prepostfe

MMEMÀSTER EN MECÀNICA DE

MATERIALS I ESTRUCTURES

Universitat de Girona

PRE AND POST-PROCESS TOOLS IN FINITE

ELEMENT ANALYSIS

M. Baena, C. Barris, N. Blanco, E.V. González, J.A. Mayugo, J.

Renart, D. Trias, A. Turon

September 2012

This document can be found at:

ftp://amade.udg.edu/mms/PrePostFE/PrePostFE.pdf

ftp://amade.udg.edu/mms/PrePostFE/PrePostFE.pdf

Contents

1 Introduction to Pre and Post-Process programs 1

1.1 Typical structure of the Finite Element programs . . . . . . . . . . . . . . . . . . . . . 1

1.2 Finite element generation, calculation and analysis of a typical mechanical problem 2

1.2.1 Generation of finite element models with the Graphical User Interface (GUI)

of ANSYSTM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.2 Generation of ANSYSTM finite element models via commands: APDL . . . . . 6

1.2.3 Generation of ANSYSTM parametrized finite element models . . . . . . . . . . 7

2 Types of finite elements 9

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 2D bar/truss elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3 3D bar/truss elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 2D beam elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5 Plane stress elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.6 Plane strain elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.7 Shell elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.8 Suggested problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3 Geometric modeling and meshing techniques 21

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Direct generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3 Solid Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.4 Assessment of the mesh quality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.4.1 Connectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.4.2 Avoid elements with high distortions, warping and inadmissible narrow or

wide angles between edges. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.5 Techniques to Import Models and Geometries . . . . . . . . . . . . . . . . . . . . . . . 44


4 Boundary conditions 47

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.2 Load application in FEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3 Constrain DOF of FE model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3.1 Constrained displacements and rotations . . . . . . . . . . . . . . . . . . . . . . 56

4.3.2 Symmetry conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3.3 Antisymmetry conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.4 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.5 Structural analysis with temperature change . . . . . . . . . . . . . . . . . . . . . . . 61

4.6 Loadcases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63


iii

5 Linear elastic material models 67

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.2 One-dimensional elastic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.3 Bi-dimensional and Three-dimensional elastic properties . . . . . . . . . . . . . . . . 68

5.3.1 Isotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.3.2 Orthotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.3.3 Anisotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.4 Element coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72


6 Postprocessing 77

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.2 General postprocessing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

6.3 Time-history postprocessing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.4 Result analysis of combined stresses or strains . . . . . . . . . . . . . . . . . . . . . . 84

6.4.1 Combined strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.4.2 Combined stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.5 Assessment of the mesh and results quality . . . . . . . . . . . . . . . . . . . . . . . . 87

6.5.1 Energy norm to estimate the error of the elements . . . . . . . . . . . . . . . . 87


7 Coupling and constrain equations. Submodelling 95

7.1 Coupling and constrain equations. Introductory concepts . . . . . . . . . . . . . . . . 95

7.1.1 Coupled DOF sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.1.2 Constraint equations of DOF . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

7.2 Submodelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

7.2.1 The global model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7.2.2 The submodel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7.2.3 Submodelling procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104


8 Nonlinear analysis. Geometric nonlinearities 113

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

8.2 Geometric nonlinear behaviour due to large displacements . . . . . . . . . . . . . . . 114

8.3 Buckling analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

8.3.1 Eigenvalue buckling analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

8.3.2 Nonlinear buckling analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121


9 Material nonlinearities 127

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

9.2 Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

9.2.1 Bilinear Kinematic Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

9.2.2 Multiple-point Isotropic Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 132

9.3 Two parameter Mooney-Rivlin Hyperelastic . . . . . . . . . . . . . . . . . . . . . . . . 135


Chapter 1

Introduction to Pre and Post-Process

programs

1.1 Typical structure of the Finite Element programs

Typically, the Finite Element programs are divided in three parts, the Pre-processor module, the

Solution module, and the Post-processor module. In the Pre-processor, the geometry, element

type, material properties, and discretisation of the problem (mesh) is achieved. Once all of the

previous are defined, the analysis type, the boundary conditions, the assembly of the stiffness

matrices, and the solution of the system of the equations is performed in the Solution module.

Finally, once the problem has been solved, the Post-processor module helps the user to analyze

the results, plotting stress, strains, displacement or reaction forces at different load levels, for

example.

Figure 1.1: Structure of a FEM program.

It should be mentioned here that different softwares can be used for the three different parts.

For example, a specific software can be used for preprocessing the problem, and another one for

solving the problem and postprocessing the results. In most of the commercial FE programs all

the Pre-Process, Solution and Post-Process can be done either directly by commands or using the

Graphical User Interphase (GUI). Although the latter is more user-friendly, it is important to know

how to use the different commands and command scripts as it results in a much more flexible

way to use these programs. A very powerful tool using command scripts is the parametrization

of the model, which allows the modification of the model in a very easy way.

1

2 Pre and post-process tools in finite element analysis

1.2 Finite element generation, calculation and analysis of a typi-

cal mechanical problem

The common procedures for the mechanical simulation and analysis with finite elements using

ANSYSTM are described in the following examples.

Reminder: Units

Finite element analysis are dimensionless, so before starting to create the model it is conve-

nient to establish which dimensions are to be used and be coherent. If the material properties

to be used are expressed in N/mm2, the geometric units should be in mm while forces should

be expressed in N.

1.2.1 Generation of finite element models with the Graphical User Interface

(GUI) of ANSYSTM

Example 1.1. Use the ANSYSTM graphical environment to simulate and analyse the cantilever

beam shown in Figure 1.2. The beam is clamped on its left end, while on the right end a constant

1 mm displacement is imposed. The cross-section of the beam is rectangular and constant along

the span. The span of the beam is 2 m, its base is 200 mm and its height is 100 mm. The material

is steel, E = 210 GPa and ν = 0.3.

In the post-process, use the graphical interface to plot the strain and stress distributions in

the beam, show the reactions forces and represent the deformed shape of the beam. Obtain a

list of the reaction forces, the stress variation with the vertical coordinate by defining a Path-plot

and represent the temporal variation of the stress in a point with History-plot.

Figure 1.2: Cantilever beam with an imposed displacement.

Solution to Example 1.1. The ANSYSTM Environment for ANSYSTM 13.0 contains 2 windows:

the Main Window and an Output Window. Within the Main Window are 5 divisions:

1. Utility Menu. The Utility Menu contains functions that are available throughout the

ANSYSTM session, such as file controls, selections, graphic controls and parameters.

2. Input Line. The Input Line shows program prompt messages and allows you to type in

commands directly.

3. Toolbar. The Toolbar contains push buttons that execute commonly used ANSYSTM com-

mands. More push buttons can be added if desired.



Chapter 1. Introduction to Pre and Post-Process programs 3

4. Main Menu. The Main Menu contains the primary ANSYSTM functions, organized by

preprocessor, solution, general postprocessor, design optimizer. It is from this menu that

the vast majority of modelling commands are issued.

5. Graphics Window. The Graphic Window is where graphics are shown and graphical pick-

ing can be made. It is here where you will graphically view the model in its various stages

of construction and the ensuing results from the analysis.

Graphics Window

Main Menu

Tool Bar

Input Line Utility Menu

Figure 1.3: ANSYSTM 13.0 interface.

On the other hand, the Output Window shows text output from the program, such as listing

of data etc. It is usually positioned behind the main window and can be put to the front if

necessary.

Before starting with the generation of the model, it is important to know how to locate and

obtain help on the different commands within ANSYSTM . The Help for the program can be

started within the Graphic Window by different procedures. The first option is by clicking on the

Help button in the Utility Menu and selecting Help Topics. The second option is by clicking on

the question mark button next to the Input Line. The last option, only valid to obtain help on

a specific ANSYSTM command, consists on typing help,XXXXX on the Input Line, where XXXXX

is the name of the specific command. Using one of the three options, a new window appears

with four different tabs. In the first tab, Contents, all the information about the program has

been arranged in accordance to different subjects. Here you can find specific help for different

commands, elements, analysis types, theory, tutorials and examples. In the second tab, Index,

you can find specific help for any command, element or analysis types arranged in alphabetical

order. In the third tab, you can search all the relative information within ANSYSTM (theory,

command structure, analysis types, examples, etc.) related to a specific keyword. In the last tab,

Favourites, you can store all the information about frequently consulted or important Help items

you wish to locate easily.

