot sustain shear force, that is, f 1 y = 0, f 2 y = 0. Any effect of transverse displacement is ignored. Hooke’s law applies, that is, s x = E e x. No intermediate applied loads.

Two nodes: Local nodal displacements:

Node 1, Node 2 ∧ ∧ d 1x , d 2 x (inch, m, mm)

Local nodal forces:





f 1x , f 2 x (lbs., Newton)

Length Cross-sectional area Modulus of elasticity

L (inch, m, mm) A (Sq. inch, Sq. m, Sq. mm) E (psi, Pa, MPa)

The stiffness equation for a single spring element in a local coordinate system can be written as: ∧   f 1x   k11  ∧ =  f   k21  2x 

∧  k12   d 1x   ∧   ∧   ∧  ⇒  f  = k d   k22   ∧        d 2 x 

Step 2: Select a displacement function • • • •

Degree of freedom (DOF) per node = 1 Number of nodes per element = 2 Total (DOF) per element = 2 × 1 = 2 Number of coefficients = 2

Assume a linear displacement function: ∧



u = a1 + a2 x

Write in matrix form.  ∧   a1  ∧ u = 1 x      a2  ∧



Express u^ as a function of nodal displacements ( d 1x , d 2 x ) Apply boundary condition: ∧





At x∧ = 0 u∧ = d 1x ∴ u∧ ( 0) = a1 + a2 ( 0) = d 1x ∴ a1 = d 1x

Introduction to Finite Element Analysis   •  33 ∧



At x = L u = d 2 x ∴ u ( L) = a1 + a2 ( L) = d 2 x ∴ a1 = ∧









d 2 x − d 1x L

Substituting values of coefficients: ∧

∴u =



∧ ∧   d 2 x − d 1x  ∧  x = 1− L      

∧ d 1x + 

 ∧  ∧ ∧ x x ∧  x ∧ d 2 x = 1 − d 1x +  L L  L    

∧  ∧ x   d 1x    L ∧   d 2 x 

 ∧ ∧ ∴ u = [ N ] d    Where N1 = 1 −





x x and N 2 = L L

Step 3: Define the strain/displacement and stress/strain relationships From the definition of strain, the strain/displacement relationship can be derived as:  ∧ d u  ∴ {e} =  = dx    

  d   [ N1  dx  

 ∧    d 1x    dN1 N2 ]    =   d∧2 x    dx   

∧  dN 2   d 1x   = dx   ∧  d 2 x 

 1  − L

∧  1   d 1x    L   ∧  d 2 x 

 ∧ ∴ {e} = [ B ]  d    Where e is known as strain–displacement matrix. From Hooke’s law, the stress/strain relationship is: s x = E e x. Step 4: Derive the element stiffness matrix and equations To derive the stiffness equation using direct method, we employ equilibrium condition at each node. 

∧ ∧  ∧  d 2 x − d 1x  AE  ∧ = d 1x − d 2 x    L   L   



At node 1, f 1x = − T = − As x = − A( E e x ) = − AE  ∧

At node 2, f 2 x = T =

∧  AE  ∧ d 2 x − d 1x    L 

In matrix form, ∧  ∧   f 1x  AE  1 −1 d 1x   ∧ (e )   ∧ (e )   ∧ (e )   ⇒  f  =  k  d   =         f∧  L  −1 1  d∧2 x      2x 

Steps 5 to 7 are same as before.

34  •   Using ANSYS for Finite Element Analysis

1.8 Examples of Linear FEM Example 1 For the structure shown in the figure, determine the nodal displacements, the forces in each element, and the reactions.

2 1

1

2

3

2

3

15 kN

4

4

5

3m 2

k=

E = 210 GPa A = 3 × 10–4 m2

3m

EA the stiffness of element L

All elements of the previous figure have the same material and dimensions. k (1) = k ( 2) = k (3) = k ( 4) = k =

k=

EA  1 −1 L  −1 1 

210 *103 * 300  1 −1  −1 1  3000    21 −21 3 k=  *10  −21 21 

The global stiffness matrix (K) K = k1 + k2 + k3 + k4   21 −21   −21 21  3 0 K = 10 *   0  0  0   0 0 

0 0 0  0 0 0 0 0 0  0 21 −21 0 0 0  + 0 −21 21   0 0 0  0 0 0 0 0 0  0 0 0

0 0  0 0 0  0 21 0  + 0 0   0  0 −21 0  0 0

0 0 0 0

0 0 0 −21 0 0 0 0 0 0

0  0 0 0  0 21 0  + 0 0   0  0 0 0  0 −21

0  0 −21  0 0   0 0  0 21 

0 0 0 0 0 0

Introduction to Finite Element Analysis   •  35

 21  −21  3 K = 10 *  0   0  0

−21 0 0 0  84 −21 −21 −21 −21 21 0 0   −21 0 21 0  −21 0 0 21 

The nodal displacement equations: {F}=[K ]{d}  F1x   d1x  F  d   2 x   2 x   F3 x  = [ K ]  d3 x  F  d   4x   4x   F5 x   d5 x   F1x   21 15000   −21    3  F3 x  = 10 *  0   F   0  4x   0  F5 x 

0 0  0  −21 0 84 −21 −21 −21 d 2 x    −21 21 0 0  0   −21 0 21 0  0    −21 0 0 21   0 

F1x = −21*103 * d 2 x (1) 15000 = 84 *103 * d 2 x (2) F3 x = −21*103 * d 2 x (3) F4 x = −21*103 * d 2 x (4) F5 x = −21*103 * d 2 x (5) From equation (2), d2x = 0.17857 mm From equations (1–3–4–5), the reactions, F1x = F3x = F4x = F5x = –3.75kN The force in each element, {f} = [k]{d} The first element between nodal 1, 2:  f1(x1)  −21 d1x = 0  3  21   (1)  = 10 *    −21 21   d 2 x   f 2 x 

36  •   Using ANSYS for Finite Element Analysis

f1(x1) = −21*103 * d 2 x f1(x1) = −3.75 kN f 2(1x) = − f1(x1) = 3.75 kN The second element between nodal 2, 3:  f 2(x2)  −21  d 2 x  3  21    ( 2)  = 10 *  21 21  d3 x = 0  −   f3 x  f 2(x2) = 21*103 * d 2 x f 2(x2) = 3.75 kN f3(x2) = − f 2(x2) = −3.75 kN The third element between nodal 2, 4:  f 2(x2)  −21  d 2 x  3  21   ( 2)  = 10 *    −21 21  d 4 x = 0   f 4 x  f 2(x2) = 21*103 * d 2 x f 2(x2) = 3.75 kN f 4(x2) = − f 2(x2) = −3.75 kN The fourth element between nodal 2, 5:  f 2(x2)  −21  d 2 x  3  21    ( 2)  = 10 *  21 21  d5 x = 0  −   f5 x  f 2(x2) = 21*103 * d 2 x f 2(x2) = 3.75 kN f5(x2) = − f 2(x2) = −3.75 kN

Introduction to Finite Element Analysis   •  37

Example 2 Derive a finite element to solve this problem. 2

1

1000 lb 60°

3

2

1000 lb

1

30° 3 4

L

To derive this problem must be tack the rotational effect by multiplying the local stiffness matrix in rotational matrix (R). The rotational matrix in this problem can be derived as:  c2   cs = 2  −c   − cs

cs

−c 2

s2

−cs

−cs

c2

−s2

cs

−cs   −s2   cs   s2 

Where: c2 = cos2 (β), s2 = sin2 (β), cs = cos (β)*sin (β). β = The angle between x-axis and the element C.C.W. c2 β1 = 120° β2 = 180° β3 = 210°

cs

s2

0.25

−0.433

0.75

1

0

0

0.75

0.433

0.25

The property of this matrix is a symmetric matrix. This matrix is ­multiplied with the local stiffness matrix of each element. k (i ) =

Ei Ai [ Ri ] i = 1, 2, 3 Li

38  •   Using ANSYS for Finite Element Analysis

Where: [Rj]= [R] when derived with βi . i =3

[ K ] = ∑k (i ) i = 1, 2, 3 i =1

And, the next steps as the same in example 1. Example 3 Analyze a thin plate of uniform thickness t = 5 mm using an appropriate finite element formulated in part (a) and (b). The recommended mesh for the plate is shown in the figure. Determine the displacement at point (250, 125) and stresses in the element A only. Use E = 210 GPa and ʋ = 0.30.

4 (1), A 1

(4) 5 (2)

3 (3)

250 mm 2

500 mm (a)

The governing global matrix equation is: {F} = [K]{d} Expanding matrices in global matrix equation, we obtain  F1x   R1x   d1x   0  F   d    0   1 y   R1 y   1y     F2 x  20000  d2 x  d2 x           F2 y   34641 d 2 y  d 2 y  F   R  d   0   3x   3x   3x  K = [ ]     = [K ]  =   0   d3 y   F3 y   R3 y   0  d  F   R    4x   4x   4x    0  d 4 y   F4 y   R4 y  d         5x   d5 x   F5 x   0   d5 y   d5 y   F5 y   0     

30° 40 kN

Introduction to Finite Element Analysis   •  39

Where [K] is an 10 × 10 before deleting rows and column to account for the fixed boundary support at nodes 1, 3, and 4. Assemblage of stiffness matrix The stiffness matrix for element is: [k] = tA[B]T [D] [B] Element (1) m=4

(1)

j=5

i=1

For element (1), we have coordinates and xi = 0, yi = 0, xm = 0, ym = 250, xj = 250 and yj = 125 because the global axes are set up at node 1, and

1 bh 2 1 A = ( 250)( 250) = 31250 mm 2 2 t = 5 mm

A=

The matrix [B] is given by:  bi 1  [B] = 2 A  0   gi

0

bj

0

bm

gi

0

gj

0

bi

gj

bj

gm

Where: bi = y j − ym = 125 − 250 = − 125

b j = ym − yi = 250 − 0 = 250 bm = yi − y j = 0 − 125 = −125 gi = xm − x j = 0 − 250 = −250 g j = xi − xm = 0 − 0 = 0 gm = x j − xi = 250 − 0 = 250

0  gm   bm 

40  •   Using ANSYS for Finite Element Analysis

Therefore, substituting in [B] matrix, 0 250 0 −125 0   −125 1  0 250  [ B ] = 62500  0 −250 0 0  −250 −125 0 250 250 −125 For plane stress, the [D] matrix is conveniently expressed here as:   1 u 0   E  u 1 0  [D] = 2  (1 − u )  1− u  0 0   2  With ʋ = 0.3 and E = 210 GPa, we obtain

[D] =

0   1 0.3 210000  0 3 1 0  . 0.91   0 0 0.35

0 −250   −125  0 −250 −125   0   1 0.3 0 0  210000  250 T Then, [ B ] [ D ] = 0   0.3 1  0 250   62500(0.91)  0  0 0 0.35  −125 0 250    250 −125   0

 −125 −37.5 −87.5   −75 −250 −43.75    250 75 0  T ∴ [ B ] [ D ] = 3.6923   0 87.5   0  −125 −37.5 87.5    250 −43.75  75

Introduction to Finite Element Analysis   •  41

The stiffness matrix for element (1) is:  −125 −37.5 −87.5   −75 −250 −43.75   0 250 0 −125 0   −125  250 75 0  1  0 −250 0 0 0 250  k = 5 31250 3 . 6923 ( )( )( ) [ ]  ×  0 87.5  62500  0  −250 −125 0 250 250 −125  −125 −37.5 87.5    250 −43.75  75

20313 −31250 −21875 −6250 1563   37500  20313 67969 −18750 −10938 −1563 −57031    −31250 −18750 6250 0 −31250 18750  [k ] = 9.2308  −21875 −10938  0 21875 21875 −10938   −6250 −1563 −31250 21875 37500 −20313   −57031 18750 −10938 −20313 67969   1563 Element (2) m=5 (2)

i=1

j=2

A=

1 bh 2

A=

1 (500)(125) = 31250 mm2 2

For element (2), we have coordinates xi = 0, yi = 0, xm = 250, ym = 125, xj = 250 and yj = 0 because the global axes are set up at node 1, and

bi = y j − ym = −125 b j = ym − yi = 125 bm = yi − y j = 0 gi = xm − x j = −250 g j = xi − xm = −250 gm = x j − xi = 500

42  •   Using ANSYS for Finite Element Analysis

Therefore, substituting in [B] matrix, 0 125 0 0 0   −125 1  [ B ] = 62500  0 −250 0 −250 0 500  −250 −125 −250 125 500 0  0   1 0.3 210000  [ D ] = 0.91 0.3 1 0   0 0 0.35 0 −250   −125  0 −250 −125  0   1 0.3 0 −250   210000  125 T Then, [ B ] [ D ] = 0    0.3 1 −250 125   62500(0.91)  0  0 0 0.35  0 0 500    500 0   0  −125 −37.5 −87.5   −75 −250 −43.75    125 37.5 −87.5  T ∴ [ B ] [ D ] = 3.6923    −75 −250 43.75   0 0 175    500 0   150 The stiffness matrix for element (2) is:  −125 −37.5 −87.5   −75 −250 −43.75   0 125 0 0 0   −125  125 37.5 −87.5  1 [k ] = (5)(31250)(3.6923)  −75 −250 43.75  × 62500  0 −250 0 −250 0 500    −250 −125 −250 125 500 0   0 0 175    500 0   150

−1560 −43750 −18750  20313 6250  37500  20313 67970 1560 57030 −21880 −125000    6250 1560 37500 −20310 −43750 18750  [k ] = 9.2308  −1560 57030 −20310 67970 21880 −125000    −43750 −21880 −43750 21880  87500 0   0 250000   −18750 −1125000 18750 −125000

Introduction to Finite Element Analysis   •  43

Element (3) j=3

m=5

(3)

i=2

1 A = bh 2 1 A = ( 250)( 250) = 31250 mm 2 2 For element (3), we have coordinates xi = 500, yi = 0, xm = 250, ym = 125, xj = 500 and yj = 250 because the global axes are set up at node 1, and

bi = y j − ym =125 b j = ym − yi =125 bm = yi − y j = − 250 gi = xm − x j = − 250 g j = xi − xm = 250 gm = x j − xi = 0 Therefore, substituting in [B] matrix, 0 125 0 −250 0   125 1  0  [ B ] = 62500  0 −250 0 250 0  −250 125 250 125 0 −250 

[D] =

0   1 0.3 210000  0 3 1 0  . 0.91   0 0 0.35

0 −250   125  0 − 250 125   0   1 0.3 0 250   210000  125 T [ B ] [ D ] = 62500(0.91)  0 250 125  0.3 1 0    0 0 0.35   −250 0 0    0 −250   0

44  •   Using ANSYS for Finite Element Analysis

 125 37.5  −75 −250   125 37.5 T ∴ [ B ] [ D ] = 3.6923  250  75  −250 −75  0  0

−87.5 43.75  87.5   43.775  0   −87.5

Then, The stiffness matrix for element (3) is:  125 37.5  −75 −250   125 37.5 [k ] = (5)(31250)(3.6923)  75 250   −250 −75  0  0

−87.5 43.75  0  0 125 0 −250  125 87.5  1   0 250 0 250 − 0 0 ×  43.75  62500   −250 125 250 125 0 −250  0   −87.5

 37500 −20313 −6250  −20313 67979 1563   −6250 1563 37500 [k ] = 9.2308  −1563 −57031 −20313   −31250 18750 −31250   21875 −10938 −21875

−1563 −31250 21875  −57031 18750 −10938 −20313 −31250 −21875  67979 −18750 −10938 −18750 62500 0   −10938 0 21875 

Element (4) j=3

m=4 (4) i=5

1 A = bh 2 1 A = (500)(125) = 31250 mm 2 2 For element (4), we have coordinates xi = 250, yi = 125, xm = 0, ym = 250, xj = 500 and yj = 250, because the global axes are set up at node 1, and

bi = y j − ym = 0 b j = ym − yi =125

Introduction to Finite Element Analysis   •  45

bm = yi − y j = −125 gi = xm − x j = − 500 g j = xi − xm = 250 gm = x j − xi = 250 Therefore, substituting in [B] matrix,

[B] =

0 125 0 −125 0   0 1  0 500 0 250 0 250  − 62500   −500 0 250 125 250 125 

0   1 0.3 210000  [ D ] = 0.91 0.3 1 0   0 0 0.35 0 −500   0  0 − 500 0   0   1 0.3 0 250   210000  125 T Then, [ B ] [ D ] = 0    0.3 1 250 125   62500(0.91)  0  0 0 0.35  −125 0 250    250 −125  0 0 −175   0  −150 −500 0    125 37.5 87.5  T ∴ [ B ] [ D ] = 3.6923   250 43.75   75  −125 −37.5 87.5    250 −43.75  75 The stiffness matrix for element (4) is: 0 −175   0  −150 −500 0   0 125 0 −125 0   0  125 37.5 87.5  1 [k ] = (5)(31250)(3.6923)  75 250 43.75  × 62500  0 −500 0 250 0 250    −500 0 250 125 250 125   −125 −37.5 87.5    250 −43.75  75

46  •   Using ANSYS for Finite Element Analysis

0 −43750 −21880 −43750 21880   87500  0 250000 −18750 −125000 18750 −125000    −43750 −18750 37500 20310 6250 −1560  [k ] = 9.2308  −21880 −125000 20310 67979 1560 57030     −43750 18750 6250 1560 37500 −20310    57030 −20310 67970   21880 −125000 −1560

The stiffness matrix 20313  37500  20313 67969   0 0  0 0   0 0  k (1)  = 9.2308 ∗    0 0   −6250 −1563  −57031  1563  31250 18750 − −   −21875 −10938

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1563 −31250 −21875 −57031 −18750 −10938 0 0 0   0 0 0 0  0 0 0 0 0   0 0 0 0 0  0 37500 −20313 −31250 21875   0 −20313 67969 18750 −10938  0 −31250 18750 6250 0  0 21875 −10938 0 21875  0 0 0 0

−6250 −1563 0

20313 6250 −1560  37500  20313 67 9 70 1560 57030   6250 1560 37500 −20310  57030 −20310 67970  −1560  0 0 0 0 2 ( )  k  = 9.2308 ∗    0 0 0  0  0 0 0 0  0 0 0  0  − 43750 − 21880 − 43750 21880   −18750 −125000 18750 −125000

0 0  0  0 0 3 ( )  k  = 9.2308 ∗    0 0  0  0 0

0 0 0

0 0 37500

0 0 −20313

0 0 0 0 −43750 −18750  0 0 0 0 −21880 −125000  0 0 0 0 −43750 18750   0 0 0 0 21880 −125000   0 0 0 0 0 0  0 0 0 0 0 0   0 0 0 0 0 0  0 0 0 0 0 0   0 0 0 0 87500 0  0 0 0 0 0 250000 

0 0 −6250

0 0 −1563

0 −20313 67979 1563 0 −6250 1563 37500 0 −1563 −57031 −20313 0 0 0 0 0 0 0 0 0 −31250 18750 −31250 0 21875 −10938 −21875

−57031 −20313 67979 0 0 −18750 −10938

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

   −31250 21875   18750 −10938 −31250 −21875  −18750 −10938 0 0   0 0   62500 0  0 21875  0 0

0 0

Introduction to Finite Element Analysis   •  47 0 0  0  0 0 4 ( )  k  = 9.2308 ∗    0 0  0  0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

    0 0 0 0 0 0  0 0 0 0 0 0  37500 20310 6250 −1560 −43750 −18750   20310 67979 1560 57030 −2 21880 −125000  6250 1560 37500 −20310 −43750 18750   −1560 57030 −20310 67970 21880 −125000   −43750 −21880 −43750 21880 0 87500  −18750 −125000 18750 −125000 0 250000  0 0

0 0

0 0

0 0

0 0

0 0

The total stiffness matrix can be obtained:

[ K ] =  K (1)  +  K (2)  +  K (3)  +  K (4)  [ K ] = 9.2308 ∗ 40626 6250 −1560 0 0 −6250 1563  75000  40626 135939 1560 57030 0 0 − 1563 − 57031   6250 1560 75000 −40623 −6250 −1563 0 0  57030 −40623 135949 1563 −57031 0 0  −1560  0 0 1563 75000 0 6250 −6250 −1560  0 −1563 −57031 0 135958 1560 57030  0  −6250 −1563 0 0 6250 1560 75000 −40623  −57031 0 0 −1560 57030 −40623 135939  1563   −75000 −40630 −75000 40630 −75000 −40630 −75000 40630  −40625 −135938 40625 −135938 −40625 −135938 40625 −135938

−75000 −40625  −40630 −135938 −75000 40625   40630 −135938 −75000 −40625   −40630 −135938 −75000 40625   40630 −135938  243750 0  0 543750 

By applying the governing global matrix equation, we get: {F} = [K]{d}  R1x   75000 40626 6250 −1560 0 0 −6250 1563  R   939 1560 57030 0 0 −1563 −57031  1 y   40626 1359 20000   6250 1560 75000 −40623 −6250 −1563 0 0    57030 −40623 135949 1563 −57031 0 0  34641  −1560  R   0 0 −6250 1563 75000 0 6250 −1560  3x    =  0 −1563 −57031 0 135958 1560 57030  R3 y   0  R   −6250 −1563 0 0 6250 1560 750000 −40623  4x   −57031 0 0 −1560 57030 −40623 135939  R4 y   1563   −75000 −40630 −75000 40630 −75000 −40630 −75000 40630 0    −40625 −135938 40625 −135938 −40625 −135938 40625 −135938  0  

−75000 −40625    −40630 −135938   −75000 40625    40630 −135938  −75000 −40625    −40630 −135938  −75000 40625    40630 −135938    243750 0  0 543750  

0  0   2x   2y  0   0  0   0   5x  5y  

The previous governing global matrix equation has 10 equations with 10 unknowns, which are (R1x, R1y, R3x, R3y, R4x, R4y, d2x, d2y, d5x, d5y); we can get them by solving the following equations.

48  •   Using ANSYS for Finite Element Analysis

By using MATLAB, we can solve them and the results are as follows: R1x = − 7.6722 kN , R1 y = 13.675 kN , R1 = 15.680 kN R3 x = − 13.483 kN , R3 y = − 42.210 kN , R3 = 44.311 kN R4 x = − 4.4911 kN , R4 y = − 6.1025 kN , R4 = 7.5770 kN d 2 x = 0.0632 mm, d 2 y = 0.0513 mm d5 x = 0.0109 mm, d5 y = 0.0081 mm The displacement at point (250, 125) = (0.0109, 0.0081) mm We can determine the stresses in each element by using equation:

{s} = [ D ][ B ]{d } The stress for element A, we then have

{s} =

E

(1 − u ) 2

  1 u 0   b1  1   0 × 0 u 1 2A     g1 1− u  0 0 2  

0

g1 b1

b5 0 g5

0

g5 b5

b4 0 g4

 d1x  d   1y  0   d5 x  g4    d5 y b4    d   4x  d 4 y 

Substituting numerical values for matrix [D], [B] given by the analysis of element A and the appropriate part of {d}, we can obtain:  0   0   0   −125 0 250 2 −125 0   1 0.3 210000   0.0109  × 0 0 . 3 1 0 − 250 0 0 0 250 {s} =   0.0081   0.91( 62500)    0 0 0.35  −250 −125 0 250 250 −125   0     0   s  10.0615  x     s y  =  3.0185  Mpa    2.6169   txy  

1.9 FEA: Modeling, Errors, and Accuracy Modeling is the simulation of a physical structure or physical process by means of a substitute analytical or numerical construct. It is not simply preparing a mesh of nodes and elements. Modeling requires that the physical action of the problem be understood well enough to choose suitable

Introduction to Finite Element Analysis   •  49

A problem must be solved

Is FE analysis required?

Yes

Anticipate physical behavior plan how FE results will be checked to see if they are reasonable

A problem must be solved Plan revised FE model using insight provided by the current FE model

Analytical or experimental solution

Stop

Preprocess prepare the FE model

Solve equations of the FE model

No

Yes

Are results reasonable? Are error estimates small? Does model revision do little to alter computed results?

Postprocess display FE results Computer software

Figure 1.11.  Outline of an FE analysis project.

kinds of elements, and enough of them, to represent the physical action adequately. Figure 1.11 gives an outline of an FEA project. 1.9.1  Modeling Error Whatever the analysis method, we do not analyze the actual physical problem; rather, we analyze a mathematical model of it. Thus, we introduce modeling error. For example, in the elementary beam theory, we represent a beam by a line (its axis) and typically ignore deformations associated with transverse shear. This is an excellent approximation for slender beams, but not for very short beams. Or, for axial load problem of ­Figure 1.12, we would probably assume that a state of uniaxial stress prevails throughout the bar, which is proper if taper is slight, but improper if taper is pronounced. Real structures are not so easily classified, as they are often built of parts that would be idealized mathematically in different ways and have cutouts, stiffeners, and connectors whose behavior is uncertain. The foregoing considerations must be addressed in order to decide what types of elements to use and how many of them. If a beam is deep, transverse shear deformation may become important and should be included in beam elements. If a beam is very deep, two- or three-­dimensional elements are more appropriate than beam elements. If a beam has a wide cross-section, the plate theory may be more appropriate than beam theory (then, of course, choose plate elements rather than beam elements). If an

50  •   Using ANSYS for Finite Element Analysis hO hO + ∆h x

P

LT

(a)

x, u

[email protected] = LT

(b)

Figure 1.12.  (a) A tapered bar loaded by axial force P, (b) Discretization of the bar into four uniform two-node elements of equal length.

axisymmetric pressure vessel has a thick wall, one should regard it as a solid of revolution, rather than a shell of revolution, and choose axisymmetric solid elements, rather than axisymmetric shell elements. 1.9.2 Discretization Error Let us now consider the axially tapered bar of Figure 1.4 in more detail and describe how the FEM implements the mathematical model. We will assume that a satisfactory mathematical model is based on a state of uniaxial stress. An analytical solution is then rather easy, but we pretend not to know it and ask for an FE solution instead. We discretize the mathematical model by dividing it into two node elements of constant cross-section, as shown in Figure 1.4b. Each element has length L, accounts only for a constant uniaxial stress along its length, and has an axial deformation given by the elementary formula PL/AE. For each element, A may be taken as constant and equal to the cross-sectional area of the tapered bar at an x coordinate corresponding to the element center. The displacement of load P is equal to the sum of the element deformations. Intuitively, we expect that the exact displacement be approached as more and more elements are used to span the total length LT. However, even if many elements are used there is an error, known as discretization error, which exists because the physical structure and the mathematical model each has infinitely many degrees of freedom (DOF) (namely, the displacements of infinitely many points), while the FE model has a finite number of DOF (the axial ­displacements of its nodes). How Many Elements Are Enough? Imagine that we carry out two FEAs, the second time using a more refined mesh. The second FE model will have lesser discretization error than the first, and will also represent the geometry better if the physical object has

Introduction to Finite Element Analysis   •  51

curved surfaces. If the two analyses yield similar solutions, we suspect that results are not much in error. Or, we might establish a sequence of solutions by solving the problem more than twice, using a finer mesh each time. By studying how the sequence converges, we may be able to state with some confidence that the results from the finest mesh are in error by less than, say, 5 percent. 1.9.3 Numerical Error After the analyst has introduced modeling error and discretization error, the computer introduces numerical error by rounding or truncating ­numbers as it builds matrices and solves equations. Usually, the numerical error is small, but some modeling practices can greatly increase it. FE computer programs have become widely available, easier to use, and can display results with attractive graphics. Even an inept user can produce some kind of answer. It is hard to disbelieve FE results because of the effort needed to get them and the polish of their presentation. But, any model, good or bad, can produce smooth and colorful stress contours. It is possible that most FEAs are so flawed that they cannot be trusted. Even a poor mesh, inappropriate element types, incorrect loads, or improper supports may produce results that appear reasonable on casual inspection.

200 mm

20 mm

1-ms presuure pulse

4 mm

10 3c 3c

Displacement, mm

5

2

2

9 5

7 0

3b

2

7 1

–5

–10 0

10

9

10

7

8

5

1

9

3b

8

5

8

1 10

1

2

3

4

5

6

7

Figure 1.13.  Lateral midpoint displacement versus time for a beam loaded by a pressure pulse. The material is elastic-perfectly plastic. Plots were generated by various users and various codes [2].

52  •   Using ANSYS for Finite Element Analysis

A poor model may have defects that are not removed by refinement of the mesh. A responsible user must understand the physical nature of the ­problem and the behavior of finite elements well enough to prepare a suitable model and evaluate the quality of the results. Competence in using FE for stress analysis does not imply competence in using FE for (say) magnetic field problems. The engineer who uses the software, not the software vendor, even if the results are affected by the errors in the software, takes responsibility for the results produced. Figure 1.13 is an example of discrepancies that may appear. A pressure pulse is applied to a straight beam with hinge supports. The loading causes the material to yield and the beam to vibrate. Analysis seeks to track the lateral displacement of the midpoint as a function of time. The results plotted come from 10 reputable analysis codes and were obtained by users regarded as expert. Yet, if any of the curves is correct, we c­ annot tell which one it is. Admittedly, the problem is difficult. The results ­indicate strong sensitivities of both physical and computational nature. This example reminds us that any analysis program is based on theory and approximation, and that a user may push the program beyond its range of validity.

Chapter 2

Static Analysis Using ANSYS 2.1 Overview of Structural Analysis A brief overview of structural analysis as carried out by ANSYS is ­provided in the following sections. 2.1.1 Definition of Structural Analysis Structural analysis is probably the most common application of the finite element method. The term structural (or structure) implies not only civil engineering structures, such as bridges and buildings, but also naval, aeronautical, and mechanical structures such as ship hulls, aircraft bodies, and machine housings, as well as mechanical components, such as pistons, machine parts, and tools. 2.1.2 Definition of Static Analysis A static analysis calculates the effects of steady loading conditions on a structure, while ignoring inertia and damping effects, such as those caused by time-varying loads. A static analysis can, however, include steady inertia loads (such as gravity and rotational velocity) and time-varying loads that can be approximated as static equivalent loads (such as the static equivalent wind and seismic loads commonly defined in many building codes). Static analysis is used to determine the displacements, stresses, strains, and forces in structures or components caused by loads that do not induce significant inertia and damping effects. Steady loading and response conditions are assumed, that is, the loads and the structure’s response are assumed to vary slowly with respect to time. The kinds of loadings that can be applied in a static analysis include:

54  •   Using ANSYS for Finite Element Analysis

• • • • •

externally applied forces and pressures, steady-state inertial forces (such as gravity or rotational velocity), imposed (nonzero) displacements, temperatures (for thermal strain), fluences (for nuclear swelling).

2.1.3 Linear versus Nonlinear Static Analyses A static analysis can be either linear or nonlinear. All types of nonlinearities are allowed—large deformations, plasticity, creep, stress stiffening, contact (gap) elements, hyperelastic elements, and so on. This chapter focuses on linear static analyses, with brief references to nonlinearities. Details of how to handle nonlinearities are described in Nonlinear Structural Analysis notes.

2.2 Static Analysis Procedure

2.2.1  Preprocessing: Building a Model Building a finite element model requires more of your time than any other part of the analysis. First, you specify a jobname and analysis title. Then, you use the PREP7 preprocessor to define the element types, element real constants, material properties, and the model geometry. 2.2.1.1  Defining the Jobname The jobname is a name that identifies the ANSYS job. When you define a jobname for an analysis, the jobname becomes the first part of the name of all files the analysis creates. (The extension or suffix for these files’ names

Static Analysis Using ANSYS   •  55

is a file identifier such as .DB.) By using a jobname for each analysis, you insure that no files are overwritten. If you do not specify a jobname, all files receive the name FILE or file, depending on the operating system. You can change the default jobname as follows: by using the initial jobname entry option when you enter the ANSYS program, either via the launcher or on the ANSYS execution command. From within the ANSYS program, you can use either of the following: Command(s):/FILNAME or GUI: Utility Menu > File > Change Jobname. The /FIL NAME command is valid only at the Begin level. It lets you change the jobname.

2.2.1.2  Defining an Analysis Title The /TITLE command (Utility Menu > File > Change Title), defines a title for the analysis. ANSYS includes the title on all graphics displays and on the solution output. You can issue the /STITLE command to add subtitles; these will appear in the output, but not in graphics displays. 2.2.1.3  Defining Units The ANSYS program does not assume a system of units for your analysis. Except in magnetic field analyses, you can use any system of units so long as you make sure that you use that system for all the data you enter. (Units must be consistent for all input data.) For micro-electro ­mechanical ­systems (MEMS), where dimensions are on the order of microns, see the conversion factors in System of Units in the ANSYS Coupled-Field Analysis Guide. Using the /UNITS command, you can set a marker in the ANSYS database, indicating the system of units that you are using. This command does not convert data from one system of units to another; it simply serves as a record for subsequent reviews of the analysis.

56  •   Using ANSYS for Finite Element Analysis

2.2.1.4  Defining Element Types The ANSYS element library contains more than 150 different element types. Each element type has a unique number and a prefix that ­identifies the element category: BEAM4, PLANE77, SOLID96, and so on. The f­ ollowing element categories are available: Beam HF (high frequency) Mass Prets ­(pretension) Target

Circuit Combination Contact Hyperelastic Infinite Interface

Fluid Link

Matrix Shell

Mesh Solid

Pipe Source

Plane Surface

Transducer

User

Viscoelastic or viscoplastic

1. Main Menu > Preprocessor > Element Type > Add/Edit/Delete

2. Add an element type.

Static Analysis Using ANSYS   •  57

3. Structural shell family of elements. 4. Choose the eight-noded layered shell element. (SHELL99). 5. OK to apply the element type and close the dialog box.