Once you have identified the different parts of the ANSYSTM environment, start with the

creation of the new model.

1. First, the model has to be identified with the simulated problem. Change the name of the

model to PrePostFE_Ex101 (File>Change Jobname).




2. Save the generated model. The menu command File>Save As Jobname.db saves the model

as PrePostFE_Ex101.db.

3. Start with the preprocessor module (»PreProcessor in the Main Menu).

• Define the type of the element you are going to use throughout the analysis: »Pre-

processor>Element Type>Add/Edit/Delete>Add...>Solid>Brick 8node 185. ANSYSTM

13.0 recommends the use of the current-technology SOLID185 element instead of

SOLID45. However, for the SOLID185 element to have the same behaviour as SOLID45,

the element technology should be stated to be "simplified enhanced element formula-

tion", which stands as keyoption(2)=3. Edit the options of the element and choose this

property. Spend a while looking at the different type of elements available in ANSYSTM

13.0

• Define the material properties of the model: »Preprocessor>Material Props>Material

Models>Structural>Linear>Elastic>Isotropic. Enter the required material properties

for the model, Young modulus and Poisson ratio, using the appropriate units. Again,

take a while to examine the different material models available.

• Generate the geometry: »Preprocessor>Modelling>Create>Volumes>Block>By Di-

mensions. Note that on the top left corner of the prompt there is the name of the

associated command, [BLOCK], and a short description. Use one of the three HELP

options to obtain further information on this command. Enter the dimensions of the

model in order to align the span of the beam with x-axis. Use the graphical tools on

the right of the Graphic Window to obtain different views of the created geometry.

• Mesh the model. First define the controls to guide the mesh, for example, 40 divisions

in X-direction, 4 in Y direction and 8 in Z direction. Open the mesh module »Prepro-

cessor>Meshing>MeshTool and apply the mesh controls to the model with the option

Size Controls>Lines>Set and then NDIV equal to 40, 4, 8 for the corresponding lines.

Define the shape of the element to be hexahedral Hex. Then click on the Mesh button

to select the block and mesh it (Mesh>Volumes>Mapped>4 to 6 sided ).

4. Continue with the solution module, »Solution in the Main Menu.

• Define the type of the analysis: »Solution>Analysis Type>New Analysis>Static.

• Define the parameters of the analysis »Solution>Analysis Type>Sol’n Controls. Time

at the end of load step=1, Number of substeps=10 and Write Items to Results file>

Frequency>Write every Nth substep. With these choices the computation will be per-

formed in 10 substeps, and the results file will be updated every Nth substep.

• After that, define the boundary conditions of the model. Apply a 1mm vertical displace-

ment at the end of the beam »Solution>Define Loads>Apply> Structural> Displace-

ment>On Lines. Make sure the displacement is applied in the correct direction. In

order to select the appropriate lines, it might result convenient to plot only the lines in

the model. In the Utility Menu »Plot>Lines. To display the number of the lines »PlotC-

trls>Numbering...>LINE Line numbers. The same procedure can be used for areas,

volumes, nodes, elements, etc. Clamp the other side of the beam »Solution>Define

Loads>Apply>Structural>Displacement>On Areas selecting ALL DOF.

• Finally, solve the model »Solution>Solve>CurrentLS

5. Continue with the postprocess: General Postproc in the Main Menu.

• Plot the deformed shape: »General Postproc>Plot Results>Deformed Shape




• Plot Stress or Strain quantities: »General Postproc>Plot Results>Contour Plot

• List Reactions: »General Postproc>List Results>Reaction Solu

• List any nodal quantity: »General Postproc>List Results>Nodal Solution. Repeat the

previous instructions for another substep. To do that you must move to another substep

with the command »General Postproc>Read Results>By Pick

Reminder: Results Viewer

Alternatively, you can open the Results viewer tool »General Postproc>Results Viewer to plot

contour plots, list nodal or element quantities, easily change substeps and animate the results

among many other utilities.

Figure 1.4: Contour plot of stresses in X-direction.

6. Create a PATH PLOT. You can define a path through the mesh and then plot a quantity on

that PATH. To do this follow the next instructions.

• Define the PATH, »General Postproc>Path Operations>Define Path>By nodes, select-

ing a line of nodes. For example, select the nodes on the vertical edge of the clamped

area and name it.

• Plot the PATH:»General Postproc>Path Operations>Plot Paths.

• Create the path plot mapping the desired quantity onto the pathGeneral Postproc>Path

Operations>Map onto Path>Stress>X-direction SX.

• Plot the path onto the geometry »General Postproc>Path Operations>Plot Path Item>On

Graph>SX




7. HISTORY PLOT. Plot the evolution of a quantity on a node/element with step time or another

quantity.

• Open the Time History Postprocess tool: »TimeHist Postpro.

• Add Time-History Variable, for exampleNodal Solution>Stress>X-Component of Stress.

• Select the desired node.

• Plot the Time-History variable.

• As an exercise plot the stress-strain evolution of a selected node.

1.2.2 Generation of ANSYSTM finite element models via commands: APDL

In ANSYSTM the generation, solving and analysis of the results can be done via commands using

an specific programming language called ANSYS Parametric Design Language (APDL). Using

this language the finite models can be defined as a sequence of commands in an external file,

which results to be very convenient for execution and modification of the models.

Example 1.2. Generate, simulate and analyse the model for the beam presented in Ex. 1.1 by

means of a command file.

Solution to Example 1.2. The ANSYSTM command sequence for this example is listed below.

You can either type these commands on the command window, or you can type them on a file,

then, on the command window enter /input, file, ext or just use copy and paste.

FINISH !close all previous modules

/CLEAR !clear all previous models

/TITLE, Cantilever beam

/PREP7 !start PRE-PROCESSOR MODULE

ET,1,SOLID185 !element type #1: SOLID185 (3 DoF)

KEYOPT,1,2,3 !element #1: keyoption 2=3

MP,EX,1,210000 !Young’s modulus for material #1

MP,NUXY,1,0.3 !Poisson modulus for material #1

BLOCK,0,2000,0,100,0,200 !create a solid of 2mx100mmx200mm

LESIZE,5,,,40 !40 divisions in line 5 (2m)

LESIZE,6,,,4 !4 division in line 6 (100mm)

LESIZE,10,,,8 !8 divisions in line 10 (200mm)

MSHAPE,0,3D !hexaedral-shape: Element shape to be used for meshing

VMESH,1 !mesh volume

FINISH !end of PRE-PROCESSOR MODULE

/SOLU !start SOLUTION MODULE

ANTYPE,0 !static analysis

NSUBST,10,0,0 !perform the analysis in 10 sub-steps

OUTRES,ALL,1 !write results for each sub-step

TIME,1 !end time of the analysis

DL,10,,UY,-1 !applied 1mm displacement in vertical direction at x=2000mm.

!bottom line of the area

DA,5,ALL !clamped area at x=0

SOLVE !solve the current load case

FINISH !end of the SOLUTION MODULE

This file can be found at:

ftp://amade.udg.edu/mms/PrePostFE/input_files/T1/PrePostFE_Ex102.dat

The ANSYSTM command sequence for postprocessing this example is listed below.





/POST1 !start Post-Processor module

PLDISP,1 !plot the deformed shape

PLESOL,S,EQV,0,1. !contour plot of Von Mises stress

PLESOL,S,X,0,1. !contour plot of stress in X-direction

PLESOL,EPEL,X,0,1. !contour plot of strain in X-direction

PLESOL,EPEL,1,0,1. !contour plot of principal strain

PRRSOL !list of reaction forces

PATH,cami,2,1,1 !definition of PATH "cami"

PPATH,1,414 !start node on PATH "cami"

PPATH,2,708 !end node on PATH "cami"

/PBC,PATH,1 !show path

/REPLOT

PDEF,defcami,EPEL,X,AVG !project x-strain on path, name it ’defcami’

PLPATH,defcami !plot defined "defcami"

/POST26 !history plot module

ANSOL,2,866,EPEL,X,EPELX_2 !create evolution of X-strain at node 866

PLVAR,2 !plot x-strain evolution

PRVAR,2 !list x-strain evolution


ftp://amade.udg.edu/mms/PrePostFE/input_files/T1/PrePostFE_Ex102post.dat

1.2.3 Generation of ANSYSTM parametrized finite element models

Example 1.3. Modify the input file generated in Ex.1.2 to parametrize the geometry, loading

and material properties of the model.



then, on the command window enter /input, file, ext. For post-processing the command sequence

is the same as in the previous example.