6. Close the element type dialog box.

2.2.1.5  Defining Element Real Constants Element real constants are properties that depend on the element type, such as cross-sectional properties of a beam element. For example, real constants for BEAM3, the 2D beam element, are area (AREA), moment of inertia (IZZ), height (HEIGHT), shear deflection constant (SHEARZ), initial strain (ISTRN), and added mass per unit length (ADDMAS). Not all element types require real constants, and different elements of the same type may have different real constant values. 1. Main Menu > Preprocessor > Element Type > Add/Edit/Delete

58  •   Using ANSYS for Finite Element Analysis

2. Add a real constant set.

3. OK for PLANE42

4. Enter the value of thickness

Static Analysis Using ANSYS   •  59

5. OK to define the real constant and close the dialog box.

2.2.1.6  Defining Material Properties Most element types require material properties. Depending on the application, material properties can be linear (see Linear Material Properties) or nonlinear (see Nonlinear Material Properties). As with element types and real constants, each set of material properties has a material reference number. The table of material reference numbers versus material property sets is called the material table. Within one analysis, you may have multiple material property sets (to correspond with multiple materials used in the model). ANSYS identifies each set with a unique reference number. While defining the elements, you point to the appropriate material reference number using the MAT command. 2.2.1.7 Linear Material Properties Linear material properties can be constant or temperature-dependent, and isotropic or orthotropic. To define constant material properties

60  •   Using ANSYS for Finite Element Analysis

(either ­isotropic or orthotropic), use one of the following: Command(s): MPGUI:Main Menu > Preprocessor > Material Props > Material Models. (See Material Model Interface for details on the GUI.) You also must specify the appropriate property label; for example, EX, EY, EZ for Young’s ­modulus; KXX, KYY, KZZ for thermal conductivity; and so forth. For ­isotropic material, you need to define only the X-direction property; the other directions default to the X-direction value. For example: MP, EX,1,2E11 ! Young’s modulus for material ref. no. 1 is 2E11 MP, DENS,1,7800 ! Density for material ref. no. 1 is 7800 MP, KXX,1,43 ! Thermal conductivity for material ref. no 1 is 43

2.2.1.8 Nonlinear Material Properties Nonlinear material properties are usually tabular data, such as plasticity data (stress–strain curves for different hardening laws), magnetic field data (B–H curves), creep data, swelling data, hyperelastic material data, and so on. The first step in defining a nonlinear material property is to ­activate a data table using the TB command (see Material Model Interface for the GUI equivalent). For example, TB,BH,2 activates the B–H table for material reference number 2. To enter the tabular data, use the TBPT command. For example, the following commands define a B–H curve: TBPT,DEFI,150,.21 TBPT,DEFI,300,.55 TBPT,DEFI,460,.80 TBPT,DEFI,640,.95 TBPT,DEFI,720,1.0 TBPT,DEFI,890,1.1 TBPT,DEFI,1020,1.15 TBPT,DEFI,1280,1.25 TBPT,DEFI,1900,1.4

Static Analysis Using ANSYS   •  61

You can verify the data table through displays and listings using the TBPLOT or TBLIST commands.

2.2.2 Creating the Model Geometry Once you have defined material properties, the next step in an analysis is generating a finite element model—nodes and elements—that adequately describes the model geometry. There are two methods to create the finite element model: solid modeling and direct generation. With solid modeling, you describe the geometric shape of your model, then instruct the ANSYS program to automatically mesh the geometry with nodes and elements. You can control the size and shape in the elements that the program creates. With direct generation, you manually define the location of each node and the connectivity of each element. Several convenience operations, such as copying patterns of existing nodes and elements, symmetry reflection, and so on, are available. Details of the two methods and many other aspects related to model generation—coordinate systems, working planes, coupling, constraint equations, and so on, are described in the ANSYS Modeling and Meshing Guide. 2.2.2.1 Apply Loads and Obtain the Solution In this step, you use the SOLUTION processor to define the analysis type and analysis options, apply loads, specify load step options, and initiate the finite element solution. You also can apply loads using the PREP7 preprocessor. 2.2.2.2  Defining the Analysis Type and Analysis Options You choose the analysis type based on the loading conditions and the response you wish to calculate. For example, if natural frequencies and

62  •   Using ANSYS for Finite Element Analysis

mode shapes are to be calculated, you would choose a modal analysis. You can perform the following analysis types in the ANSYS program: static (or steady-state), transient, harmonic, modal, spectrum, buckling, and substructuring. Not all analysis types are valid for all disciplines. Modal analysis, for example, is not valid for a thermal model. The analysis guide manuals in the ANSYS documentation set describe the analysis types available for each discipline and the procedures to do those analyses. Analysis options allow you to customize the analysis type. Typical analysis options are the method of solution, stress stiffening on or off, and Newton–Raphson options. To define the analysis type and analysis options, use the ANTYPE command (Main Menu > Preprocessor > Loads > Analysis Type > New Analysis or Main Menu > Preprocessor > Loads > Analysis Type > Restart) and the appropriate analysis option commands (TRNOPT, HROPT, MODOPT, SSTIF, NROPT, etc.). For GUI equivalents for the other commands, see their descriptions in the ANSYS Elements Reference. If you are performing a static or full transient analysis, you can take advantage of the Solution Controls dialog box to define many options for the analysis. For details about the Solution Controls dialog box, see Solution. You can specify either a new analysis or a restart, but a new analysis is the choice in most cases. A single frame restart that allows you to resume a job at its end point or abort point is available for static (steady-state), harmonic (2D magnetic only), and transient analyses. A multi-frame restart that allows you to restart an analysis at any point is available for static or full transient structural analyses. See Restarting an Analysis for complete information on performing restarts. The various analysis guides discuss additional details necessary for restarts. You cannot change the analysis type and analysis options after the first solution. A sample input listing for a structural transient analysis is shown next. Remember that the discipline (structural, thermal, magnetic, etc.) is implied by the element types used in the model. ANTYPE, TRANS TRNOPT, FULL NLGEOM, ON Once you have defined the analysis type and analysis options, the next step is to apply loads. Some structural analysis types require other items to be defined first, such as master degrees of freedom and gap conditions. The ANSYS Structural Analysis Guide describes these items where necessary. 2.2.2.3 Applying Loads The word loads as used in ANSYS documentation includes boundary conditions (constraints, supports, or boundary field specifications), as well as

Static Analysis Using ANSYS   •  63

other externally and internally applied loads. Loads in the ANSYS program are divided into six categories: DOF Constraints, Forces, Surface Loads, Body Loads, Inertia Loads, and Coupled-field Loads. You can apply most of these loads either on the solid model (keypoints, lines, and areas) or the finite element model (nodes and elements). For details about the load categories and how they can be applied on your model, see Loading in this book. Two important load-related terms you need to know are load step and substep. A load step is simply a configuration of loads for which you obtain a solution. In a structural analysis, for example, you may apply wind loads in one load step and gravity in a second load step. Load steps are also useful in dividing a transient load history curve into several segments. Substeps are incremental steps taken within a load step. You use them mainly for accuracy and convergence purposes in transient and nonlinear analyses. Substeps are also known as time steps—steps taken over a period of time. 2.2.2.4 Specifying Load Step Options Load step options are options that you can change from load step to load step, such as number of substeps, time at the end of a load step, and output controls. Depending on the type of analysis you are doing, load step options may or may not be required. The analysis procedures in the analysis guide manuals describe the appropriate load step options as necessary. See Loading for a general description of load step options. 2.2.2.5 Initiating the Solution To initiate solution calculations, use either of the following: Command(s): SOLVE GUI: Main Menu > Solution > Solve > Current LS. When you issue this command, the ANSYS program takes model and loading information from the database and calculates the results. The results are written to the results file (Jobname.RST, Jobname.RTH, Jobname.RMG, or Jobname.RFL) and also to the database. The only difference is that only one set of results can reside in the database at one time, while you can write all sets of results (for all substeps) to the results file. 2.2.3 Review the Results Once the solution has been calculated, you can use the ANSYS postprocessors to review the results. Two postprocessors are available: POST1

64  •   Using ANSYS for Finite Element Analysis

and POST26. You use POST1, the general postprocessor, to review results at one substep (time step) over the entire model or selected portion of the model. The command to enter POST1 is /POST1 (Main Menu > General Postproc), valid only at the Begin level. You can obtain contour displays, deformed shapes, and tabular listings to review and interpret the results of the analysis. POST1 offers many other capabilities, including error estimation, load case combinations, calculations among results data, and path operations. You use POST26, the time history postprocessor, to review results at specific points in the model over all time steps. The command to enter POST26 is /POST26 (Main Menu > TimeHist Postpro), valid only at the Begin level. You can obtain graph plots of results data versus time (or frequency) and tabular listings. Other POST26 capabilities include arithmetic calculations and complex algebra.

Chapter 3

Geometric Modeling The ultimate purpose of finite element analysis is to recreate mathematically the behavior of an actual engineering system. In other words, the analysis must be an accurate mathematical model of a physical ­prototype. In the broadest sense, this model comprises all the nodes, elements, material properties, real constants, boundary conditions, and other features, including geometry, that are used to represent the physical system. The ANSYS program offers you the following approaches to geometric modeling: • Creating a solid model within ANSYS. • Using direct generation. • Importing a model created in a computer-aided design (CAD) ­system.

3.1 Typical Steps Involved in Model Generation Within ANSYS 1. Begin by planning your approach. Determine your objectives, decide what basic form your model will take, choose appropriate element types, and consider how you will establish an appropriate mesh density. You will typically do this general planning before you initiate your ANSYS session. 2. Enter the preprocessor (PREP7) to initiate your model building ­session. Most often, you will build your model using solid m ­ odeling procedures. 3. Establish a working plane.

66  •   Using ANSYS for Finite Element Analysis

4. Generate basic geometric features using geometric primitives and Boolean operators. 5. Activate the appropriate coordinate system. 6. Generate other solid model features from the bottom up. That is, ­create keypoints, and then define lines, areas, and volumes as needed. 7. Use more Boolean operators or number controls to join separate solid model regions together as appropriate. 8. Create tables of element attributes (element types, real constants, material properties, and element coordinate systems). 9. Set element attribute pointers. 10. Set meshing controls to establish your desired mesh density if desired. This step is not always required because default element sizes exist when you enter the program. (If you want the program to refine the mesh automatically, exit the preprocessor at this point and activate adaptive meshing.) 11. Create nodes and elements by meshing your solid model. 12. After you have generated nodes and elements, add features such as surface-to-surface contact elements, coupled degrees of freedom, and constraint equations 13. Save your model data to Jobname.DB. 14. Exit the preprocessor.

3.2 Importing Solid Models Created in CAD systems As an alternative to creating your solid models within ANSYS, you can create them in your favorite CAD system and then import them into ANSYS for analysis, by saving them in the IGES file format or in a file format supported by an ANSYS connection product. Creating a model using a CAD package has the following advantages: • You avoid a duplication of effort by using existing CAD models to generate solid models for analysis. • You use more familiar tools to create models. However, models imported from CAD systems may require extensive repair if they are not of suitable quality for meshing.

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68  •   Using ANSYS for Finite Element Analysis

3.3 Solid Modeling The purpose of using a solid model is to relieve you of the time-consuming task of building a complicated finite element model by direct generation. Let us take a brief look at some of the solid modeling and meshing operations that you can use to speed up the creation of your final analysis model. Building your model from the bottom up: Keypoints, the points that define the vertices of your model, are the lowest-order solid model entities. If, in building your solid model, you first create your keypoints, and then use those keypoints to define the higher-order solid model entities (that is, lines, areas, and volumes), you are said to be building your model from the bottom up. Keep in mind that models built from the bottom up are defined within the currently active coordinate system. Building your model from the top down: The ANSYS program also gives you the ability to assemble your model using geometric primitives, which are fully defined lines, areas, and volumes. As you create a primitive, the program automatically creates all the lower entities associated with it. If your modeling effort begins with the higher primitive entities,

Geometric Modeling   •  69

you are said to be building your model from the top down. You can freely combine bottom-up and top-down modeling techniques, as appropriate, in any model. Remember that geometric primitives are built within the working plane, while bottom-up techniques are defined against the active coordinate system. If you are mixing techniques, you may wish to consider using the CSYS, WP or CSYS, 4 command to force the coordinate system to follow the working plane. Using Boolean operators: You can sculpt your solid model using intersections, subtractions, and other Boolean operations. Booleans allow you to work directly with higher solid model entities to create complex shapes. (Both bottom-up and top-down creations can be used in Boolean operations.) Dragging and rotating: Boolean operators, although convenient, can be computationally expensive. Sometimes, a model can be constructed more efficiently by dragging or rotating. Moving and copying solid model entities: A complicated area or volume that appears repetitively in your model needs only be constructed once; it can then be moved, rotated, and copied to a new location on your model. You might also find it more convenient to place geometric primitives in their proper location by moving them, rather than by changing the working plane.

3.4 Tutorial 1: Solid Modeling using 2D Primitives In this tutorial, the bottom-up solid modeling of a retaining ring is presented. In the ring diagram shown, build a solid model of the retaining ring using extrusion command. When you are done, save the database. Rather than drawing two circles and subtracting one from the other, circular annulus will be used to create the main ring. A 2D cross-section of the retaining ring will be first drawn, having the complete details (including fillet radius) shown. Solid Boolean operations will be performed to make a single object from the various entities. The 2D object, together with its mesh, will be extruded to obtain the final retaining ring.

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Program Options Boolean operations, add, subtract circle, annulus modeling, reflect ­(mirror) modeling, extrude Step-by-Step ANSYS Solution 1. Start ANSYS; save file with your given job name: File -> Save As -> Retaining-Ring -> OK 2. Create the ring by using the annulus command: Preprocessor -> Modeling -> Create -> Areas -> Circle -> Annulus -> In dialog box, enter WP X = 0, WP Y = 0, Rad-1 = 8, Rad-2 = 6.5 -> OK

3. Create a rectangle: Preprocessor -> Modeling -> Create -> Areas -> Rectangle -> By Dimensions -> In dialog box, enter X1, X2 X-coords = -2, 2 and Y1, Y2 Y-coords = -7.5, -9.5 -> OK

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4. Add the two areas (ring and rectangle) by Boolean operation to make a single area: Preprocessor -> Modeling -> Operate -> Booleans -> Add -> Areas -> Using the mouse, pick the ring (point to the ring and left-click), then pick the rectangle, or select pick all -> OK Save _DB

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5. Create another rectangle: Preprocessor -> Modeling -> Create -> Areas -> Rectangle -> By Dimensions -> In dialog box, enter X1, X2 X-coords = -0.25, 0.25 and Y1, Y2 Y-coords = -6, -10 -> OK

6. Subtract the new rectangle from the existing area to get middle slot of the retaining ring: Preprocessor -> Modeling -> Operate -> Booleans -> Subtract -> Areas -> Using the mouse, pick the existing area -> Apply -> pick rectangle -> OK Neglect warning; close. Save _DB

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7. Create a circle to get one of the holes in the retaining ring: Preprocessor -> Modeling -> Create -> Areas -> Circle -> Solid Circle -> In dialog box, enter WP X = 1, WP Y = -8, Radius = 0.6 -> OK

8. Make a mirror copy of the circle to get the second hole in the retaining ring: Preprocessor -> Reflect -> Areas -> Pick the circle -> Apply -> In dialog box, enter KINC Keypoint Increment = 0 -> OK The hole on the right will be mirrored on the left side.