Hint 1: Parameter definition in ANSYSTM

Basically, there are two different ways to define parameters in ANSYSTM scripts:

• Using the command *SET followed by the name and value for the parameter

*SET,L,2000 (defines the parameter L with a value of 2000)

• Directly defining the parameter and its value

H=100 (defines the parameter H with a value of 100)



/TITLE, Parametrized cantilever beam


!Parameters

H=100 !beam height in mm

L=2000 !beam span in mm

B=200 !beam base in mm

Young=210000 !Young modulus in MPa

Pois=0.3 !Poisson



ftp://amade.udg.edu/mms/PrePostFE/input_files/T1/PrePostFE_Ex102post.dat


ndiv=L/H*2 !number of divisions


KEYOPT,1,2,3 !element #1: keyoption 2=3

MP,EX,1,Young !Young’s modulus for material #1

MP,NUXY,1,Pois !Poisson modulus for material #1

BLOCK,0,L,0,H,0,B !create a solid block

LESIZE,5,,,ndiv !divisions in line 5

LESIZE,6,,,4 !4 division in line 6

LESIZE,10,,,8 !8 divisions in line 10

MSHAPE,0,3D !hexaedral-shape: Element shape to be used for meshing

VMESH,1 !mesh volume

FINISH !end PRE-PROCESSOR MODULE






DL,10,,UY,-1 !applied 1mm displacement in vertical direction at x=L.

!bottom line of the area

DA,5,ALL !clamped area at x=0


FINISH !end SOLUTION MODULE






Chapter 2

Types of finite elements

2.1 Introduction

Finite element programs usually have an element library that contains different element types.

In addition, there are several different physical problems that can be modeled and simulated by

means of the finite element method and every field of application needs a different element type.

For instance, the type of finite element used to simulate and obtain displacements and stresses

of the beam in Fig. 1.2 cannot be used to simulate and obtain the gradients of temperature in

the same beam when a heat focus is acting on the top surface. The element type determines the

element formulation used, the degrees of freedom, interpolation functions, wheter the element

is 2D or 3D, etc. Thus, before starting the simulation of any real problem with finite elements it

is necessary to stablish the nature of the problem and determine the type of finite elements that

fit the simulation.

In this course, most of the analyses and simulations will be of structural nature. Table 2.1

summarises some of the ANSYSTM structural elements that are more commonly used.

9


Element nodes DOFs Element description Element Order

Linear

LINK1 2 ux uy line bar/truss, 2D space linear

LINK8 2 ux uy uz line bar/truss, 3D space linear

LINK180 2 ux uy uz line bar/truss, 3D space linear

BEAM3 2 ux uy θz line beam in 2D space linear

BEAM4/44 2 ux uy uz θx θy θz line beam in 3D space linear

BEAM188 2 ux uy uz θx θy θz ω Timoshenko beam (includes

shear deformation)

linear(2 nodes)/

quadratic(2+1 nodes) /

cubic (2+2 nodes)

BEAM189 3 ux uy uz θx θy θz ω 3-node quadratic version of

BEAM188

quadratic

2D Solid

PLANE42 4 ux uy solid 4-node quadrilateral in

2D space

linear

PLANE82 8 ux uy solid 8-node quadrilateral in

2D space

quadratic

PLANE182 4 ux uy Similar to PLANE42 linear

PLANE183 8 ux uy Similar to PLANE82 quadratic

3D Solid

SOLID45 8 ux uy uz Basic linear brick. 8-node hex-

ahedral 3D space

linear

SOLID95 20 ux uy uz 20-node version of SOLID45 quadratic

SOLID92 8 ux uy uz 10-node tetrahedral 3D space quadratic

SOLID185 8 ux uy uz Similar to SOLID45 linear

SOLID186 20 ux uy uz Similar to SOLID85 quadratic

3D Shell

SHELL63 4 ux uy uz θx θy θz shell 4-node quadrilateral in

3D space

linear

SHELL93 8 ux uy uz θx θy θz shell 8-node quadrilateral in

3D space

quadratic

SHELL91 4 ux uy uz θx θy θz nonlinear layered shell 4-node

quadrilateral in 3D space

linear

SHELL99 4 ux uy uz θx θy θz linear layered shell 4-node

quadrilateral in 3D space

linear

SHELL181 4 ux uy uz θx θy θz Similar to SHELL63 linear

Table 2.1: Some linear structural elements available in ANSYSTM . Try to stick with the 18x series

of elements when possible because they incorporate the latest technology.

Each element type has different options. For example, the planar solid element has an option

to choose between plane strain and plane stress. On the other hand, some elements require the

determination of real constants. These are properties that depend on the element type, such as

cross-sectional properties of a beam, the ply sequence in a laminated shell element, etc.

In the following sections, different structural problems are modelled using one of the suitable

finite element types in order to illustrate the selection of the most appropriate finite element for

each problem.

2.2 2D bar/truss elements

Example 2.1. Compute the maximum vertical displacement in the 2D overhead hoist repre-

sented in Figure 2.1, where P = 100 kN, b = 10 cm, h = 20 cm, t = 2 mm, L = 3 m, and a = 1.5 m.

The material of the trusses is linear elastic with E = 210 GPa and ν = 0.3. Use TRUSS elements.



then, on the command window enter /input, file, ext.



/TITLE, Overhead hoist, 2D truss




Chapter 2. Types of finite elements 11

Figure 2.1: Overhead hoist.

!Parameters

P=100000 !applied load in N

b=100 !section base in mm

h=200 !section height in mm

t=2 !section thickness in mm

Ll=3000 !span in mm

a=1500 !hoist height in mm

S=(b*h)-(b-2*t)*(h-2*t) !section in mm

ET,1,LINK180 !element type #180: 2D/3D truss (T-C)


MP,PRXY,1,0.3 !Poisson modulus for material #1

R,1,S !real constant #1, element section

!Geometry

K,1,0,0,0 !keypoint, number and coordinates (x,y,z)

K,2,Ll,0,0

K,3,2*Ll,0,0

K,4,Ll/2,a,0

K,5,3*Ll/2,a,0

L,1,2,1 !line, initial and final keypoint and divisions

L,2,3,1

L,3,5,1

L,4,5,1

L,1,4,1

L,2,4,1

L,2,5,1

LMESH,1,7,1 !mesh from line 1 to 7 any i_th line

FINISH


!Boundary and load conditions

DK,1,UX !displacement x for keypoint 1

DK,1,UY

DK,3,UY

FK,2,FY,-P !concentrated y-load keypoint 2, value -P

FK,4,FY,-P

FK,5,FY,-P

/PBC,all !to show BC’s when solve

SOLVE

FINISH

/POST1 !start POST-PROCESSOR MODULE

PLDISP,1 !deformed shape

PRNSOL,U,Y !list y-displacement

ETABLE,STAXIL,LS,1 !obtain axial stress

PLETAB,STAXIL !plot axial stress

PRETAB,STAXIL !list axial stress







Observe the distribution of the axial stress in the different bars and how this, as expected, is

uniform for every particular element. Try to plot the different components of the stress using the

Post-Processor menu. Note that the value for all the components in all the elements is zero. The

only way to obtain the axial stress is by using the ETABLE command. Use the program help for

the LINK180 element and ETABLE command to understand how they work.

2.3 3D bar/truss elements

Example 2.2. The pin-articulated structure shown in Figure 2.2 is used to support a tank con-

taining 1 tone of water. All the structural elements are hollow-circular rods with an exterior

diameter D = 30 mm and thickness t = 2.5 mm. All the rods are made of steel, E = 210 GPa and

ν = 0.3. The geometry of the structure is given by H = 5 m and B = 4 m and L = 6 m. Generate

the input file to model the structure with ANSYSTM . For the simulation consider an horizontal

wind on the tank resulting in two horizontal forces in the upper part of the structure: FB = 5000

N and FL = 4000 N.