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9. Subtract the two circles from the main body to get the two holes: Preprocessor -> Modeling -> Operate -> Booleans -> Subtract -> Areas -> Using the mouse, pick main body -> Apply -> pick one circle, pick other circle -> OK Save _DB

10. Change view to make the lines visible, so that you can perform the fillet operation: Menu bar at the top -> Plot -> Lines

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11. Round-off on the two sides of the bottom part of the retaining ring using the fillet command: Preprocessor -> Modeling -> Create -> Lines -> Line Fillet -> Using the mouse, pick the arc portion of the right main body, then pick the straight line on the right bottom part -> Apply -> In the dialog box, enter RAD Fillet radius = 1 -> Apply Similarly, pick the arc portion of the left main body, then pick the straight line on the left bottom part -> Apply -> In the dialog box, enter RAD Fillet radius = 1 -> OK Save _DB

12. Go back to the area view mode; lines are no more visible: Menu bar at the top -> Plot -> Areas Note: Before generating the mesh, we need to define two different element types: a 2D element for the cross-sectional view of the retaining ring that we have just created, and a 3D element to get the actual solid retaining ring having the required thickness. The 2D area will be first meshed, then the area will be extruded to create a 3D volume. The mesh will be extruded along with the geometry, and the 3D mesh elements will be automatically generated in the volume. 13. Define the two element types: Preprocessor -> Element Type -> Add/Edit/Delete -> Add -> In the dialog box, click on Structural-Solid on the left, then click on Quad 4 node 42 on the right -> Apply -> Again, in the dialog box, click on Structural-Solid on the left, then click on Brick 8node 45 on the right -> OK -> Close Save _DB

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14. Perform meshing on the 2D cross-sectional view of the retaining ring: Preprocessor -> Meshing -> Mesh -> Areas -> Free -> Pick All Neglect the warning; close Save _DB

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15. Extrude the meshed area into a meshed volume: Preprocessor -> Modeling -> Operate -> Extrude -> Elem Ext Opts -> In the dialog box, [TYPE] Element type number: click on 2 SOLID45 in the drop-down menu, then enter [VAL1] No. Element divs = 10 -> OK

Preprocessor -> Modeling -> Operate -> Extrude -> Areas -> By XYZ Offset -> Pick All -> In the dialog box, enter DX = 0, DY = 0, DZ = 1 -> OK Neglect warning; close. SAVE_DB

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16. Change the view to 3D isometric view to see the complete retaining ring: Menu bar at the top -> Plot_Ctrls -> Pan Zoom Rotate … -> Click on ISO -> Close

17. Change the view to un-meshed one so that you can see the solid drawing of the retaining ring: Menu bar at the top -> Plot -> Volumes

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3.5 Tutorial 2: Solid Modeling using 3D Primitives In this tutorial, top-down solid modeling of a pillow block will be addressed. The pillow block diagram shown, build a half symmetry solid model of this pillow block. When you are done, save the database. Bracket 1.5 R, 0.75

Bushing Counterbore 1.0R, 01875 Web, 0.15

Four 0.75D holes, 0 75 offset

1.7 Base 6x3x1

All dimensions in inches

Program Options APDL, scalar parameters Boolean operations, GLUE Boolean operations, subtract graphics, hidden-line options graphics, viewing angle primitives, volumes working plane. Step-by-Step ANSYS Solution 1. Enter ANSYS in the working directory specified by your instructor using p-block as the job name. 2. Switch to isometric view: Utility Menu > PlotCtrls > Pan, Zoom, Rotate … [ISO] Or issue: /VIEW,1,1,1,1 3. Create the base of the pillow block: Main Menu > Preprocessor > -Modeling- Create > -Volumes- Block > By Dimensions ...X1 = 0, X2 = 3, Y1 = 0, Y2 = 1, Z1 = 0, Z2 = 3, then [OK] Or issue: /PREP7 BLOCK,0,3,0,1,0,3

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4. Offset working plane to location X = 2.25, Y = 1.25, Z = 0.75: Utility Menu > WorkPlane > Offset WP by Increments … Set X,Y,Z Offsets = 2.25, 1.25, 0.75 Set XY, YZ, ZX Angles = 0, -90, 0, then [OK] Or issue: WPOFF, 2.25, 1.25, 0.75 WPROT, 0, -90, 0

5. Create a solid cylinder having a diameter of 0.75 inches and a depth of -1.5 inches: Main Menu > Preprocessor > -Modeling- Create > -Volumes- Cylinder > Solid Cylinder + Radius = 0.75/2 Depth = -1.5, then [OK] Or issue: CYL4, , ,0.75/2, , , ,-1.5

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6. Copy the solid cylinder to new location with: DZ=1.5: Main Menu > Preprocessor > Copy > Volumes +Pick the cylinder volume (Vol. Number 2), then [OK] DZ = 1.5, then [OK] Or issue: VGEN,2,2, , , , ,1.5, ,0

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7. Subtract the two solid cylinders from the base: Main Menu > Preprocessor > -Modeling- Operate > -Booleans- Subtract > Volumes + Pick the base volume (Vol. 1), then [OK] Pick the two cylinder volumes (Vols. 2 and 3), then [OK] Or issue: VSBV, 1, ALL

8. Align the working plane with the Global Cartesian origin: Utility Menu > WorkPlane > Align WP with > Global Cartesian Or issue: WPCSYS,-1,0 VPLOT

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9. Create the base of the bushing bracket:

Main Menu > Preprocessor > -Modeling- Create > -Volumes- Block > By 2 Corners & Z + WP X = 0 WP Y = 1 Width = 1.5 Height = 1.75 Depth = 0.75, then [OK] Or issue: BLC4,0,1,1.5,1.75,0.75 10. Offset working plane to the front face of the bushing bracket: Utility Menu > WorkPlane > Offset WP to > Keypoints + Pick keypoint at the top left corner of the front face, then [OK] Or issue: KWPAVE, 16

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11. Create the arch of the bushing bracket: Main Menu > Preprocessor > -Modeling- Create > -Volumes- Cylinder > Partial Cylinder + WP X = 0 WP Y = 0 Rad-1 = 0 Theta-1 = 0 Rad-2 = 1.5 Theta-2 = 90 Depth = -0.75, then [OK] Or issue: CYL4,0,0,0,0,1.5,90,-0.75

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12. Create cylinders for the counterbore and the through hole in the bushing bracket: Main Menu > Preprocessor > -Modeling- Create > -Volumes- Cylinder > Solid Cylinder + WP X = 0 WP Y = 0 Radius = 1 Depth = -0.1875, then [Apply] WP X = 0 WP Y = 0 Radius = 0.85 Depth = -2, then [OK] Or issue: CYL4,0,0,1, , , ,-0.1875 CYL4,0,0,0.85, , , ,-2

13. Subtract the two solid cylinders to form the counterbore and bushing through-hole: Main Menu > Preprocessor > -Modeling- Operate > -Booleans- Subtract > Volumes + Pick the two volumes that form the base and the arch of the bushing bracket [Apply] Pick the counterbore cylinder [Apply] Pick the same two base volumes [Apply] Pick the through-hole cylinder [OK]

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14. Merge coincident keypoints: Main Menu > Preprocessor > Numbering Ctrls > Merge Items … Set Label to “Keypoints”, then [OK] Or issue: NUMMRG,KP 15. Create the web: (a)  Create a keypoint in the middle of the front top edge of the base: Main Menu > Preprocessor > -Modeling- Create > Keypoints > KP between KPs + Pick the two upper front corner keypoints of the base, then [OK] RATI = 0.5, then [OK] Or issue: KBETW,7,8,0,RATI,0.5

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(b)  Create a triangular area: Main Menu > Preprocessor > -Modeling- Create > -Areas- Arbitrary > Through KPs + Pick the 1st keypoint where the base of the bushing bracket intersects the base of the pillow block at X=1.5 Pick the 2nd keypoint where the base of the bushing bracket intersects the bottom arch surface at X=1.5 Pick the 3rd keypoint that was created in step 14a at X=1.5, Y=1, Z=3 [Ok] Or issue: A,14,15,9

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(c)  Extrude area along area normal: Main Menu > Preprocessor > -Modeling- Operate > Extrude > -Areas- Along Normal + Pick the triangular area created in step 14b, then [OK] DIST = -0.15, then [OK] Or issue: VOFFST,3,-0.15

16. Glue the volumes together: Main Menu > Preprocessor > -Modeling- Operate > -Booleans- Glue > Volumes + [Pick All] Or issue: VGLUE,ALL

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17. Turn volume numbers on and then plot volumes: Utility Menu > PlotCtrls > Numbering … Set Volume numbers on, then [OK] Or issue: /PNUM,VOLU,1 VPLOT

18. Save and exit ANSYS: Pick the “SAVE_DB” button in the Toolbar Pick the “QUIT” button in the Toolbar Select “Quit—No Save!” [OK] Or issue: FINISH /EXIT,ALL

Chapter 4

Static Analysis Using Line Elements 4.1 Tutorial 3: Static Analysis using Truss Elements In this tutorial, analysis of a two-dimensional (2D) truss is presented. The truss diagram shown, determine the nodal deflections, reaction forces, and stress for the truss system shown as follows (E = 200 GPa, A = 3250 mm2). 2

4 3

280 kN 1 1 60º

2

3.6 m

210 kN 5

60º

3

8

4

7 6

3.6 m

280 kN

5

6 11 360 kN

9 10

7

3.118 m

R

3.6 m

(Modified from Chandrupatla and Belegunda, Introduction to Finite Elements in Engineering, p. 123) Step-by-Step ANSYS Solution 1. Give the simplified version a title (such as Bridge Truss Tutorial): In the Utility menu bar, select File > Change Title:

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The following window will appear:

Enter the title and click OK. This title will appear in the bottom-left corner of the Graphics window once you begin. Note: to get the title to appear immediately, select Utility Menu > Plot > Replot 2. Enter keypoints. The overall geometry is defined in ANSYS using keypoints, which specify various principal coordinates to define the body. For this example, these keypoints are the ends of each truss. We are going to define seven keypoints for the simplified structure as given in the following table (these keypoints are depicted by numbers in the preceding figure).

keypoint 1 2 3 4 5 6 7

Coordinate x y 0 0 1800 3118 3600 0 5400 3118 7200 0 9000 3118 10800 0

From the ANSYS Main Menu select: Preprocessor > Modeling > Create > Keypoints > In Active CS

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The following window will then appear:

To define the first keypoint, which has the coordinates x = 0 and y = 0: Enter keypoint number 1 in the appropriate box, and enter the x, y coordinates: 0, 0 in their appropriate boxes (as shown earlier). Click Apply to accept what you have typed. Enter the remaining keypoints using the same method. Note: When entering the final data point, click on OK to indicate that you are finished entering keypoints. If you first press Apply and then OK for the final keypoint, you will have defined it twice! If you did press Apply for the final point, simply press Cancel to close this dialog box. The units of measure (i.e., mm) were not specified. It is the responsibility of the user to ensure that a consistent set of units is used for the problem; thus making any conversions where necessary.

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Correcting mistakes When defining keypoints, lines, areas, volumes, elements, constraints, and loads, you are bound to make mistakes. Fortunately, these are easily corrected so that you do not need to begin from scratch every time an error is made! Every Create menu for generating these various entities also has a corresponding Delete menu for fixing-up things. 3. Form lines; the keypoints must now be connected. We will use the mouse to select the keypoints to form the lines. In the main menu, select: Preprocessor > Modeling > Create > Lines > Lines > In Active Coord. The following window will then appear: Use the mouse to pick keypoint #1 (i.e., click on it). It will now be marked by a small yellow box. Now move the mouse toward keypoint #2. A line will now show on the screen joining these two points. Left-click and a permanent line will appear.

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Connect the remaining keypoints using the same method. When you are done, click on OK in the Lines in Active Coord window. Disappearing Lines: Please note that any lines you have created may disappear throughout your analysis. However, they have most likely NOT been deleted. If this occurs at any time, from the Utility Menu, select: Plot > Lines

4. Define the type of element It is now necessary to create elements. This is called meshing. ANSYS first needs to know what kind of elements to use for our problem. From the Preprocessor Menu, select: Element Type > Add/Edit/Delete. Click on the Add... button.

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The following window will appear:

For this example, we will use the 2D spar element as selected in the preceding figure. Select the element shown and click OK. You should see Type 1 LINK1 in the Element Types window. Click on Close in the Element Types dialog box. 5. Define geometric properties. We now need to specify geometric properties for our elements: In the Preprocessor menu, select: Real Constants > Add/Edit/Delete

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Click Add... and select Type 1 LINK1 (actually it is already selected). Click on OK. The following window will appear:

As shown in the preceding window, enter the cross-sectional area (3250 mm): Click on OK. Set 1 now appears in the dialog box. Click on Close in the Real Constants window.

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6. Element material properties.

You then need to specify material properties: In the Preprocessor menu, select: Material Props > Material Models Double-click on Structural > Linear > Elastic > Isotropic We are going to give the properties of steel. Enter the following fields: EX 200000 Set these properties and click on OK. Note: You may obtain the note PRXY will be set to 0.0. This is Poisson’s ratio and is not required for this element type. Click OK on the window to continue. Close the Define Material Model Behavior window by clicking on the X box in the upper right-hand corner.

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7. Mesh size.

The last step before meshing is to tell ANSYS what size the elements should be. There are a variety of ways to do this, but we will just deal with one method for now. In the Preprocessor menu, select: Meshing > Size Cntrls > ManualSize > Lines > All Lines In the size NDIV field, enter the desired number of divisions per line. For this example, we want only one division per line; therefore, enter 1 and then click OK. Note that we have not yet meshed the geometry, we have simply defined the element sizes.

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8. Mesh. Now the frame can be meshed. In the Preprocessor menu, select: Meshing > Mesh > Lines And, click Pick All in the Mesh Lines Window. Your model should now appear as shown in the window. 9. Plot numbering. To show the line numbers, keypoint numbers, node numbers..., from the Utility Menu (top of screen), select: PlotCtrls > Numbering... Fill in the window as shown next and click OK. Now you can turn numbering on or off at your discretion.

10. Saving your work. Save the model at this time, so if you make some mistakes later on, you will at least be able to come back to this point. To do this, on the Utility Menu, select File > Save as.... Select the name and location where you want to save your file. It is a good idea to save your job at different times

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throughout the building and analysis of the model to back-up your work in case of a system crash or what have you. Solution phase: assigning loads and solving You have now defined your model. It is now time to apply the load(s) and constraint(s) and solve the resulting system of equations. Open up the Solution menu (from the same ANSYS Main Menu). 11. Define analysis type. First, you must tell ANSYS how you want it to solve this problem: From the Solution Menu, select: Analysis Type > New Analysis.

Ensure that Static is selected, that is, you are going to do a static analysis on the truss as opposed to a dynamic analysis, for example. Click OK. 12. Apply constraints. It is necessary to apply constraints to the model; otherwise, the model is not tied down or grounded and a singular solution will result. In mechanical structures, these constraints will typically be fixed, pinned and roller-type connections. As shown earlier, the left end of the truss bridge is pinned, while the right end has a roller connection. In the Solution menu, select: Define Loads > Apply > Structural > Displacement > On Keypoints

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Select the left end of the bridge (Keypoint 1) by clicking on it in the  Graphics window and click on OK in the Apply U, ROT on KPs window.

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This location is fixed, which means that all translational and rotational degrees of freedom (DOFs) are constrained. Therefore, select All DOF by clicking on it and enter 0 in the Value field and click OK. You will see some blue triangles in the Graphics window indicating the displacement constraints. Using the same method, apply the roller connection to the right end (UY constrained). Note that more than one DOF constraint can be selected at a time in the Apply U, ROT on KPs window. Therefore, you may need to deselect the All DOF option to select just the UY option. 13. Apply loads. As shown in the diagram, there are four downward loads of 280kN, 210kN, 280kN, and 360kN at keypoints 1, 3, 5, and 7, respectively. Select: Define Loads > Apply > Structural > Force/Moment > on Keypoints. Select the first keypoint (left end of the truss) and click OK in the Apply F/M on KPs window. Select FY in the Direction of force/mom. This indicate that we will be applying the load in the y direction. Enter a value of −280000 in the Force/moment value box and click OK.

Note that we are using units of N here, this is consistent with the previous values input. The force will appear in the Graphics window as a red arrow.

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Apply the remaining loads in the same manner. The applied loads and constraints should now appear as shown as follows: 14. Solving the system. We now tell ANSYS to find the solution: In the Solution menu, select Solve > Current LS. This indicates that we desire the solution under the current load step (LS).

The preceding windows will appear. Ensure that your solution options are the same as shown earlier and click OK.

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Once the solution is done, the following window will pop up. Click Close and close the /STATUS Command window.

Post-processing: viewing the results (a) Hand calculations: We will first calculate the forces and stress in element 1 (as labeled in the problem description).

∑ M1 = 0 = 210 * 3.6 − 280 * 7.2 − 360 *10.8 + F7 *10.8 ∴ F7 =

−210 * 3.6 + 280 * 7.2 + 360 * 10.8 = 617kN 10.8

∑ Fy = 0 = −280 − 210 − 280 − 360 + 617 + F1 ∴ F1 = 280 + 210 + 280 + 360 − 617 = 513 kN Element (1) forces/stress:

FE1 =

sE 1 =

513 − 280 = 269 kN cos(30)

FE 1 269 = = 82.8 MPa 3250 A

(b) Results using ANSYS:

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15. Reaction forces A list of the resulting reaction forces can be obtained for this element. From the Main menu, select: General Postproc > List Results > Reaction > Solu.

Select All struc forc F as shown and click OK. These values agree with the reaction forces calculated by hand. 16. Deformation: In the General Postproc menu, select: Plot Results > Deformed Shape.

The following window will appear. Select Def + undef edge and click OK to view both the deformed and the undeformed object. Observe the value of the maximum deflection in the upper lefthand corner (DMX = 7.409). One should also observe that the constrained DOFs appear to have a deflection of 0 (as expected!).

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17. Deflection: For a more detailed version of the deflection of the beam, from the General Postproc menu, select:

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Plot results > Contour Plot > Nodal Solution. The following window will appear. Select DOF solution and USUM as shown in the preceding window. Leave the other selections as the default values. Click OK. Looking at the scale, you may want to use more useful intervals. From the Utility Menu, select:

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Plot Controls > Style > Contours > Uniform Contours... Fill in the following window as shown and click OK. You should obtain the following.

18. Listing of result: The deflection can also be obtained as a list as shown next. General Postproc > List Results > Nodal Solution Select DOF Solution and ALL DOFs from the lists in the List Nodal Solution window and click OK. This means that we want to see a listing of all DOFs from the solution.

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Are these results what you expected? Note that all the DOFs were constrained to zero at node 1, while UY was constrained to zero at node 7. If you wanted to save these results to a file, select File within the results window (at the upper left-hand corner of this list window) and select Save as. 19. Axial stress: For line elements (i.e., links, beams, spars, and pipes), you will often need to use the Element Table to gain access to derived data (i.e., stresses, strains). For this, example we should obtain axial stress to compare with the hand calculations. The Element Table is different for each element; therefore, we need to look at the help file for LINK1 (Type help link1 into the Input Line). From Table 1.2 in the Help file, we can see that SAXL can be obtained through the ETABLE, using the item LS,1. From the General Postprocessor menu, select: Element Table > Define Table Click on Add....