Figure 2.2: Pin-articulated structure supporting a tank for fluids.






/TITLE, Structure for fluid tank, 3D truss


!Parameters

B=4000 !base in mm

L=6000 !width in mm

H=5000 !height in mm

t=2.5 !thickness in mm

D=30 !outer diameter in mm

W=10000 !weigth in N

FB=5000 !horitzontal force base in N

FL=4000 !horitzontal force width in N

PI=acos(0)*2 !pi

D1=D-(2*t) !internal diameter

A=(PI*(D*D-D1*D1)/4) !rod area




W1=W/4 !loads per upper node

FB1=FB/4

FL1=FL/4

ET,1,LINK180 !element type #1: 2D/3D truss (T-C)



R,1,A !real constant #1, element section

!Geometry

K,1,0,0,0 !keypoint, number and coordinates

K,2,B,0,0

K,3,0,H,0

K,4,B,H,0

KGEN,2,ALL,,,,,L,4,1 !generate-copy all keypoints, disp. Z=L, increment number=4


L,3,4,1

L,2,4,1

L,1,4,1

L,2,3,1

L,5,7,1

L,7,8,1

L,6,8,1

L,5,8,1

L,6,7,1

L,4,8,1

L,6,4,1

L,2,8,1

L,3,7,1

L,1,7,1

L,3,5,1

LMESH,ALL !mesh all lines

FINISH



NSEL,S,LOC,Y,0 !select nodes located y=0

D,ALL,ALL !fix all DOF of all selected nodes

NSEL,ALL !select all model nodes

NSEL,S,LOC,Y,H !select nodes located y=H

F,ALL,FX,FB1 !apply x-load all selected nodes

F,ALL,FY,-W1

F,ALL,FZ,FL1

NSEL,ALL

SOLVE

FINISH



PRNSOL,U,COMP !list all displacements and sum

ETABLE,STAXIL,LS,1 !obtain axial stress

PLETAB,STAXIL !plot axial stress



Observe that in this case the model has to take into account nodes of the bars can be displaced

in the three directions of the space. Thus, a 3D element is required.





2.4 2D beam elements


sented in Figure 2.3, where P = 100 kN, q = 100 N/mm, b = 10 cm, h = 20 cm, t = 2 mm, L = 3

m, and a = 1.5 m. The material of the beams is linear elastic with E = 210 GPa and ν = 0.3. Use

BEAM elements. Plot the diagram of internal moments of each beam. Compare the results with

those obtained in Ex. 2.1.

Figure 2.3: Overhead hoist with bending loads.






/TITLE, Overhead hoist, 2D beam


!Parameters

P=100000 !applied load in N

q=100 !distribuited load N/mm

b=100 !section base in mm

h=200 !section height in mm

t=2 !section thickness in mm

Ll=3000 !span in mm

a=1500 !hoist height in mm

ET,1,BEAM188 !element type #1: BEAM188 (6 DoF)

KEYOPT,1,3,3 !element #1 keyoption 3=3 (cubic shape function)



SECTYPE,1,BEAM,HREC !Beam hollow rectangular section type for material #1

SECOFFSET,CENT !section centered (offset=0)

SECDATA,h,b,t,t,t,t !section data, with base, height and four lateral thickness

!Geometry

K,1,0,0,0 !keypoint, number and coordinates

K,2,Ll,0,0

K,3,2*Ll,0,0

K,4,Ll/2,a,0

K,5,3*Ll/2,a,0


L,2,3,10

L,3,5,10

L,4,5,10

L,1,4,10




L,2,4,10

L,2,5,10

LMESH,1,7,1 !mesh from line 1 to 7 any i_th line

FINISH



DK,1,UX !displacement x for keypoint 1

DK,1,UY !displacement y for keypoint 1

DK,1,UZ !displacement z for keypoint 1

DK,1,ROTX !rotations x for keypoint 1

DK,1,ROTY !rotations y for keypoint 1

DK,3,UY !displacement y for keypoint 3

DK,3,UZ !displacement z for keypoint 3

DK,3,ROTX !rotations x for keypoint 3

DK,3,ROTY !rotations y for keypoint 3

FK,2,FY,-P !concentrated y-load keypoint 2, value -P

FK,4,FY,-P

FK,5,FY,-P

LSEL,S,,,4 !select line 4 (top line)

ESLL,S !select attached elements to line

SFBEAM,ALL,2,PRES,q !apply pressure on face 2 (y-normal direction)

ESEL,ALL !select all elements


SOLVE

FINISH



PRNSOL,U,Y !list y-displacement

ETABLE,AXI,SMISC,1 !table of axial force FX of node i

ETABLE,AXJ,SMISC,14 !table of axial force FX of node j

PLLS,AXI,AXJ,1,0 !diagram of axial forces FX, scale 1, undeformed

ETABLE,TAI,SMISC,6 !table of section shear force SFY of node i

ETABLE,TAJ,SMISC,19

PLLS,TAI,TAJ,1,0 !diagram of section shear force SFY, scale 1, undeformed

ETABLE,MOI,SMISC,3 !table of bending moment MZ of node i

ETABLE,MOJ,SMISC,16

PLLS,MOI,MOJ,1,0 !diagram of bending moment MZ, scale 1, undeformed

/ESHAPE,1,1 !display real initial shape elements

PLNSOL,S,X



Observe that in this case the elements are subjected to bending moment, as indicated in the

moment diagram. Use the program help for the BEAM188 element and ETABLE command to

understand how they work.

2.5 Plane stress elements

Example 2.4. Generate the ANSYSTM input file to model and simulate the stress concentration

coupon represented in Figure 2.4. The geometry of the coupon, which is symmetric, is given by

H = 40 mm, h = 20 mm, B = 100 mm and b = 40 mm. The thickness of the coupon, t = 2 mm, is

constant. An horizontal displacement is applied on the right vertical edge, as shown in the figure,

where u = 5×10−2 mm. The material of the coupon is steel, E = 210 GPa and ν = 0.3.






Figure 2.4: Stress concentration coupon.





/TITLE, Plane stress coupon


!Parameters

B=100 !base in mm

B1=40 !base triangle in mm

H=40 !width in mm

H1=20 !small widht in mm

t=2 !thickness in mm

u=0.05 !displacement in mm

ET,1,PLANE182 !element type #1: plane solid

KEYOPT,1,1,3 !element #1, keyopt 1=3, simplified enhanced strain formulation

KEYOPT,1,3,3 !element #1, keyopt 3=3, plane stress w/ thickness

R,1,t !real constant #1, element thickness



!Geometry

K,1,0,0 !keypoint, number and coordinates

K,2,(B-B1)/2,0

K,3,B/2,(H-H1)/2

K,4,((B-B1)/2)+B1,0

K,5,B,0

K,6,0,H

K,7,(B-B1)/2,H

K,8,B/2,((H-H1)/2)+H1

K,9,((B-B1)/2)+B1,H

K,10,B,H


L,2,3,10

L,3,4,10

L,4,5,15

L,6,7,15

L,7,8,10

L,8,9,10

L,9,10,15

L,1,6,20

L,5,10,20

L,2,7,20

L,3,8,20

L,4,9,20

AL,1,11,5,9 !define areas by lines

AL,2,12,6,11




AL,3,13,7,12

AL,4,10,8,13

AGLUE,ALL !sum all areas

AMESH,ALL !mesh area

FINISH



NSEL,S,LOC,X,0 !select left nodes

D,ALL,UX !fix x-displacement

NSEL,S,LOC,X,0 !select left bottom node

NSEL,R,LOC,Y,0

D,ALL,UY !fix y-displacement

NSEL,S,LOC,X,B !select right nodes

D,ALL,UX,u !apply displacement

NSEL,ALL !select all nodes


SOLVE

FINISH



PLNSOL,S,EQV !Von Mises stress



Although the thickness of the coupon has been considered in the model (R,1,t), this has

no influence in the results as the displacement is applied. Check the help about the element

PLANE182 to see how the element can be defined without considering the thickness of the

coupon. Note that in this case, if a load was applied instead of a displacement, the load should

be normalized by the thickness of the coupon.