As shown, enter SAXL in the Lab box. This specifies the name of the item you are defining. Next, in the Item, Comp boxes, select By sequence number and LS. Then enter 1 after LS, in the selection box. Click on OK and close the Element Table Data window. Plot the stresses by selecting Element Table > Plot Elem Table. The following window will appear. Ensure that SAXL is selected and click OK.

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Because you changed the contour intervals for the Displacement plot to User Specified, you need to switch this back to Auto calculated to obtain new values for VMIN/VMAX.

Utility Menu > PlotCtrls > Style > Contours > Uniform Contours ... Again, you may wish to select more appropriate intervals for the contour plot. 20. List the stresses. From the Element Table menu, select List Elem Table. From the List Element Table Data window that appears, ensure SAXL is highlighted. Click OK.

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Note that the axial stress in Element 1 is 82.9MPa as predicted analytically. 21. Quitting ANSYS: To quit ANSYS, select QUIT from the ANSYS Toolbar or select: Utility Menu/File/Exit.... In the dialog box that appears, click on Save Everything (assuming that you want to) and then click on OK.

4.2 Tutorial 4 (a): Static Analysis Using BEAM Elements In this tutorial, the effect of self-weight on a cantilever beam will be addressed. The loads will not be applied to the beam shown next in order to observe the deflection caused by the weight of the beam itself. The beam is to be made of steel with a modulus of elasticity of 200GPa.

Step-by-Step ANSYS Solution Preprocessing: Defining the Problem 1. Give example a title: Utility Menu > File > Change Title... /title, effects of self-weight for a cantilever beam 2. Open Preprocessor menu: ANSYS Main Menu > Preprocessor /PREP7 3. Define keypoints: Preprocessor > Modeling > Create > Keypoints> In Active CS... K,#,x,y,z We are going to define two keypoints for this beam, as given in the following table: Keypoint 1 2

Coordinates (x,y,z) (0,0) (1000,0)

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4. Create lines. Preprocessor > Modeling > Create > Lines > Lines > In Active Coord L,1,2 Create a line joining keypoints 1 and 2 5. Define the type of element Preprocessor > Element Type > Add/Edit/Delete... For this problem, we will use the BEAM3 (Beam 2D elastic) element. This element has three degrees of freedom (translation along the X and Y axes, and rotation about the Z axis). 6. Define real constants Preprocessor > Real Constants... > Add... In the Real Constants for BEAM3 window, enter the following geometric properties: Cross-sectional area AREA: 500 Area moment of inertia IZZ: 4166.67 Total beam height: 10 This defines a beam with a height of 10 mm and a width of 50 mm. 7. Define element material properties. Preprocessor > Material Props > Material Models > Structural > Linear > Elastic > Isotropic In the window that appears, enter the following geometric properties for steel: Young’s modulus EX: 200000 Poisson’s Ratio PRXY: 0.3 8. Define element density: Preprocessor > Material Props > Material Models > Structural > Linear > Density In the window that appears, enter the following density for steel: Density DENS: 7.86e-6 9. Define mesh size: Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > All Lines... For this example, we will use an element edge length of 100 mm. 10. Mesh the frame: Preprocessor > Meshing > Mesh > Lines > Pick All Solution phase: assigning loads and solving 11. Define analysis type: Solution > Analysis Type > New Analysis > Static ANTYPE,0 12. Apply constraints: Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Fix keypoint 1 (i.e., all DOF constrained) 13. Define gravity:

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It is necessary to define the direction and magnitude of gravity for this problem. Select: Solution > Define Loads > Apply > Structural > Inertia > Gravity... The following window will appear. Fill it in as shown to define an acceleration of 9.81m/s2 in the y direction. Note: Acceleration is defined in terms of meters (not mm as used throughout the problem). This is because the units of acceleration and mass must be consistent to give the product of force units (Newton in this case). Also note that a positive acceleration in the y direction stimulates gravity in the negative Y direction. There should now be a red arrow pointing in the positive y direction. This indicates that acceleration has been defined in the y direction: DK,1, ALL,0, ACEL,9.8. The applied loads and constraints should now appear as shown in the figure as follows:

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14. Solve the system: Solution > Solve > Current LS Solve: Postprocessing: Viewing the Results Hand Calculations Hand calculations were performed to verify the solution found using ANSYS. The maximum deflection was shown to be 5.777 mm. 15. Show the deformation of the beam: General Postproc >Plot Results > Deformed Shape ... > Def + undef edge PLDISP,2

As observed in the upper left-hand corner, the maximum displacement was found to be 5.777 mm. This is in agreement with the theoretical value.

4.3 Tutorial 4 (b): Static Analysis Using Beam Elements with Distributed Load In this tutorial, the effect of distributed loads on beam elements is addressed. A distributed load of 1000 N/m (1 N/mm) will be applied to a solid steel beam with a rectangular cross-section as shown in the ­following

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figure. The cross-section of the beam is 10 mm × 10 mm, while the modulus of elasticity of the steel is 200GPa.

Step-by-Step ANSYS Solution Preprocessing: Defining the Problem 1. Open preprocessor menu /PREP7. 2. Give example a title: Utility Menu > File > Change Title ... /title, Distributed Loading 3. Create keypoints: Preprocessor > Modeling > Create > Keypoints > In Active CS K,#,x,y We are going to define two keypoints (the beam vertices) for this structure as given in the following table: Keypoint 1 2

Coordinates (x,y,z) (0,0) (1000,0)

4. Define lines: Preprocessor > Modeling > Create > Lines > Lines > Straight Line L,K#,K# Create a line between keypoint 1 and keypoint 2. 5. Define element types: Preprocessor > Element Type > Add/Edit/Delete... For this problem, we will use the BEAM3 element. This element has three degrees of freedom (translation along the X and Y axis’s, and rotation about the Z axis). With only three degrees of freedom, the BEAM3 element can only be used in 2D analysis.

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6. Define real constants: Preprocessor > Real Constants... > Add... In the Real Constants for BEAM3 window, enter the following geometric properties: Cross-sectional area AREA: 100 Area Moment of Inertia IZZ: 833.333 Total beam height HEIGHT: 10 This defines an element with a solid rectangular cross section 10 mm × 10 mm. 7. Define element material properties: Preprocessor > Material Props > Material Models > Structural > Linear > Elastic > Isotropic In the window that appears, enter the following geometric properties for steel: Young’s modulus EX: 200000 Poisson’s Ratio PRXY: 0.3 8. Define mesh size: Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > All Lines... For this example, we will use an element length of 100 mm. 9. Mesh the frame: Preprocessor > Meshing > Mesh > Lines > click Pick All 10. Plot elements: Utility Menu > Plot > Elements You may also wish to turn on element numbering and turn off keypoint numbering Utility Menu > PlotCtrls > Numbering ...

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Solution phase: assigning loads and solving 11. Define analysis type: Solution > Analysis Type > New Analysis > Static ANTYPE, 0 12. Apply constraints: Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Pin keypoint 1 (i.e., UX and UY constrained) and fix keypoint 2 in the y direction (UY constrained). 13. Apply loads: We will apply a distributed load, of 1000N/m or 1N/mm, over the entire length of the beam. Select: Solution > Define Loads > Apply > Structural > Pressure > On Beams Click Pick All in the Apply F/M window. As shown in the following figure, enter a value of 1 in the field VALI Pressure value at node I, then click OK.

The applied loads and constraints should now appear as shown in the following figure.

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Note: To have the constraints and loads appear each time you select Replot, you must change some settings. Select Utility Menu > ­PlotCtrls > Symbols.... In the window that appears, select Pressures in the ­pull-down menu of the Surface Load Symbols section. 14. Solve the system: Solution > Solve > Current LS Solve: Postprocessing: Viewing the Results 15. Plot deformed shape: General Postproc > Plot Results > Deformed Shape PLDISP.2

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Plot principle stress distribution: As shown previously, we need to use element tables to obtain principle stresses for line elements. Select: General Postproc > Element Table > Define Table Click Add.... In the window that appears, enter SMAXI in the User Label for Item section. In the first window in the Results Data Item section, scroll down and select By sequence num. In the second window of the same section, select NMISC. In the third window, enter 1 anywhere after the comma click Apply. Repeat steps 2 to 4, but change SMAXI to SMAXJ in step 3a and change 1 to 3 in step 3b. Click OK. The Element Table Data window should now have two variables in it. Click Close in the Element Table Data window. Select: General Postproc > Plot Results > Line Elem Res... Select SMAXI from the LabI pull-down menu and SMAXJ from the LabJ pull-down menu. Note: ANSYS can only calculate the stress at a single location on the element. For this example, we decided to extract the stresses from the I and J nodes of each element. These are the nodes that are at the ends of each element. For this problem, we wanted the principal stresses for the elements. For the BEAM3 element, this is categorized as NMISC, 1 for the I nodes and NMISC, 3 for the J nodes. A list of available codes for each element can be found in the ANSYS help files (i.e., type help BEAM3 in the ANSYS Input window). As shown in the following plot, the maximum stress occurs in the middle of the beam with a value of 750MPa.

Chapter 5

Static Analysis Using Area Elements 5.1 Tutorial 5: Static Analysis Using Area Elements: Plane Problem (Bracket) In this tutorial, we will discuss the plane stress. The problem to be modeled in this example is a simple bracket shown in the following figure. This bracket is to be built from a 20 mm-thick steel plate. A figure of the plate is shown as follows:

This plate will be fixed at the two small holes on the left and has a load applied to the larger hole on the right. Verification Example The first step is to simplify the problem. Whenever you are trying out a new analysis type, you need something (i.e., analytical solution or experimental data) to compare the results to. This way, you can be sure that you have gotten the correct analysis type, units, scale factors, and so

122  •   Using ANSYS for Finite Element Analysis

on. The simplified version that will be used for this problem is that of a flat rectangular plate with a hole shown in the following figure.

Step-by-Step ANSYS Solution Preprocessing: Defining the Problem 1. Give the simplified version a title: Utility Menu > File > Change Title 2. Form geometry: Boolean operations provide a means to create complicated solid models. These procedures make it easy to combine simple geometric entities to create more complex bodies. Subtraction will used to create this model; however, many other Boolean operations can be used in ANSYS. Create the main rectangular shape. Instead of creating the geometry using keypoints, we will create an area (using GUI): Preprocessor > Modeling > Create > Areas > Rectangle > By 2 Corners

Static Analysis Using Area Elements   •  123

Fill in the window as shown in the preceding figure. This will create a rectangle where the bottom-left corner has the coordinates 0,0,0 and the top-right corner has the coordinates 200,100,0. (Alternatively, the command line code for the preceding command is BLC4,0,0,200,100.) Create the circle: Preprocessor > Modeling > Create > Areas > Circle > Solid Circle Fill in the window as shown. This will create a circle where the center has the coordinates 100,50,0 (the center of the rectangle) and the radius of the circle is 20 mm. (Alternatively, the command line code for the preceding command is CYL4,100,50,20.) Subtraction: Now we want to subtract the circle from the rectangle. Prior to this operation, your image should resemble the following:

To perform the Boolean operation, from the Preprocessor menu, select: Modeling > Operate > Booleans > Subtract > Areas At this point, a Subtract Areas window will pop up and the ANSYS Input window will display the following message: [ASBA] Pick or enter base areas from which to subtract (as shown in the following figure).

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Therefore, select the base area (the rectangle) by clicking on it. Note: the selected area will turn pink once it is selected. The following window may appear because there are two areas at the location you clicked.

Ensure that the entire rectangular area is selected (otherwise click Next) and then click OK. Click OK on the Subtract Areas window. Now you will be prompted to select the areas to be subtracted; select the circle by clicking on it and then click OK. You should now have the following model: (Alternatively, the command line code for the above step is ASBA,1,2)

Static Analysis Using Area Elements   •  125

3. Define the type of element: Preprocessor Menu > Element Type > Add/Edit/Delete Add the following type of element: Solid (under the Structural heading) and the Quad 82 element, as shown in the preceding figure.

PLANE82 is a higher order version of the two-dimensional, fournode element (PLANE42). PLANE82 is an eight-node quadrilateral element, which is better-suited to model curved boundaries. For this example, we need a plane stress element with thickness; therefore, click on the Options... button. Click and hold the K3 button, and select Plane strs w/thk, as shown next.

(Alternatively, the command line code for the preceding step is ET,1,PLANE82 followed by KEYOPT,1,3,3.) 4. Define the geometric properties: As in previous examples, Preprocessor menu > Real Constants > Add/Edit/Delete Enter a thickness of 20 as shown in the following figure. This defines a plate thickness of 20 mm. (Alternatively, the command line code for the preceding step is R,1,20.)

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5. Element material properties: As shown in the previous examples, select: Preprocessor > Material Props > Material models > Structural > Linear > Elastic > Isotropic We are going to give the properties of steel. Enter the following when prompted: EX 200000 PRXY 0.3 (Alternatively, the command line code for the above step is MP,EX,1,200000 followed by MP,PRXY,1,0.3.) 6. Mesh size: To tell ANSYS how big the elements should be, Preprocessor > Meshing > Size Cntrls > Manual Size > Areas > All Areas Select an element edge length of 25. We will return later to determine whether this was adequate for the problem. (Alternatively, the command line code for the above step is AESIZE,ALL,25.) 7. Mesh: Now the frame can be meshed. In the Preprocessor menu, select: Meshing > Mesh > Areas > Free And, select the area when prompted. (Alternatively, the command line code for the above step is AMESH,ALL.) You should now have the following:

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8. Saving your job: Utility Menu > File > Save as... Solution phase: assigning loads and solving: You have now defined your model. It is now time to apply the load(s) and constraint(s) and solve the resulting system of equations. 9. Define analysis type: Ensure that a static analysis will be performed: Solution > Analysis Type > New Analysis (Alternatively, the command line code for the preceding step is ANTYPE,0.) 10. Apply constraints: As shown previously, the left end of the plate is fixed. Solution > Define Loads > Apply > Structural > Displacement > On Lines Select the left end of the plate and click on Apply in the Apply U,ROT on Lines window. Fill in the window as shown next. This location is fixed, which means that all DOFs are constrained. Therefore, select All DOF by clicking on it and enter 0 in the Value field as shown in the preceding figure. You will see some blue triangles in the graphics window indicating the displacement constraints.

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(Alternatively, the command line code for the preceding step is DL,4,,ALL,0.) 11. Apply loads: As shown in the diagram, there is a load of 20 N/mm distributed on the right-hand side of the plate. To apply this load: Solution > Define Loads > Apply > Structural > Pressure > On Lines When the window appears, select the line along the right-hand edge of the plate and click OK. Calculate the pressure on the plate end by dividing the distributed load by the thickness of the plate (1MPa). Fill in the Apply PRES on lines window as shown next. Note: The pressure is uniform along the surface of the plate; therefore, the last field is left blank. The pressure is acting away from the surface of the plate, and is therefore defined as a negative pressure. The applied loads and constraints should now appear as shown in the following figure.

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12. Solving the system: Solution > Solve > Current LS Postprocessing: Viewing the Results Hand Calculations Now, as the purpose of this exercise was to verify the results, we need to calculate what we should find. Deflection: The maximum deflection occurs on the right hand side of the plate and was calculated to be 0.001 mm—neglecting the effects of the hole in the plate (i.e., just a flat plate). The actual deflection of the plate is therefore expected to be greater but in the same range of magnitude. Stress: The maximum stress occurs at the top and bottom of the hole in the plate and was found to be 3.9MPa. Convergence Using ANSYS At this point, we need to find whether or not the final result has converged. We will do this by looking at the deflection and stress at particular nodes while changing the size of the meshing element. As we have an analytical solution for the maximum stress point, we will check the stress at this point. First, we need to find the node ­corresponding to the top of the hole in the plate. First, plot and number the nodes: Utility Menu > Plot > Nodes Utility Menu > PlotCtrls > Numbering...

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The plot should look similar to the one shown as follows. Make a note of the node closest to the top of the circle (i.e., #49).

13. List the stresses: (General Postproc > List Results > Nodal Solution > Stress, ­Principals SPRIN)

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And, check the SEQV (equivalent stress / von Mises stress) for the node in question. (as shown in red as follows). The equivalent stress was found to be 2.9141MPa at this point. We will use smaller elements to try to get a more accurate solution. 14. Resize elements: To change the element size, we need to go back to the Preprocessor menu Preprocessor > Meshing > Size Cntrls > Manual Size > Areas > All Areas Now decrease the element edge length (i.e., 20). Now remesh the model: Preprocessor > Meshing > Mesh > Areas > Free Once you have selected the area and clicked OK, the following ­window will appear:

Click OK. This will remesh the model using the new element edge length. Solve the system again (note that the constraints need not be reapplied). Solution Menu > Current LS Repeat steps a through d until the model has converged. (Note: the number of the node at the top of the hole has most likely changed. It is essential that you plot the nodes again to select the appropriate node.) Plot the stress/deflection at varying mesh sizes shown as follows to confirm that convergence has occurred. Note the shapes of both the deflection and stress curves. As the number of elements in the mesh increases (i.e., the element edge length decreases), the values converge toward a final solution. The von Mises stress at the top of the hole in the plate was found to be approximately 3.8MPa. This is a mere 2.5 percent difference between the analytical solution and the solution found using ANSYS. The approximate maximum displacement was found to be 0.0012 mm; this is 20 percent greater than the analytical solution. However, the analytical solution does not account for the large hole in the center of the

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0.000650

4.00

0.000640

3.75

0.000630

3.50

0.000620

3.25 SEQV

0.000610

Stress (Mpa)

Delection (mm)

plate, which was expected to significantly increase the deflection at the end of the plate.