2.6 Plane strain elements

Example 2.5. Generate the ANSYSTM input file to model and simulate the central cross-section

of the rectangular pressure conduct represented in Figure 2.5. Consider that the material of the

conduct is made of an homogeneous, linear and isotropic plastic with E = 27 GPa and ν = 0.25.

The geometry of the element is given by H = 75 mm and B = 50 mm. The thickness of the section

is t = 5 mm. The pressure in the interior of the conduct is p = 10 kg/cm2.

Figure 2.5: Cross-section of the hydraulic dam.










/TITLE, Plane strain square pipe


!Parameters

B=50 !base in mm

H=75 !height in mm


p=1 !pressure in N/mm2


KEYOPT,1,1,3 !element #1, keyopt 1=3, simplified enhanced strain formulation

KEYOPT,1,3,2 !element #1, keyopt 3=3, plane strain



!Geometry


K,2,B,0

K,3,B,H

K,4,0,H

K,5,t,t

K,6,B-t,t

K,7,B-t,H-t

K,8,t,H-t


L,2,3,30

L,3,4,20

L,4,1,30

L,5,6,20

L,6,7,30

L,7,8,20

L,8,5,30

L,1,5,4

L,2,6,4

L,3,7,4

L,4,8,4


AL,2,6,10,11

AL,3,7,11,12

AL,4,8,9,12

AGLUE,ALL !sum all areas


FINISH




NSEL,A,LOC,Y,0 !add bottom nodes to selection

D,ALL,ALL !fix displacement

LSEL,S,LINE,,5,8 !select lines 5 to 8

NSLL,S,1 !select all nodes on lines

SF,ALL,PRES,p !apply pressure on all selected nodes

NSEL,ALL

/PBC,ALL !to show BC’s when solve

SOLVE

FINISH









Although the length of the conduct is not known, this parameter is not required for the simu-

lation as plane strain condition is considered.

2.7 Shell elements

Example 2.6. Write the ANSYSTM or ABAQUSTM input file to model a simply supported rectan-

gular plate with dimensions L = B = 1 m and thickness t = 10 mm. The material of the plate is

steel and it is loaded with a uniform pressure p = 0.1 N/mm2. Obtain the stress distribution on

the x-direction at bottom, middle and top locations in the thickness direction.






/TITLE, 3-D simply supported shell plate


!Parameters

B=1000 !base in mm

L=1000 !width in mm


p=0.1 !pressure in N/mm2

ET,1,SHELL181 !element type #1: shell

KEYOPT,1,3,2 !element #1: keyoption 3=2 (full integration incompatible modes)

KEYOPT,1,8,2 !element #1, keyopt 8=2, store data top, bottom and mid surfaces

SECTYPE,1,SHELL !section #1 = shell

SECDATA,t !thickness



!Geometry


K,2,B,0

K,3,0,L

K,4,B,L


L,1,3,40

L,3,4,40

L,2,4,40

AL,ALL !define areas by lines


FINISH



NSEL,S,LOC,X,B/2 !select central node

NSEL,R,LOC,Y,L/2

D,ALL,UX,,,,,UY,ROTX,ROTY,ROTZ !fix DOF’s

NSEL,S,LOC,X,0

NSEL,R,LOC,Y,L/2

D,ALL,UY

NSEL,S,LOC,X,0

D,ALL,UZ

NSEL,S,LOC,X,B

D,ALL,UZ

NSEL,S,LOC,Y,0





D,ALL,UZ

NSEL,S,LOC,Y,L

D,ALL,UZ

NSEL,ALL

SFE,ALL,,PRES,,p !apply pressure on all elements

/PBC,ALL !to show BC’s when solve

SOLVE

FINISH



PLNSOL,S,X !x-stress in top and bottom surfaces

SHELL,MID !select results on mid surface (TOP,MID,BOT)

PLNSOL,S,X !s-stress in midle surface



Observe that the value of the stress in the x-direction varies from the top to the mid and

bottom surfaces. This is due to the fact that although a shell element corresponds to a planar

representation, it is a 3D element and takes into account the thickness of the system.

2.8 Suggested problems

Problem 2.1. Generate the ANSYSTM input file for the simulation of the beam represented in

Figure 1.2 using the TRUSS, BEAM, PLANE STRESS and SHELL element types. Analyze the

differences and limitations in the use of the element types listed above and the solid elements

used in example 1.1.

Problem 2.2. Modify the input file generated in the previous problem for shell elements and

the input file in Ex. 1.3 for solid elements and simulate different ratios of H/L (height and span

of the beam, respectively). Analyze for which values of the ratio, the results obtained with shell

elements are similar to those obtained with solid elements.

Problem 2.3. Modify the input file generated in Ex. 2.2 to simulate the structure in Figure 2.2

without the inclined bars. Consider different types of elements and boundary conditions and use

the most appropriate.

Problem 2.4. Modify the input file generated in Ex. 2.4 to obtain the variation of the maximum

Von Mises stress versus the size of the finite element. Redefine the mesh to consider 10, 20, 40,

60, 80 and 100 elements in the vertical direction and plot the obtained value of stress versus the

number of divisions.




Chapter 3

Geometric modeling and meshing

techniques

3.1 Introduction

The goal of the meshing process is to convert a problem in the continuum real into a problem

which can be solved by a computer and so, a discrete problem. This process involves two different

aspects:

• The discretization of the geometry. The mesh must be fine enough to avoid sharp corners

in curves or curved surfaces which may lead to unreal stress concentrations.

• The discretization of the solution. The result is obtained by the FE code only in a discrete

set of locations so the mesh should also ensure that the number of solution points is enough

to express the variation of the solution fields along the domain.

On the other hand, a coarse mesh needs less computational resources than a fine mesh, so

the user generally should establish an equilibrium position, that is, to obtain a mesh fine enough

to discretize the geometry and the solution and coarse enough for the computational resources

available.

The meshing process creates elements and nodes. The elements are the portions (lines, sur-

faces or volumes) in which the continuum is decomposed. These elements are delimited by nodes

which are those locations in which the solution (the variables corresponding to the degrees of

freedom such as displacements, rotations and temperatures) of the system of equations is ob-

tained. From the degrees of freedom, other relevant magnitudes which cannot be expressed in a

point may be obtained and expressed in the elements. These are generally tensorial magnitudes

(which need a volume face to be expressed) or fluxes (which need an area to be computed and

expressed). However, this magnitudes are computed by integrating the shape function of the

element at the integration points of the element which guarantee a minimum error numerical

integration.

In this chapter some general mesh strategies and some rule-of-the-thumb recommendations

are given.

Meshing strategies

There exist different methods to model and mesh the geometry of a structure. They can be

classified in two categories: direct generation of mesh and solid modeling.

21


Hint 2: Direct generation vs Solid modeling

Generally, commercial FE codes allow the direct definition of finite element entities (nodes,

elements, and boundary conditions on nodes and elements) or the definition of geometric en-

tities (points, lines, surfaces, or volumes) which are meshed later in order to obtain the nodes

and elements.

Considerations

Direct generation Convenient for simple models

Direct control of nodes and elements

Program skills are needed

Allows the translation of the model between different commer-

cial codes or definition using standard codes

Solid modeling Convenient for large models

Requires larger computational resources

It is easy to modify the geometry

No direct control over the mesh entities

Meshing recommendations

• The mesh should ensure connectivity through the nodes to guarantee a correct load distri-

bution, that is, each element must share some nodes with its neighboring elements and all

nodes, except from those nodes located at corners of the global geometry, must belong to

at least two elements.

• Curves should be meshed with, at least, 12 elements for each circumference quarter to

guarantee a smooth geometric discretization.

• Commercial FE codes generally provide re-meshing tools which in an iterative process re-

fine the mesh in those places with large gradients of the solution.