3.00

USUM 2.75

0.000600 0

500

1000

1500

2000

Number of elements

Therefore, the results using ANSYS were determined to be appropriate for the verification model. 15. Deformation: General Postproc > Plot Results > Deformed Shape > Def + undeformed To view both the deformed and the undeformed object. Observe the locations of deflection.

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16. Deflection: General Postproc > Plot Results > Nodal Solution... Then select DOF solution, USUM in the window.

Alternatively, obtain these results as a list. General Postproc > List Results > Nodal Solution... Are these results what you expected? Note that all translational degrees of freedom were constrained to zero at the left end of the plate. 17. Stresses: General Postproc > Plot Results > Nodal Solution... Then select Stress, von Mises in the window.

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You can list the von Mises stresses to verify the results at certain nodes: General Postproc > List Results Select: Stress, Principals SPRIN Bracket Example Now we will return to the analysis of the bracket. A combination of GUI and the command line will be used for this example. The problem to be modeled in this example is a simple bracket shown in the following figure. This bracket is to be built from a 20 mm-thick steel plate. A figure of the plate is shown as follows. This plate will be fixed at the two small holes on the left and have a load applied to the larger hole on the right.

Preprocessing: Defining the Problem 1. Give the bracket example a title: Utility Menu > File > Change Title 2. Form the geometry: Again, Boolean operations will be used to create the basic geometry of the bracket. Create the main rectangular shape. The main rectangular shape has a width of 80 mm, a height of 100 mm, and the bottom-left corner is located at coordinates (0,0). Ensure that the Preprocessor menu is open. (Alternatively, type /PREP7 into the command line window.) Now instead of using the GUI window, we are going to enter code into the command line. Now, I will explain the line required to create a rectangle: BLC4, XCORNER, YCORNER, WIDTH, HEIGHT BLC4, X coord (bottom left), Y coord (bottom left), width, height. Therefore, the command line for this rectangle is BLC4,0,0,80,100. Create the circular end on the right-hand side. The center of the circle is located at (80,50) and has a radius of 50 mm. The following code is used to create a circular area:

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CYL4, XCENTER, YCENTER, RAD1 CYL4, X coord for the center, Y coord for the center, radius; therefore, the command line for this circle is CYL4,80,50,50. Now create a second and third circle for the left-hand side using the following dimensions: Parameter XCENTER YCENTER RADIUS

Circle 2 0 20 20

Circle 3 0 80 20

Create a rectangle on the left hand end to fill the gap between the two small circles. XCORNER YCORNER WIDTH HEIGHT

−20 20 20 60

Your screen should now look like the following:

Boolean Operations: Addition We now want to add these five discrete areas together to form one area. To perform the Boolean operation, from the Preprocessor menu, select:

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Modeling > Operate > Booleans > Add > Areas In the Add Areas window, click on Pick All. (Alternatively, the command line code for the above step is AADD,ALL.) Parameter WP X WP Y Radius

Circle 1 80 50 30

Circle 2 0 20 10

Circle 3 0 80 10

Create the Bolt Holes We now want to remove the bolt holes from this plate. Create the three circles with the parameters given as follows: Now select Preprocessor > Modeling > Operate > Booleans > Subtract > Areas Select the base areas from which to subtract (the large plate that was created). Next, select the three circles that we just created. Click on the three circles that you just created and click OK. (Alternatively, the command line code for the preceding step is ASBA,6,ALL.) Now you should have the following:

3. Define the type of element: As in the verification model, PLANE82 will be used for this example: Preprocessor > Element Type > Add/Edit/Delete Use the Options... button to get a plane stress element with thickness.

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(Alternatively, the command line code for the preceding step is ET,1,PLANE82 followed by KEYOPT,1,3,3.) Under the Extra Element Output K5, select nodal stress. 4. Define the geometric contents: Preprocessor > Real Constants > Add/Edit/Delete Enter a thickness of 20 mm. (Alternatively, the command line code for the preceding step is R,1,20.) 5. Element material properties: Preprocessor > Material Props > Material Library > Structural > Linear > Elastic > Isotropic We are going to give the properties of steel. Enter the following when prompted: EX 200000 PRXY 0.3 (The command line code for the preceding step is MP,EX,1,200000 followed by MP,PRXY,1,0.3.) Mesh Size Preprocessor > Meshing > Size Cntrls > Manual Size > Areas > All Areas Select an element edge length of 5. Again, we will need to make sure the model has converged. (Alternatively, the command line code for the preceding step is AESIZE,ALL,5.) 6. Mesh: Preprocessor > Meshing > Mesh > Areas > Free And, select the area when prompted. (Alternatively, the command line code for the preceding step is AMESH,ALL.)

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7. Saving your job: Utility Menu > File > Save as... Solution Phase: Assigning Loads and Solving You have now defined your model. It is now time to apply the load(s) and constraint(s) and solve the resulting system of equations. 8. Define analysis type: Solution > New Analysis And select Static. (Alternatively, the command line code for the above step is ANTYPE,0.) 9. Apply constraints. As illustrated, the plate is fixed at both of the smaller holes on the left-hand side. Solution > Define Loads > Apply > Structural > Displacement > On Nodes Instead of selecting one node at a time, you have the option of creating a box, polygon, or circle, of which all the nodes in that area will be selected. For this case, select circle as shown in the following window. You may want to zoom in to select the points: Utility Menu / PlotCtrls / Pan, Zoom, Rotate...

Click on the center of the bolt hole and drag the circle out so that it touches all of the nodes on the border of the hole.

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Click on Apply in the Apply U,ROT on Lines window and constrain all DOFs in the Apply U,ROT on Nodes window. Repeat for the second bolt hole. 10. Apply loads: As shown in the diagram, there is a single vertical load of 1000N, at the bottom of the large bolt hole. Apply this force to the respective keypoint: Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints Select a force in the y direction of -1000 The applied loads and constraints should now appear as shown next:

11. Solving the system: Solution > Solve > Current LS Post-Processing: Viewing the Results We are now ready to view the results. We will take a look at the deflected shape and the stress contours once we determine whether convergence has occurred. Convergence using ANSYS At this point, we need to find whether or not the final result has converged. Because we cannot solve for a solution analytically, we must try. 12. Deformation: General Postproc > Plot Results > Def + undeformed

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To view both the deformed and the undeformed object, the graphic should be similar to the following:

Observe the locations of deflection. Ensure that the deflection at the bolt hole is indeed 0. 13. Deflection: To plot the nodal deflections use:

Static Analysis Using Area Elements   •  141

General Postproc > Plot Results > Contour Plot > Nodal Solution Then, select DOF Solution—USUM in the window. Alternatively, obtain these results as a list: General Postproc > List Results > Nodal Solution... Are these results what you expected? Note that all translational degrees of freedom were constrained to zero at the bolt holes. 14. Stresses: General Postproc > Plot Results > Nodal Solution... Then, select von Mises Stress in the window.

You can list the von Mises stresses to verify the results at certain nodes: General Postproc > List Results. Select Stress, Principals SPRIN 15. Quitting ANSYS To quit ANSYS, click QUIT on the ANSYS toolbar or select: Utility Menu > File > Exit... In the window that appears, select Save Everything (assuming that you want to) and then click OK.

5.2 Tutorial 6: Static Analysis using Area Elements: Plane Problem (Wrench) In this tutorial, the stress analysis of a bicycle wrench is presented. The problem to be found is the von Mises stresses for the bicycle wrench shown under the given distributed and boundary condition.

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Material Properties: Modulus of elasticity: E = 200GPa, Poisson’s ratio: ν = 0.32 Geometry: Lengths and radii as shown, Thickness: 3 mm Loading: Distributed load: 88 N/cm, Constraints: ux, uy, uz = 0 at left hexagon

Solution Methodology As the thickness (3 mm) is quite small compared to the overall length of about 12 cm, the problem can be reasonably treated as a plane stress problem. In ANSYS, it will be treated as a plane stress with thickness problem. The geometry will be created through rectangle, circle and hexagon commands, together with Boolean operations of add (union) and subtract (difference). Step-by-Step ANSYS Solution 1. Start ANSYS: File -> Save As -> Filename -> OK 2. Use the structural solid element PLANE82 for FEM modeling: Preprocessor -> Element Type -> Add/Edit/Delete -> Add -> Structural Mass-Solid -> Select 8node 82 -> OK -> Options -> Element Behavior: Select Plane Stress w/thk -> OK -> Close 3. Enter real constants for the element type chosen: Preprocessor -> Real Constants -> Add/Edit/Delete -> Add -> OK -> Enter Thickness THK = 0.3 -> OK -> Close 4. Enter material property data for specified steel: Preprocessor -> Material Props -> Material Model -> Structural -> Linear -> Elastic -> Isotropic -> Enter Young’s modulus EX = 200e9 and Poisson’s ratio PRXY = 0.32 -> OK -> Close Note: After each significant step, click SAVE_DB on ANSYS toolbar.

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5. Create the geometry for two similar rectangles 1.5 cm by 3 cm at locations (2.25, 0.5) and (7.25, 0.5): Preprocessor -> Modeling -> Create -> Areas-Rectangle -> By 2 Corners -> In dialogue box enter WP X = 2.25, WP Y = 0.5, Width = 3, Height = 1.5 -> Apply -> Enter values for next rectangle: WP X = 7.25, WP Y = 0.5, Width = 3, Height = 1.5 -> OK

Note: Click on Plot Controls -> Pan, Zoom, Rotate -> Click on small circle until both rectangles fit the screen. 6. Create the geometry for three circles, all of 1.25 cm radius: Preprocessor -> Modeling -> Create -> Areas -> Circle -> Solid Circle -> In the dialog box, enter WP X = 1.25, WP Y = 1.25, Radius = 1.25 -> Apply -> Enter values for the next circle -> WP X = 6.25, WP Y = 1.25, Radius = 1.25 -> Apply -> Enter values for the next circle -> WP X = 11.25, WP Y = 1.25, Radius = 1.25 -> OK

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7. Perform the Boolean operation add to union the areas together: Preprocessor -> Modeling -> Operate -> Booleans -> Add -> Areas -> Pick All 8. Create the geometry for three hexagons, two of 7 mm side (at the ends) and one of 9 mm side (in the center): Preprocessor -> Modeling -> Create -> Areas -> Polygon -> Hexagon -> In the dialog box, enter WP X = 1.25, WP Y = 1.25, Radius = 0.7 -> Theta = 120 -> Apply -> Enter values for the next hexagon -> WP X = 6.25, WP Y = 1.25, Radius = 0.9 -> Theta = 120 -> Apply -> Enter values for the next hexagon -> WP X = 11.25, WP Y = 1.25, Radius = 0.7 -> Theta = 120 -> OK 9. Perform the Boolean operation subtract to get the hexagonal holes in the wrench body: Preprocessor -> Modeling -> Operate ->Booleans -> Subtract -> Areas -> Click on the solid portion of the wrench -> Apply -> One by one, pick the three hexagonal areas -> OK

10. Now, create a mesh in the final wrench shape, first refining the mesh size: Preprocessor -> Meshing -> Size Controls -> ManualSize -> Size -> Enter Size = 0.1 -> OK Preprocessor -> Meshing -> Mesh -> Areas -> Free -> Click on the wrench -> OK

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11. Apply the boundary conditions and the load: Preprocessor -> Loads -> Analysis Type -> New Analysis -> Static -> OK Preprocessor -> Loads -> Define Loads -> Apply -> Structural -> Displacement -> On Key Points -> Click on the six corner points of the left hexagon -> OK -> Select All DOF -> OK Preprocessor -> Loads -> Define Loads -> Apply -> Structural -> Pressure -> On Lines -> Pick the line indicated in problem statement (top line of the right arm) -> OK -> Enter VALUE = 88 -> OK 12. Perform the solution: Solution -> Solve -> Current LS -> OK 13. Start post-processing: Check the deformed shape: General Post Proc -> Plot Results -> Deformed Shape -> Def + undef edge -> OK

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14. Examine the stresses: find von Mises stresses: General Postproc -> Plot Results -> Contour Plot -> Nodal Solution -> Stress -> von Mises SEQV -> Def + undef edge -> OK

Chapter 6

Static Analysis Using Volume Elements 6.1 Tutorial 7: Static Analysis using Volume Elements: Component Design In this tutorial, the analysis of an open-ended cylinder with internal pressure is presented. The unknown to be found is the stresses in a thick open-ended steel cylinder (shown) with an internal pressure (such as a pipe discharging to the atmosphere). Material properties: Young’s modulus: E = 30 × 106 psi, Poisson’s ratio: ν = 0.3 Geometry: Internal radius: ri = 5 in, external radius: ro = 11 in, length: arbitrary, for segment of a long, open-ended cylinder Loading: Internal pressure: p = 1,000 psi. Constraints: uz = 0 on a radial surface parallel to the x-y plane, ux = 0 on a radial surface parallel to y-z plane, uy = 0 on top/bottom surface (prevent rigid body motion in y-direction)

Y X

5 in

11 in

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Solution Methodology The y-axis is the axis of symmetry. The cylinder can be generated by revolving a rectangle 6in. wide and of arbitrary height 360° about the y-axis. For the 3D analysis, we will make use of symmetry and analyze one quadrant of the cylinder. If y-axis is the axis of revolution of the cylinder, both x-y and y-z planes are planes of symmetry. As the height of the segment considered is arbitrary, we will use a segment 1in. in height for the finite element model. The geometry is shown as follows:

5 in

1 in 11 in

Step-by-Step ANSYS Solution 1. Start ANSYS: File → Save As → Filename → OK 2. Use a tetrahedron element for the FEM modeling: Preprocessor → Element Type → Add/Edit/Delete → Add → Structural Mass → Solid → Select Tet 10node 92 → OK → Close 3. Enter material property data for steel: Preprocessor → Material Props → Material Modes → Structural → Linear → Elastic → Elastic → Enter EX = 3e7 and PRXY = 0.3 → OK → Close window 4. Create the geometry for rectangle 1 × 6in. starting 5in. from the y-axis. This rectangular area will be revolved 90° about the y-axis to produce the desired volume: Preprocessor → Modeling → Create → Areas → Rectangle →By 2 Corners → In the dialog box, enter WP X = 5, WP Y = 0, Width = 6, Height = 1 → OK

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5. Define two keypoints on the y-axis for revolving purposes: In the command dialog box, enter: K, 10, 0, 0, 0 (Enter) Defines keypoint number 10 at (0, 0, 0) K, 11, 0, 5, 0 (Enter) Defines keypoint number 11 at (0, 5, 0) Keypoints 1,2,3,4 are already used by ANSYS to define the rectangle, so we used numbers 10 and 11. 6. Revolve the area: Preprocessor → Modeling → Operate → Extrude → Areas → About Axis → Pick the area → OK → Pick the two keypoints on the y-axis → OK → Arc length in degrees = 90 → OK Go to the oblique view: Plot Controls → Pan, zoom, rotate → Click on Obliq

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7. Now create a mesh of tetrahedral elements using the default settings. Preprocessor → Meshing → Mesh → Volumes → Free: Pick the solid just created → OK

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8. Define loads and boundary conditions. Apply an internal pressure of 1,000 psi and displacement constraints that prevent points m ­ oving across planes of symmetry and also restrain rigid body movement in the y-direction. Preprocessor → Loads → Define Loads → Apply → Structural → Displacement → Symmetry B.C. → On Areas → Pick the radial surface parallel to x-y plane → Apply → Pick the radial surface parallel to y-z plane → OK Preprocessor → Loads → Define Loads → Apply → Structural → Displacement → On Areas → Pick top (or bottom) surface → OK → Select degree of freedom constraint UY → OK

Use Cntl + Right Mouse button to rotate the solid so as to get a better view of the area you want to pick. It may also help to activate the area numbers and plot them. Then, pick the area number of interest, for example: PlotCntls → Numbering . . . turn area numbers ON Plot → Areas Preprocessor → Loads → Define Loads → Apply → Structural → Pressure → On Areas: Pick the inner area of the cylinder. Enter a ­pressure  of 1000 → OK

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9. Perform the solution: Solution → Solve → Current LS → OK → Close → File → Close 10. Start post-processing. Check the deformed shape to see whether it is reasonable (the dotted line is the undeformed shape): General Post Processor → Plot Results → Deformed Shape → Def + undef edge → OK

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11. Examine the stresses: General Postprocessor → Plot Results → Contour Plots → Element Solu → Stress → X- direction SX → OK

Solution Comparison The theoretical solution for this problem is: Inside surface of cylinder: SX = -1,000 psi, SZ = 1,520.8 psi Outside surface of cylinder: SX = 0 psi, SZ = 520.8 psi. In the ANSYS solution shown, the upper arrow shows Sx normal to the y-z plane, which corresponds to the cylinder hoop stress. The scale at the bottom indicates a maximum value of 1,528 psi, which is close to the theoretical value of 1,521 psi. The lower arrow shows Sx in a direction normal to the inner surface of the cylinder with a scale value of -1,006, again close to the internal pressure of 1,000 psi. Thus the ANSYS-calculated results seem to agree pretty well with the theoretical results.