• Elements in a mesh may be triangles (or tetrahedral) or quadrilateral (or bricks). Generally,

triangular elements provide constant values for the integrated variables (stress) but can

be used to mesh any geometry easily. On the other hand, quadrilateral elements offer a

slightly better approximation of the distribution of the integrated variables through the

element, although they may not be able to mesh some geometries accurately. Except for

those cases with complicated geometries quadrilaterals are preferable.

• There are some element types with internal nodes which improve the discretization of the

solution and are useful when large gradients are present. These are elements such as

9-node quadrilaterals, 12-node bricks, etc.

3.2 Direct generation

The process of creating elements out of nothing, without starting from any geometric entity is

called direct generation or direct mesh. In this process the user imposes explicitly the location

of the nodes and the size of the elements. In this process the user has a total control on the node

location and element size. This approach is generally used for simple geometries.



Chapter 3. Geometric modeling and meshing techniques 23

Direct generation consists on defining first the node distribution in the geometry. The position

of each node is defined according to a coordinate system. Some commercial codes allow to use

different types of coordinate systems (cartesian, cylindrical, spherical, as so on).

Hint 3: Useful ANSYS commands to directly define nodes

Command Main Parameters

Define individual node N x,y,z,thxy,thyz,thxz

Generate nodes from a pattern of nodes NGEN ntime,inc,n1,n2,ninc,dx,dy,dz

Generate line nodes between two nodes FILL node1,node2,nfill

Generate a reflected set of nodes NSYM ncomp,inc,node1,node2,ninc

List nodes NLIST node1,node2,ninc

Display nodes NPLOT

Delete nodes NDELE node1,node2,ninc

Once the nodes have been generated, the elements can be defined. The element is defined

using the connectivity with its nodes. The set of elements defined should define the whole geom-

etry.

Hint 4: Useful ANSYS commands to define elements


Define individual element E i,j,k,l,m,n,o,p

Generate elements from a pattern EGEN ntime,ninc,elem1,elem2,elinc,

Generate elements by reflection ESYM ,--,ninc,elem1,elem2,elinc

Modify element attributes EMODIF iel,stloc,i1,i2,i3,i4,...

List elements ELIST elem1,elem2,elinc

Display elements EPLOT

Delete elements EDELE elem1,elem2,elinc

The following examples show the use of these different commands to define nodes and ele-

ments.

Example 3.1. Let us model the bi-pinned structure shown in Figure 3.1. A vertical concentrated

force F = 10 kN is applied in the horizontal span, as shown. The beam has a squared cross section

of 10×10 mm, and is made of wood with E = 3000 GPa. Use the direct generation method in

ANSYSTM to define the structure with beam elements.

Solution to Example 3.1. Beam elements of length 20 mm have been used to model the struc-

ture. For this reason, node number 1 has been created at point A, node number 11 where the

concentrated force is applied, point B. Thus, between these two nodes 10 elements of length 20

mm have been defined. Node number 16 has been created at point C and node 20 at the bottom

end, point D. Nodes #1, #11, #16 and #20 have been defined with the N command. The distances

between these nodes have been filled using the FILL command, obtaining nodes every 20 mm

(see Figure 3.2).

A consecutive node numeration has been obtained. Therefore, the elements can be defined

easily with E and EGEN commands.




8 cm

F

Figure 3.1: Bi-pinned structure.

The ANSYSTM command sequence for this example is listed below. You can either type these

commands on the command window, or you can type them on a file, then, on the command

window enter /input, file, ext or just use copy and paste.



/TITLE, Bi-pinned structure using BEAM188 element

!Parameters

L1=200 !first horizontal distance (mm)

L2=300 !second horizontal distance (mm)

L3=80 !vertical distance (mm)

b=10 !element width (mm)

h=10 !element height (mm)




SECTYPE,1,BEAM,RECT !section #1 = rectangular beam

SECOFFSET,CENT !centered section (offset=0)

SECDATA,h,b !section width and height

MP,EX,1,3000e3 !material #1: E=300e3 MPa, if Poisson ratio is not defined 0.3 is assumed

N,1 !define node #1 on coordinates (0,0)

N,11,L1 !define node #11 on coordinates (L1,0)

N,16,L2 !define node #16 on coordinates (L2,0)

N,20,L2,-L3 !define node #20 on coordinates (L2,-L3)

FILL,1,11 !fill nodes between #1 and #11



E,1,2 !define element by nodes #1, #2

EGEN,19,1,1 !generate 18 elements incrementing 1 node like element 1 (total 19 elements)

FINISH !finish PRE-PROCESSOR MODULE



Before solving the model it is necessary to define the pins at both ends (nodes #1 and #20)

and apply the vertical force in node #11.

Example 3.2. Model using shell elements the 80 mm thick steel bending plate shown in Figure

3.3 . The plate has a trapezoidal shape with two right angles in B and C and is subjected to a

bending load. Note that there is a simple support at 1000 mm from the clamped edge.

Solution to Example 3.2. With N and FILL commands 21 rows of 13 nodes are defined to com-

plete the whole geometry (see Figure 3.4). Between each row there is an increment in numeration





Figure 3.2: Nodes and beam elements for the bi-pinned structure.

0.9kN

1000

1500

1000800

A

B

C

90º

90º

Figure 3.3: Schema of the bending plate (dimensions in mm).

of 1, meanwhile between each column the increment in numeration is 100. With this regular dis-

tribution of the node numbers it is easy to define all elements only using the EGEN command twice.

The simple support can be defined in the central row of nodes.

Figure 3.4: Finite element model of bending plate.




FINISH !close all previous models


/TITLE,Clamped Plate under Bending


ET,1,SHELL181 !element type #1: SHELL181 (6 DoF)






SECDATA,8 !thickness

MP,EX,1,210e3 !material #1: Young modulus E=210e3 MPa

MP,PRXY,1,0.3 !material #1: Poisson coeffient=0.3

N,1 !define node #1 on coordinates (0,0)

N,21,,2000 !define node #21 on coordinates (0,2000) mm

N,1201,1500 !define node #1201 on coordinates (1500,0) mm

N,1221,800,2000 !define node #1221 on coordinates (800,2000) mm



FILL,1,1201,11,,100,21,1 !use command fill to generate rest of nodes

E,1,101,102,2 !define element #1

EGEN,20,1,1 !generate the first element column from element #1

EGEN,12,100,1,20 !generate the rest of element columns




To generate more complex geometries with the direct method it is possible to change between

different coordinate systems and also create new coordinate systems defined as "local".

Hint 5: Useful ANSYS commands to manage and define coordinate systems


Active the coordinate systems CSYS ncsys: 0 - cartesian, 1 - cylindri-

cal, 2 - spherical,...

Define local coordinate system LOCAL newcsys,ncsys,xc,yc,zc,thxy,...

The following example uses different coordinate systems to define the geometry of a curved

beam.

Example 3.3. Model the geometry shown in Figure 3.5 using 2-D solids and the direct generation

of nodes and elements. The element thickness is 4 mm and the material used is stainless steel

(with a Young’s modulus of 190 GPa and a Poisson coefficient of 0.3).

Figure 3.5: Geometry of a curved beam (dimensions in mm).





Solution to Example 3.3. The curved part of the beam is defined using a cylindrical (polar)

coordinate system (where the first coordinate corresponds to the radius R and the second to the

angle θ). The rectangular part of the beam is defined with a cartesian coordinate system.

In this case the FILL command has been used with a tailored ratio aiming to increase the

density of the mesh in the critical part (see Figure 3.6).

Figure 3.6: Nodes to define the mesh and its numeration.






/TITLE, Curved Beam - Direct generation


ET,1,PLANE182 !element type #1: PLANE182 (2 DoF)

KEYOPT,1,1,3 !element type #1: keyoption 1=3 (simplified enhanced strain formulation)

KEYOPT,1,3,3 !element type #1: keyoption 3=3 (plane stress w/thickness)

R,1,4 !real constant set #1: th=4 mm

MP,EX,1,190e3 !material #1: E=190000 MPa, if Poisson ratio is not defined 0.3 is assumed

CSYS,1 !activate cylindrical coordinate system

N,1,25,180 !define node #1 on coordinates R=25 mm and Angle=180º

N,15,25,90 !define node #15 on coordinates R=25 mm and Angle=90º

FILL,1,15,,,,,,3 !fill nodes between #1 and #15, last spacing 3 times larger than the initial one

NGEN,9,20,1,15,1,3 !generate nodes repeating the row: 8 new rows, increasing the radius

!a distance of 3 mm in each row

CSYS,0 !activate cartesian coordinate system



NGEN,9,20,16,20,1,,3 !generate nodes repeating the row: 8 new rows, increasing the Y position

!a distance of 3 mm in each row

E,1,2,22,21 !define element by nodes #1, #2, #22, and #21

EGEN,19,1,1 !generates 19 elements of first row

EGEN,8,20,1,19 !generates rest of elements repeating the row








Models can include different types of elements, real constants sets, material behaviours, etc.

The ANSYS commands ET, R and MP allow to define database tables with the different attributes of

the elements used in the model. Each table uses reference numbers that associate the attribute

with the element, each element keep these kind of information with the nodal connectivity.

As it is shown in the Figure 3.7, the ANSYS commands TYPE, REAL and MAT define the element

type, real constant set and material, respectively. The elements generated after this selection

will have the previous selected attributes. However, the change of this attributes a posteriori is

also possible using the command EMODIF.

{

Element Types

ET,1,BEAM188

ET,4,MASS21

ET,3,SOLID185

ET,2,SHELL181

ET,5, ...Real constants

R,1,A,Izz,h

R,4,mass

R,3,th2

R,2,th1

R,5,...Materials

MP,EX,1,210e3

MP,PRXY,2,0.29

MP,EX,2,70e3

MP,PRXY,1,0.3

MP,DENS,2,2.7

MAT,2

REAL,3

TYPE,2

Figure 3.7: Element attributes selection.


sented in Figure 3.8, where P = 100 kN, q = 100 N/mm, b = 10 cm, h = 20 cm, t = 2 mm, L = 3

m, and a = 1.5 m. The material of the beams is linear elastic with E = 210 GPa and ν = 0.3. Use

BEAM elements on the top element and TRUSS elements for the rest. Compare the results with

those obtained in 2.3.

Figure 3.8: Overhead hoist with bending loads.

Solution to Example 3.4. Two element types are defined (BEAM188 and LINK180). A section set

is defined associated to BEAM188, to compute the section Area and Inertia. A real constant set is




also defined associated to LINK180 defining the Area.

The commands TYPE and REAL allow selecting the element type and the constant set for the

elements defined next. With the ELIST command a list of defined elements are shown and the

assigned type element and constant set can be checked.


commands on the command window, or you can type them on a file, then, on the command win-

dow enter /input, file, ext.



/TITLE, Overhead hoist - Including beam and truss elements


!Parameters

P=100000 !applied load

q=100 !distributed load

b=100 !section base

h=200 !section height

t=2 !section thickness

Ll=3000 !span

A=1500 !hoist height

S=(b*h)-(b-2*t)*(h-2*t) !cross section area

ET,1,BEAM188 !element type #1: 3D beam (6 DoF)


SECTYPE,1,BEAM,HRECT !section #1 = hollow rectangular beam


SECDATA,h,b,t,t,t,t !section width, height and thicknesses

ET,2,LINK180 !element type #2: 3D truss (3 DoF)

R,2,S !geometric properties section #2: for truss elements

MP,EX,1,210000 !material #1: Young’s modulus

MP,PRXY,1,0.3 !material #1: Poisson modulus

N,1,0,0,0 !define node #1 on coordinates (0,0,0)

N,2,Ll,0,0 !define rest of nodes

N,3,2*Ll,0,0

N,4,Ll/2,A,0

N,14,3*Ll/2,A,0

FILL,4,14

TYPE,2 !select TRUSS element type #2

REAL,2 !select real constant set #2

E,1,2 !define TRUSS elements

E,2,3

E,3,14

E,1,4

E,2,4

E,2,14

TYPE,1 !select BEAM element type #1

E,4,5 !define BEAM elements

EGEN,10,1,7



ANTYPE,STATIC

D,1,ALL !displacement X, Y and Z for node #1

D,3,UY,,,,,UZ !displacement Y and Z for node #3

D,9,UZ !displacement Z for node #9

D,4,ROTX,,,,,ROTY !rotations X and Y for node #4

D,14,ROTX,,,,,ROTY !rotations X and Y for node #14

F,2,FY,-P !concentrated Y-load nodes #2, #4 and #14

F,4,FY,-P

F,14,FY,-P




ESEL,S,TYPE,,1 !select BEAM elements (type #1)

SFBEAM,ALL,2,PRES,q !apply distributed load on element

ESEL,ALL !select all elements

/PBC,ALL !show BC’s when solve

SOLVE !solve current load state

FINISH !finish SOLUTION MODULE


PLDISP,1 !plot deformed shape

/ESHAPE,1,1 !plot element with real shape

PLNSOL,S,X !contour plot of stress X distribution



Note that both BEAM188 and LINK180 are 3D elements and as we are considering a 2D

analysis some displacements and rotations must be constrained to avoid displacements and de-

formations in the Z-direction.

Example 3.5. Model the beam shown in Figure 3.9 using 2D solid elements. The beam is com-

posed by a stainless steel bar reinforced with two aluminium plates perfectly bonded.

80

100

8 N/mm2

Figure 3.9: Two materials beam example (dimensions in mm).

Solution to Example 3.5. The PLANE182 element can be used to define a 2D solid with differ-

ent thicknesses, using KEYOPTION number 3 (plain stress with thickness). Two thicknesses (16

mm thickness constant set #1 and 10 mm thickness constant set #2) and two materials (Stain

Steel material #1, and Aluminium material #2) are defined. Before defining the elements with

REAL and MAT commands the thickness and the material is selected.






/TITLE, Two materials beam





R,1,16 !real constants set #1: th=16 mm,

R,2,10 !real constants set #2: th=10 mm,

MP,EX,1,190000 !material #1: E=190000 MPa

MP,PRXY,1,0.29 !material #1: Poisson=0.29










NGEN,5,200,1,101,1,,1.25 !generate 5 rows of nodes with y=+1.25 mm each row respect to pattern

NGEN,3,200,811,891,1,,1 !generate 3 rows of nodes with y=+1 mm each row respect to pattern

NSYM,Y,2000,ALL !generates nodes by symmetry to plane normal to Y axis

REAL,1 !select constant set #1

MAT,1 !select material #1

E,1,2,202,201 !define element #1

EGEN,100,1,1 !generate a row of elements

EGEN,4,200,1,100,1 !generete 3 more rows of elements

REAL,2 !select constant set #2

MAT,2 !select material #2

E,811,812,1012,1011 !define element #301

EGEN,80,1,401 !generate a row of elements

EGEN,2,200,401,480 !generate 1 more row of elements

ENSYM,2000,,2000,ALL !generate elements by symmetry

NUMMRG,NODES !merge coincident nodes



ANTYPE,STATIC !select Static analysis (default option)

D,1,ALL !articulation on node #1

D,101,UY !constrain vertically node #101

NSEL,S,LOC,X,20,80 !select nodes of elements to apply surface load

NSEL,R,LOC,Y,5.75,7.25

ESLN,S,1 !select elements attached to the selected nodes

SFE,ALL,3,PRES,,8 !apply surface load on elements: q=8 N/mm2

ALLSEL !select everythhing

/PBC,ALL




PLDISP,2 !plot displaced model




Direct generation commands can be also used when the geometry and the mesh can be de-

fined with mathematical functions. With the help of a programming language (Python, FORTRAN,

C/C++, MatLab, etc.) the functions can be used to define the nodes position, the element con-

nectivity and the applied loads. After using simple commands (N, E, D, F, etc.) the whole model

can be generated. The definition of a model using a MatLab code is shown in Example 3.6.

Example 3.6. Model a sculpture that represents a Moebius tape (see Fig. 3.10). The mid radius

R of the structure is 1325 mm. The length B of the 48 cross beams is 700 mm. The structure is

made of steel (E = 210000 MPa and ν = 0.29), the perimetric beam is a 40 mm diameter circular

rod and the cross beams are plates of 10 × 60 mm2. The cross beams that support and fix the

sculpture to the floor have a cross-section of 25 × 130 mm2.