6.2 Tutorial 8: Static Analysis Using Volume Elements: Assembly Design In this tutorial, the open 3D solid analysis of an extrusion-die assembly is discussed. The unknown to be found is the von Misses stress distribution

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and the deformed shape of given extrusion-die assembly, under give pressure loading and boundary condition. Material properties: E = 5 × 106 MPa, ν = 0.15 Geometry:

Loading: Pressure = 90 Mpa on the front face of the die Boundary Cond.: UZ = 0 on back surface of sub-bolster. Solution Methodology Using the geometry given, reconstruct it in the ANSYS environment in 3D. Applying the appropriate loading and boundary conditions, solve to attain the desired stress distribution and deformed shapes. Step-by-Step ANSYS Solution 1. Start ANSYS. 2. Defining of element type to be used: Preprocessor → element type → add/del…→add … → structural solid – Tet 10 node 92 → ok → close. 3. Enter the material properties. Preprocessor → material properties → material model → structural → linear → elastic → isotropic Press OK and then close window after you have entered the following values: EX = 5e6, PRXY = 0.15 4. Creating the geometry. Sub-bolster

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Preprocessor → modeling → create → volumes → cylinder → hollow cylinder Enter the following values: WPx = 0, WPy = 0, Rad-1 = 140, rad-2 = 190, depth = 90 To get a better 3D view, go to the ANSYS top menu and click plotctrls → pan-zoom-rotate → oblique → Close To move the working plane to a new origin, type the following in the ANSYS command window: WPAVE,0,0,90

5. Creating the geometry. bolster Preprocessor → modeling → create → volumes → cylinder → hollow cylinder Enter the following values: WPx = 0, WPy = 0, Rad-1 = 0, Rad-2 = 190, Depth = 100 Now, we create the bolster extrusion passage by creating a block in the bolster and then subtracting it form the bolster volume. Preprocessor → modeling → create → volumes → block → by center,cornr… Enter the following values: WPx = 0, WPy = 0, Width= 80, height = 80, depth = 100 Preprocessor → modeling → operate → Boolean → subtract → volumes Now, pick first the volume of the bolster via mouse. Click apply. Now pick the volume of the block. Click OK. To redraw the volumes, go to ANSYS top menu and click: Plot → volumes. To move the working plane to a new origin, type the following in the ANSYS command window: WPAVE,0,0,190

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6. Creating the geometry. Die ring Preprocessor → modeling → create → volumes → cylinder → hollow cylinder Enter the following values: WPx = 0, WPy = 0, Rad-1 = 150, rad-2 = 190, depth = 150

7. Creating the geometry. Die backer Preprocessor → modeling → create → volumes → cylinder → hollow cylinder

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Enter the following values: WPx = 0, WPy = 0, Rad-1 = 0, rad-2 = 150, depth = 80 Now, we create the die-backer extrusion passage by creating a block in the die-backer and then subtracting it from its volume. Preprocessor → modeling → create → volumes → block → by center,cornr… Enter the following values: WP = 0, WPy = 0, Width= 74, height = 74, depth = 80 Preprocessor → modeling → operate → Boolean → subtract → volumes Now, pick first the volume of the die-backer via mouse. Click apply. Now pick the volume of the block. Click OK. To move the working plane to a new origin, type the following in the ANSYS command window: WPAVE,0,0,270

8. Creating the geometry. Die Preprocessor → modeling → create → volumes → cylinder → hollow cylinder Enter the following values: WPx = 0, WPy = 0, Rad-1 = 0, rad-2 = 150, depth = 70 Now we create the extrusion-die passage by creating a block in the die and then subtracting it from its volume. Preprocessor → modeling → create → volumes → block → by center,cornr… Enter the following values: WPx = 0, WPy = 0, Width= 70, height = 70, depth = 70

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Preprocessor → modeling → operate → Boolean → subtract → volumes Now pick first the volume of the die via mouse. Click apply. Now, pick the volume of the block. Click OK. In the ANSYS top menu, click plot → volumes to redraw all the volumes.

Before we mesh the volumes, we must glue all of them to restrict relative motion between the surfaces in contact. Preprocessor → modeling → operate → Booleans → glue → volumes → pick all→ Meshing Before we mesh, we will set up a start sizing tool for meshing. Preprocessor → meshing → size cntrl → smart size→ basic → (set at) 6 Now we mesh all volumes: Preprocessor → meshing→ mesh → volumes → free→ (pick all)

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Having completed the preprocessor phase, we now move to the solution phase. Go to the ANSYS main menu and click on solution. First we put the boundary conditions to the assembly. We must restrict the movement of the back face of the sub-bolster. But, first, we change the view so that we can easily manipulate the back of the assembly. From the ANSYS top menu: Plotctrls → pan-zoom-rotate → (click) back → (click) once each on “+X” “+Y” → Close Plotctrls → numbering… → (check) area numbers → ok → Plot → areas ANSYS main menu → solution → define loads → apply → structural → displacement → on areas → (select the back face of the subbolster i.e. area #1) → apply → (select ALL) in the constraints window → ok → Ok

9. Again, for convenience, change the view of the assembly from the previously shown steps back to oblique. Now we apply the pressure from the front face of the die. ANSYS main menu → solution → define loads → apply → structural → pressure → on areas → (select the front face of the die only i.e. area#43) → apply → (load pressure value) 90e6 → ok. Now we solve the problem by: ANSYS main menu → solution → solve → current LS 10. Having completed the solution phase, we now wish to look at the stresses developed and displacements if any in the assembly units.

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For viewing von Misses stress distribution as a result of the applied pressure on the assembly: ANSYS main menu → general post processing → plot result → contour plot → nodal solution → (chose) stress-von misses SEQV→ (check) deformed + undef. Edge → ok

For viewing displacements, if any, as a result of the applied pressure on the assembly: ANSYS main menu → general post proc → plot results → deformed shape → (deformed + unreformed edge)

Chapter 7

Thermal Stress Analysis 7.1 Tutorial 9: Thermal Analysis of Mechanical Structure In this section, thermal analysis of an axisymmetric pipe with fins is presented. A pipe with cooling fins is to be analyzed for temperature and thermal flux distributions under given loading conditions (see next page). The model is axisymmetric. Also, only half the fin is modeled, so the bottom of the model is a symmetry boundary.

Loads and Material Properties

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Step-by-Step ANSYS Solution Preprocessing: Defining the Problem 1. Enter ANSYS in the working directory specified by your instructor using “pipe-th” as the jobname. 2. Set the GUI Preferences to Thermal: Main Menu > Preferences … Select “Thermal”, then [OK] 3. Read input from “pipe-th.inp” to create the 2D axisymmetric model and specify mesh divisions on lines: Utility Menu > File > Read Input from … Select “pipe-th.inp”, then [OK] Or issue:/INPUT,pipe-th,inp

Note: The input file is given at the end of the tutorial. Type it using any text editor or give input directly in PREP. 4. Specify the element type to be PLANE55 and set keyopt(3) = axisymmetric: Main Menu > Preprocessor > Element Type > Add/Edit/Delete … [Add ...] Pick “Thermal Solid” and “Quad 4node 55”, then [OK] [Options ...] Set K3 = Axisymmetric, then OK] [Close] Or issue: ET, 1, PLANE55,,, 1

Thermal Stress Analysis   •  163

5. Read the material properties from the material library for 304 Stainless Steel: Main Menu > Preprocessor > Material Props > Material Library > Import Library … Select “BIN”, then [OK] Select “Stl_AISI-304.BIN_MPL”, then [OK] Review the ansuitmp window Close the ansuitmp window

6. Mesh the model using mapped meshing with 2D quad elements: Main Menu > Preprocessor > MeshTool … Pick [Set] under Size Controls: Global SIZE = 0.25/2, then [OK] Select “Mapped”, then [Mesh] [Pick All] [Close] Or issue: MSHAPE,0,2D MSHKEY,1 ESIZE,0.25/2 AMESH, ALL

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Thermal Stress Analysis   •  165

7. Apply convection loads to the solid model lines: Main Menu > Solution > -Loads- Apply > -Thermal- Convection > On Lines + Pick the outer four lines (10, 2, 7, and 6), then [OK] Set VALI = 0.69e-4 and VAL2I = 70, then [Apply] Pick the inner two lines (9 and 13), then [OK] Set VALI = 0.28e-3 and VAL2I = 450, then [OK] Utility Menu > Plot > Multi-Plots Or issue: /SOLU SFL,2,CONV,0.69e-4, ,70 SFL,6,CONV,0.69e-4, ,70 SFL,7,CONV,0.69e-4, ,70 SFL,10,CONV,0.69e-4, ,70 SFL,9,CONV,0.28e-3, ,450 SFL,13,CONV,0.28e-3, ,450 GPLOT

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8. Save the database and obtain the solution: Pick the “SAVE_DB” button in the Toolbar (or select: Utility Menu > File > Save as Jobname.db) Main Menu > Solution > -Solve- Current LS Review the “/STATUS Command” window and then close it [OK] [Close] - to close the yellow message window after the solution is completed Or issue: SAVE /SOLU SOLVE 9. Enter the general postprocessor and review the results: Main Menu > General Postproc Or issue: /POST1 (a) Plot temperatures: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “DOF solution” and “Temperature TEMP”, then [OK] Or issue: /POST1/ PLNSOL, TEMP (b) Plot thermal flux: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Flux & gradient” and “Thermal flux TFSUM”, then [OK] Or issue: PLNSOL,TF,SUM

Thermal Stress Analysis   •  167 (c) Plot the thermal gradient:

Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Flux & gradient” and “Thermal grad TGSUM”, then [OK] Or issue: PLNSOL,TG,SUM

10. Save and exit ANSYS: Pick the “QUIT” button from the toolbar (or select: Utility Menu > File > Exit...). Select “Save Everything” [OK]. Or issue: FINISH /EXIT,ALL

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11. The command log file: /TITLE, 2D AXI-SYMM THERMAL ANALYSIS W/ CONV. LOADING-ESIZE=0.125 /PREP7 ! Solid Model RECTNG,5,6,0,1 RECTNG,6,12,0,.25 AADD,all !* Create fillet LFILLT,10,7,.2, , FLST,2,3,4 FITEM,2,5 FITEM,2,8 FITEM,2,2 AL,5,8,2 AADD,all KWPAVE, 9 wpro,,,90.000000 ASBW, 2 KWPAVE, 5 wpro,,30.000000, wpro,,30.000000, wpro,,30.000000, ASBW,3 KWPAVE,5,10 wpro,,90.000000, ASBW,2 ,asbw,4 LCOMB,11,18 ,0 LCCAT,11,12 ! Prepare for meshing LESIZE,5, , ,33,.25, , , ,0 LESIZE,7, , ,33,.25, , , ,0 LESIZE,2, , ,6, , , , ,0 LESIZE,8, , ,3, , , , ,0 LESIZE,11, , ,3, , , , ,0 LESIZE,12, , ,3, , , , ,0 LESIZE,17, , ,3, , , , ,0 WPSTYLE,,,,,,,,0 lplot

7.2 Tutorial 10 (a): Thermal-Stress Analysis-Sequential Coupled Field In this tutorial, the thermal analysis of an axisymmetric pipe with fins and sequential-coupled field is introduced. Continue the axisymmetric fin problem from Tutorial 9 to do a thermal-stress analysis. The pipe has an internal pressure as shown. Also, as the top line (at Y = 1.0) represents a line of repetitive symmetry, we will couple the UY degrees of freedom of all nodes along that line.

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Loads and Material Properties

Step-by-Step ANSYS Solution 1. Enter ANSYS in the working directory specified by your instructor using “pipe-th-str” as the jobname. 2. Resume the pipe-th.db database file from Workshop 3 (or pipe-th. db1): Utility Menu > File > Resume from … Select “pipe-th.db”, then [OK] Or issue: RESUME,pipe-th,db 3. Set the GUI Preferences to Structural: Main Menu > Preferences … Select “Structural” and unselect “Thermal”, then [OK] 4. Change the title: Utility Menu > File > Change Title ... /TITLE = “2D AXI-SYMM THERMAL-STRESS ANALYSIS W/ INT. PRESS - ESIZE=0.125” [OK]

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5. Delete the convection loading on the solid model lines: Main Menu > Preprocessor > Loads > -Loads- Delete > All Load Data > All SolidMod Lds … [OK] Or issue: /PREP7 LSCLEAR,SOLID 6. Change the thermal elements to their corresponding structural elements: Main Menu > Preprocessor > Element Type > Switch Elem Type … Select “Thermal to Struc”, then [OK]. Review the warning message, then [Close] Or issue: ETCHG,TTS

7. Set the element behavior to axisymmetric: Main Menu > Preprocessor > Element Type > Add/Edit/Delete … [Options ...] Set K3 = Axisymmetric, then [OK] [Close] Or issue: KEYOPT, 1, 3, 1 8. Apply temperature load from thermal analysis in Workshop 3: Main Menu > Preprocessor > Loads > -Loads- Apply > Temperature > From Therm Analy . Select the “pipe-th.rth” results file, then [OK]. Or issue: LDREAD,TEMP, , , , ,pipe-th,rth

Thermal Stress Analysis   •  171

9. Apply symmetry boundary condition on lines at Y = 0:

Main Menu > Preprocessor > Loads > -Loads- Apply > Displacement > Symmetry B.C.- On Lines + Select lines 3, 5, 11, then [OK]. Or issue: DL,3, ,SYMM DL,5, ,SYMM DL,11, ,SYMM 10. Couple UY DOF on nodes at Y = 1: (a) Select nodes at Y = 1:

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Utility Menu > Select > Entities ... Select “Nodes” and “By Location” Select “Y coordinates” Set Min,Max to 1, then [OK] Or issue: NSEL,S,LOC,Y,1 (b) Define a UY DOF couple set on the select set of nodes: Main Menu > Preprocessor > Coupling / Ceqn > Couple DOFs + [Pick All] NSET = 1 Set Lab = UY, then [OK]. Utility Menu > Select > Everything Or issue: CP,1,UY,ALL ALLSEL,ALL

11. Apply internal constant pressure to line: Main Menu > Preprocessor > Loads > -Loads- Apply > Pressure > On Lines + Pick lines 9 and 13, then [OK] VALUE = 1000, then [OK] Or issue: SFL,9,PRES,1000 SFL,13,PRES,1000

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12. Verify the temperature load by displaying body force loads: Utility Menu > PlotCtrls > Symbols Set Body Load Symbols = “Structural temps”, then [OK] Utility Menu > Plot > Elements Or issue: /PBF,TEMP, ,1 EPLOT 13. Save the database and obtain the solution: Pick the “SAVE_DB” button in the Toolbar (or select: Utility Menu > File > Save as Jobname.db) Main Menu > Solution > -Solve- Current LS Review the “/STATUS Command” window and then close it [OK]. [Close] - to close the yellow message window after the solution is completed Or issue: SAVE /SOLU SOLVE 14. Enter the general postprocessor and review the results: Main Menu > General Postproc > Or issue: /POST1

(a) Plot the displacement: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “DOF solution” and “Translation USUM”, select “Def + undef edge”, then [OK] Or issue: PLNSOL,U,SUM,2,1

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(b) Plot von Mises stress: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Stress” and “von Mises SEQV”, select “Def shape only”, then [OK] Or issue: PLNSOL,S,EQV

(c) Plot radial stress: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Stress” and “X-direction SX”, then [OK] Or issue: PLNSOL,S,X

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(d) Expand the axisymmetric radial stress 90° about the Y axis and reflect about the x-z plane: Utility Menu > PlotCtrls > Style > Symmetry Expansion > 2D Axi-Symmetric ... Pick “1/4 expansion” and set reflection to “yes”, then [OK] Utility Menu > PlotCtrls > Pan, Zoom, Rotate … [ISO] Or issue: /EXPAND, 9,AXIS,,,10,,2,REC T,HALF,,0.00001 /VIEW,1,1,1,1 /REPLOT

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(e) Plot the longitudinal (axial) stress:

Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Stress” and “Y-direction SY”, then [OK] Or issue: PLNSOL,S,Y (f) Plot the tangential (circumferential or hoop) stress: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Stress” and “Z-direction SZ”, then [OK] Or issue: PLNSOL,S,Z

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15. Save and exit ANSYS: Pick the “QUIT” button from the Toolbar (or select: Utility Menu > File > Exit...) Select “Save Everything” [OK] Or issue: FINISH /EXIT,ALL

7.3 Tutorial 10 (b): Thermal-Stress Analysis: Direct-Coupled Field In this tutorial, the thermal analysis of axisymmetric pipe with fins and direct-coupled field will be addressed. In this workshop problem, we will rerun the previous problem using the direct coupled field method. The axisymmetric fin will again be used for this analysis, which will include the thermal and structural loads applied previously.

Loads and Material Properties Same thermal loading applied in Tutorial 9. Internal pressure applied simultaneously:

Step-by-Step ANSYS Solution 1. Enter ANSYS in the working directory specified by your instructor using “pipe-direct” as the jobname.

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2. Read input from “pipe-th.inp” to create the 2D axisymmetric model and specify mesh divisions on lines: Utility Menu > File > Read Input from … Select “pipe-th.inp”, then [OK] Or issue: /INPUT,pipe-th,inp 3. Add an axisymmetric coupled field element type (plane13): Main Menu > Preprocessor > Element Type > Add/Edit/Delete… Select “Coupled Field” and “Vector Quad 13”, then [OK] 4. Modify element options for structural or thermal, axisymmetric: Options K1 = UX UY Temp AZ K3 = Axisymmetric [OK] [Close] Or issue: ET,1,PLANE13 KEYOPT,1,1,4 KEYOPT,1,3,1

5. Mesh the model using mapped meshing with 2D quad elements: Main Menu > Preprocessor > MeshTool … Pick [Set] under Size Controls: Global SIZE = 0.25/2, then [OK] Select “Mapped”, then [Mesh] [Pick All] Or issue: MSHAPE,0,2D MSHKEY,1 ESIZE,0.25/2 AMESH,ALL 6. Read the material properties from the material library for 304 ­Stainless Steel:

Thermal Stress Analysis   •  179

Main Menu > Preprocessor > Material Props > Material Library > Library Path ... Enter path for “Path for READING files” (eg. h:\ ansys57\matlib) [OK]. Main Menu > Preprocessor > Material Props > Material Library > Import Library … Select “BIN”, then [OK]. Select “Stl_AISI-304.BIN_MPL”, then [OK] Review the ansuitmp window Close the ansuitmp window 7. Apply convection loads to the solid model lines: Main Menu > Preprocessor > Loads > - Loads- Apply > -ThermalConvection > On Lines + Pick the outer four lines, then [OK] Set VALI = 0.69e-4 and VAL2I = 70, then [Apply] Pick the inner two lines, then [OK] Set VALI = 0.28e-3 and VAL2I = 450, then [OK] Or issue: SFL,2,CONV,0.69e-4, ,70 SFL,6,CONV,0.69e-4, ,70 SFL,7,CONV,0.69e-4, ,70 SFL,10,CONV,0.69e-4, ,70 SFL,9,CONV,0.28e-3, ,450 SFL,13,CONV,0.28e-3, ,450

180  •   Using ANSYS for Finite Element Analysis

8. Apply internal constant pressure to line:

Thermal Stress Analysis   •  181

Main Menu > Preprocessor > Loads > -Loads- Apply > Pressure > On Lines + Pick lines 9 and 13, then [OK] VALUE = 1000, then [OK] Or issue: SFL,9,PRES,1000 SFL,13,PRES,1000 9. Couple UY DOF on nodes at Y = 1: (a) Select nodes at Y = 1: Utility Menu > Select > Entities ... Select “Nodes” and “By Location” Select “Y coordinates” Set Min,Max to 1, then [OK] Or issue: NSEL,S,LOC,Y,1 (b) Define a UY DOF couple set on the select set of nodes: Main Menu > Preprocessor > Coupling / Ceqn > Couple DOFs + [Pick All] NSET = 1 Set Lab = UY, then [OK] Utility Menu > Select > Everything

182  •   Using ANSYS for Finite Element Analysis

Or issue: CP,1,UY,ALL ALLSEL,ALL

10. Apply symmetry boundary condition on lines at Y = 0: Main Menu > Preprocessor > Loads > -Loads- Apply > Displacement > Symmetry B.C.- On Lines + Select lines 3, 5, 11, then [OK] Or issue: DL,3, ,SYMM DL,5, ,SYMM DL,11, ,SYMM 11. Save the database and obtain the solution: Pick the “SAVE_DB” button in the Toolbar (or select: Utility Menu > File > Save as Jobname.db) Main Menu > Solution > -Solve- Current LS Review the “/STATUS Command” window and then close it [OK] [Yes] – to continue with solve after warnings [Close] - to close the yellow message window after the solution is completed Or issue: SAVE /SOLU SOLVE

12. Enter the general postprocessor and review the results: Main Menu > General Postproc > Or issue: /POST1

Thermal Stress Analysis   •  183

(a) Plot the temperatures: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “DOF solution” and “Temperature TEMP”, then [OK] Or issue: PLNSOL,TEMP (b) Plot the displacement:: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “DOF solution” and “Translation USUM”, select “Def + undef edge”, then [OK]. Or issue: PLNSOL,U,SUM,2,1 (c) Plot von Mises stress: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “Stress” and “von Mises SEQV”, select “Def shape only”, then [OK] Or issue: PLNSOL,S,EQV (d) Expand the axisymmetric radial stress 90° about the Y axis and reflect about the x-z plane: Utility Menu > PlotCtrls > Style > Symmetry Expansion > 2D Axi-Symmetric ... Pick “1/4 expansion” and set reflection to “yes”, then [OK] Utility Menu > PlotCtrls > Pan, Zoom, Rotate …[ISO] Or issue: /EXPAND, 9,AXIS,,,10,,2,RECT,HALF,,0.00001 /VIEW,1,1,1,1 /REPLOT (e) Plot the temperatures: Main Menu > General Postproc > Plot Results > -Contour PlotNodal Solu ... Pick “DOF solution” and “Temperature TEMP”, then [OK] Or issue: PLNSOL,TEMP

184  •   Using ANSYS for Finite Element Analysis

13. Save and exit ANSYS: Pick the “QUIT” button from the Toolbar (or select: Utility Menu > File > Exit...) Select “Save Everything” [OK] Or issue: FINISH /EXIT,ALL

Summary In this volume of the book “Fundamentals of Finite Element Methods with ANSYS Tutorials and Applications for Engineering” after providing a brief introduction on finite element analysis and modeling, various guided tutorials were presented to demonstrate how simple structural analysis can be carried out by ANSYS. Most of these tutorials have been specifically developed for this book and are based on the version of ANSYS shown in the corresponding figures of the tutorial examples. For the static ­analysis of different structures, the most common finite element types introduced in Section 1.4, such as solid modeling using 2D and 3D primitives ­available in ANSYS, static structural analysis (truss, beam, 2D and 3D structures), and thermal analysis, are explained. In the second volume, tutorials demonstrate dynamic analysis capabilities of ANSYS and the topics covered include harmonic and modal analysis. In addition, analysis of composite structures, probabilistic design analysis, APDL programming, and design optimization are presented through corresponding tutorials.

Bibliography Cook, R.D. 1995. Finite Element Modeling for Stress Analysis, 1st ed. H ­ oboken, NJ: John Wiley & Sons. Incropera, F.P. 1985. Fundamentals of Heat and Mass Transfer, Example 3.7, 2nd ed., 104. Singapore: John Wiley & Sons. Reddy, J.N. 1993. An Introduction to the Finite Element Method, New York: McGrawHill. Logan, D.L. 2001. A First Course in the Finite Element Method, 3rd ed. Mason, OH: Thomas Learning Publishing. Reddy, J.N. 1972. “Exact Solutions of Moderately Thick Laminated Shells.” ­Journal Engineering Mechanics 110, no. 5, pp. 794–805. Timoshenko, S. 1956. Strength of Material, Part II, Elementary Theory and ­Problems, 3rd ed., 111. New York, NY: D. Van Nostrand Co., Inc.

Index A addition of matrices, 15 algebraic equation. See linear algebraic equations ANSYS. See also geometric modeling; static analysis model geometry analysis type and analysis options, 61–62 applying loads and obtaining solution, 61, 62–63 initiate solution, 63 load step options, 63 preprocessing analysis title, 55 element real constants, 57–59 element types, 56–57 jobname, 54–55 linear material properties, 59–60 material properties, 59 nonlinear material properties, 60–61 units, 55 results, 63–64 applied mechanics, 1 area elements, static analysis plane problem (bracket), 121–122, 134–141 analysis type, 127 Boolean operations, 122–124 constraints, 127–128 deflection, 133

deformation, 132 element material properties, 126 geometric properties, 125–126 listing the stresses, 130–132 loads, applying, 128–129 meshing, 126–127 saving, 127 solving, 129–130 stresses, 133–134 title, giving, 122 type of element, 125 plane problem (wrench) ANSYS solution, 142–146 geometry, 142 loading, 142 material properties, 142 methodology, 142 assembly design (static analysis) ANSYS solution, 154–160 boundary condition, 154 geometry, 154 loading, 154 material properties, 154 methodology, 154 B bar element formulation, 31–33 beam elements, static analysis. See also distributed loads on beam elements analysis type, 113

190  •   Index

constraints, 113 deformation, 115 element density, 113 element material properties, 113 gravity, 113–114 keypoints, 112 lines, creating, 113 meshing, 113 preprocessor menu, 112 real constants, 113 solving, 115 title, giving, 112 type of element, 113 bracket (plane problem). See area elements, static analysis buckling analysis, 11 C CAD model, 66–68 compatibility equation, 19 component design (static analysis) ANSYS solution, 148–153 constraints, 147 geometry, 147 loading, 147 material properties, 147 methodology, 148 composites, 12 computational mechanics, 1 contact problems, 11–12 Cramer’s rule, 22–23 D differentiating a matrix, 16 direct-coupled field (thermal analysis) ANSYS solution, 177–184 loads and material properties, 177 direct (equilibrium) method, 27 bar element formulation, 31–33 spring element equations, 28–31 discretization error, 50–51 displacement/stiffness method, 26

distributed loads on beam elements. See also beam elements, static analysis analysis type, 118 constraints, 118 deformation, 119–120 element material properties, 117 element types, 116 keypoints, 116 lines, 116 loads, applying, 118–119 meshing, 117 plot elements, 117–118 preprocessor menu, opening, 116 real constants, 117 solving, 119 title, giving, 116 dynamic analysis. See transient dynamic analysis E elasticity equations one dimension, 20 plane strain, 21 plane stress, 20–21 strain-displacement relationship, 19 stress equilibrium equations, 17–18 stress-strain relationships, 19–20 elementary beam theory, 49 element characteristic matrix direct method, 27 variational method, 28 weighted residual methods, 28 element real constants, 57–59 engineering analysis, 13 F fatigue, 13 fatigue analysis, 9 FEA program, 7–8 finite difference method, 4 finite element analysis (FEA)

Index   •   191

discretization error, 50–51 history of, 5 modeling error, 49–50 numerical error, 51–52 finite element method (FEM) bar element formulation, 31–33 description, 1 element characteristic matrix, 27–28 examples, 34–48 FEA, 6–8 (see also finite element analysis (FEA)) mathematical model, 1–3, 5–6 numerical methods, 4 physical, 6 spring element equations, 28–31 steps of, 26–27 structural analysis (see structural analysis of FEM) types, 27 force / flexibility method, 26 fracture mechanics, 12 G Galerkin method, 28 Gaussian elimination, 23–26 geometric modeling solid model, 68–69 steps in, 65–66 3D, 79–89 2D, 69–78 using CAD system, 66–68 H harmonic analysis, 10 heat transfer analysis, 9 I identity matrix, 16 integrating a matrix, 16 inverse of a matrix, 16 inversion, 23

L linear algebraic equations Cramer’s rule, 22–23 Gaussian elimination, 23–26 inversion, 23 linear behavior, 9–10 linear material properties, 59–60 linear vs. nonlinear static analysis, 54 load step options, 63 M material properties, 59 mathematical model, 1–3, 5–6 matrix algebra, 15–17 mechanical structure (thermal analysis) ANSYS solution, 162–168 loads and material properties, 161 modal analysis, 10 modeling, 48–49 modeling error, 49–50 multi-physics application, 14 multiplication of a matrix by a scalar, 15 multiplication of two matrices, 15 N nonlinear analysis, 11 nonlinear behavior, 9–10 nonlinear material properties, 60–61 nonstructural application, 14 numerical error, 51–52 numerical methods, 4 O one dimension elasticity equation, 20 P physical FEM, 6 plane strain elasticity equation, 21

192  •   Index

plane stress elasticity equation, 20–21 post-processing phase, 27 preprocessing phase, 26 S sequential-coupled field (thermal analysis) ANSYS solution, 169–177 loads and material properties, 169 solid modeling, 68–69 3D, 79–89 2D, 69–78 solution phase, 26 spectrum analysis, 10 spring element equations, 28–31 static analysis, 9–10 definition of, 53–54 linear vs. nonlinear, 54 loadings in, 54 using area elements (see area elements, static analysis) using beam elements (see beam elements, static analysis) using truss elements (see truss elements, static analysis) using volume elements (see volume elements, static analysis) static analysis procedure model geometry analysis type and analysis options, 61–62 applying loads and obtaining solution, 61, 62–63 initiate solution, 63 load step options, 63 preprocessing analysis title, 55 element real constants, 57–59 element types, 56–57 jobname, 54–55

linear material properties, 59–60 material properties, 59 nonlinear material properties, 60–61 units, 55 results, 63–64 steady-state transfer, 9 stiffness matrix, 27 strain-displacement relationship, 19 stress equilibrium equations, 17–18 stress-strain relationships, 19–20 structural analysis, 53 structural analysis of FEM, 8–9 advantages, 13–14 application, 14–15 buckling analysis, 11 composites, 12 contact problems, 11–12 disadvantages, 14 engineering types, 13 fatigue, 13 fracture mechanics, 12 harmonic analysis, 10 modal analysis, 10 nonlinear analysis, 11 spectrum analysis, 10 static analysis, 9–10 transient dynamic analysis, 10 structural application, 14 T theoretical mechanics, 1 thermal analysis direct-coupled field ANSYS solution, 177–184 loads and material properties, 177 mechanical structure ANSYS solution, 162–168 loads and material properties, 161

Index   •   193

sequential-coupled field ANSYS solution, 169–177 loads and material properties, 169 3D model, 8, 79–89 time history analysis. see transient dynamic analysis transient dynamic analysis, 10 transpose of a matrix, 15 truss elements, static analysis analysis type, 101 axial stress, 110–111 constraints, applying, 101–103 deflection, 107–109 deformation, 106–107 element material properties, 98 forming lines, 94–95 geometric properties, 96–97 keypoints, 92–94 listing, 109–110 listing the stresses, 111–112 loads, applying, 103–104 meshing, 95–96, 99, 100 plot numbering, 100 quitting ANSYS, 112 reaction forces, 106 saving, 100–101

solving, 104–105 title, giving, 91–92 2D model, 8, 69–78 V variational method, 28 vibrational analysis, 9 volume elements, static analysis assembly design ANSYS solution, 154–160 boundary condition, 154 geometry, 154 loading, 154 material properties, 154 methodology, 154 component design ANSYS solution, 148–153 constraints, 147 geometry, 147 loading, 147 material properties, 147 methodology, 148 W weighted residual methods, 28 wrench (plane problem). See area elements, static analysis

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A Tutorial for Engineers, Volume I Wael A. Altabey • Mohammad Noori • Libin Wang Over the past two decades, the use of finite element method as a design tool has grown rapidly. Easy to use commercial software, such as ANSYS, have become common tools in the hands of students as well as practicing engineers. The objective of this book is to demonstrate the use of one of the most commonly used Finite Element Analysis software, ANSYS, for linear static, dynamic, and thermal analysis through a series of tutorials and examples. Some of the topics covered in these tutorials include development of beam, frames, and Grid Equations; 2-D elasticity problems; dynamic analysis; composites, and heat transfer problems. These simple, yet, fundamental tutorials are expected to assist the users with the better understanding of finite element modeling, how to control modeling errors, and the use of the FEM in designing complex load bearing components and structures. These tutorials would supplement a course in basic finite element or can be used by practicing engineers who may not have the advanced training in finite element analysis. Wael A. Altabey is an assistant professor in the department of mechanical engineering, faculty of engineering, Alexandria University, Alexandria, Egypt and has been a postdoctoral researcher at the International Institute for Urban Systems Engineering, Southeast University, Nanjing, China. Mohammad Noori is a professor of mechanical engineering at California Polytechnic State University in San Luis Obispo, California, USA, and a fellow of the American Society of Mechanical Engineers. Dr Noori has over 34 years of experience as a scholar and educator. He has also been a distinguished visiting professor at the International Institute for Urban Systems Engineering, Southeast University, Nanjing, China. Libin Wang is a professor and the dean of the school of civil engineer-

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ing at Nanjing Forestry University, in Nanjing, China. He has been an educator and scholar, for over 20 years, and has taught the subject of finite element analysis both at the undergraduate and graduate level.

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Using ANSYS for Finite Element Analysis, Volume I

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Using ANSYS for Finite Element Analysis

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Using ANSYS for Finite Element Analysis A Tutorial for Engineers Volume I

Wael A. Altabey Mohammad Noori Libin Wang