Solution to Example 3.6. The position of the nodes and the forces due to wind loads are calcu-

lated using a MATLAB code. This code generates two files: ’nodes.txt’ which contents the nodal

positions, and ’wind.txt’ that contents the nodal forces due to wind.

% PrePost solution of Example 3.06

% J.A. Mayugo, 2007

% J. Renart, 2010





B

R

B

48 cross

beams

perimetral

beam

Figure 3.10: Geometry of the Moebius tape sculpture.

clear all;close all;clc; %clear all figures, data and screen

number=[1:48]; %number of cross beams

angle=[0:360/48:360*(1-1/48)]; %angle increment

c_angle=cos(angle/180*pi); %cos(angle)

s_angle=sin(angle/180*pi); %sin(angle)

rotation=[45:180/48:225-180/48]; %rotation_ini=45 degrees

c_rotation=cos(rotation/180*pi); %cos(rotation)

s_rotation=sin(rotation/180*pi); %sin(rotation)

radius=1325; %mid radius

width=700; %Moebius width

width_xy=width*c_rotation; %width in 3D

radius_mid_X=radius.*c_angle; %Moebius mid radius X component

radius_mid_Y=radius.*s_angle; %Moebius mid radius Y component

%cilindrical coordinates to define the bars of the Moebius tape

radius_bot_X=radius_mid_X+0.5*width_xy.*c_angle;

radius_bot_Y=radius_mid_Y+0.5*width_xy.*s_angle;

radius_bot_Z=width*s_rotation;

radius_top_X=radius_mid_X-0.5*width_xy.*c_angle;

radius_top_Y=radius_mid_Y-0.5*width_xy.*s_angle;

radius_top_Z=-width*s_rotation;

hold on;

plot3(radius_bot_X,radius_bot_Y,radius_bot_Z,’r’) %plot the external face of the Moebius tape

plot3(radius_top_X,radius_top_Y,radius_top_Z) %plot the internal face of the Moebius tape

for i=1:length(number);

plot3([radius_bot_X(i) radius_top_X(i)],[radius_bot_Y(i) radius_top_Y(i)],... %Moebius tape bars

[radius_bot_Z(i) radius_top_Z(i)],’k’);

plot(radius_mid_X(i),radius_mid_Y(i),’ks’); %Moebius tape load points

end

hold off

%write the results in the file nodes.txt

fid0=fopen([’nodes.txt’],’w’);

for inum=1:48;

fprintf(fid0,’n,%i,%8g,%8g,%8g \n’,inum,radius_mid_X(inum),radius_mid_Y(inum),0);

fprintf(fid0,’n,%i,%8g,%8g,%8g \n’,inum+100,radius_bot_X(inum),radius_bot_Y(inum),radius_bot_Z(inum));

fprintf(fid0,’n,%i,%8g,%8g,%8g \n’,inum+200,radius_top_X(inum),radius_top_Y(inum),radius_top_Z(inum));

end;

fclose(fid0);




wind_Kmh=150; %wind [Km/h)

wind_ms=wind_Kmh/3.6; %wind [m/s]

pressure=wind_ms^2/16; %pressure on the bars [kg/m^2]

pressure=pressure*9.81/1e6; %pressure on the bars [N/mm^2]

C=0.88; %pressure coefficient

p=C*pressure;

Area_plate=abs(width_xy)*140; %area on which de pressure is applied (plate area)

F_plate=p*Area_plate; %force applied on the plate

%write the results in the file wind.txt

fid1=fopen([’wind.txt’],’w’);

for inum=1:48;

fprintf(fid1,’f,%i,fz,%8g \n’,inum,F_plate(inum));

end;

fclose(fid1);


ftp://amade.udg.edu/mms/PrePostFE/input_files/T3/PrePostFE_Ex306.m



dow enter /input, file, ext. The files ’nodes.txt’ and ’wind.txt’ previously generated using the

MatLabTM code should be placed in the working directory.

FINISH !close all revious modules


/TITLE,Moebius Sculpture




SECTYPE,1,BEAM,CSOLID !section #1: preimetral beams, solid cylindrical


SECDATA,40 !section radius

SECTYPE,2,BEAM,RECT !section #2: internal beams, rectangular


SECDATA,60,10 !section width and height

SECTYPE,3,BEAM,RECT !section #3: bottom internal beams, rectangular




MP,PRXY,1,0.29 !material #1: 0.29 Poisson

MP,DENS,1,8e-6 !material #1: density=8 T/m^3

/INPUT,nodes,txt !input nodes from file: ’nodes.txt’

SECNUM,1 !section set #1

E,101,102 !elements to generate the perimetral circles

EGEN,47,1,1

E,148,201

E,201,202

EGEN,47,1,49

E,248,101 !define first 96 elements


E,101,1,248 !define transversal elements

E,1,201,148

E,102,2,101

E,2,202,201

EGEN,47,1,99,100

ESEL,S,,,167,172 !redefine section bottom elements

EMODIF,ALL,SECNUM,3





ALLSEL !select everything



D,236,ALL !fix sculpture to floor

D,237,ALL

D,238,ALL

ACEL,,,10 !gravity load

/PBC,ALL

SOLVE

FINISH



3.3 Solid Modeling

The mesh may also be created starting from geometric entities (which can be created in the

pre-processor or imported from other CAD software) and the software creates the mesh. The

user, however, may impose some restrictions or conditions to the meshing process. Some of this

restrictions are:

• Imposing a number of divisions in edges or lines of the primitive geometry.

• Giving some values for the element size, in terms on edge length.

• Starting the mesh from lower order geometric entities and extrude or sweep the created

elements to fill higher order geometries. For instance, to mesh a regular volume some of the

areas may be meshed first and then mesh the whole model following the pattern imposed

in the areas.





Controlling the element size

Hint 6: Useful ANSYS commands for solid modelling


Define individual keypoint K npt,x,y,z

Generate line from keypoints L p1,p2,ndiv

Generate area from keypoints A p1,p2,p3,...,p18

Generate area from lines AL l1,l2,l3,...,l10

Generate volume from keypoints V p1,p2,p3,...,

Generate volume from areas VA na1,na2,na3,...,na10

Generate circular arc lines CIRCLE pcent,rad,...

Generate circular arc lines LARC p1,p2,pc,rad

Generate surface by rotating lines AROTAT nl1,...,nl6,

pax1,pax2,arc,nseg

Generate cylindrical volume CYL4 xc,yc,rad1,theta1,rad2,

theta2,depth

Generate sphere SPH4 xc,yc,rad1,rad2

Create keypoint between two keypoints KBETW kp1,kp2,kpnew,...

Add areas to create single area AADD na1,na2,...,na9

Add volumes to create single volume VADD nv1,nv2,...,nv9

Intersect volume with line LINV nl,nv

Boolean operation options BOPT (See ANSYS Help)

Mesh a volume by sweeping a meshed

area

VSWEEP vnum,srca,trga,lsmo

Mesh a volume by sweeping a meshed

area along a path

VDRAG vdrag,na1,na2,...,na6,

nlp1,...,nlp6

Hint 7: Useful ANSYS commands which affect the element size


Control de general size of elements ESIZE size

Control element size on lines LESIZE nline,size,ndiv,

Control element size on areas AESIZE elem1,elem2,einc




Hint 8: Useful ANSYS commands to mesh entities


Mesh lines LMESH nline1,nline2,nlinc

Mesh areas AMESH narea1,narea2,nainc

Mesh volumes VMESH nvolu1,nvolu2,nvinc

Unmesh lines LCLEAR nline1,nline2,nlinc

Unmesh areas ACLEAR narea1,narea2,nainc

Unmesh volumes VCLEAR nvolu1,nvolu2,nvinc

Mesh type MSHKEY key: 0 - free mesh, 1 - mapped

mesh, 2 - mapped mesh if possi-

ble

Refine mesh around selected nodes EREFINE ne1,ne2,nicn,level,depth,...

Shape of elements MSHAPE key,dimension

Example 3.7. Mesh the geometry considered in Example 3.3 using solid generation and meshing

tools.






/TITLE, Curved Beam - Solid Modeling





R,1,4 !real constant set #1: th=4 mm

MP,EX,1,190e3 !material #1: E=190000 MPa, if Poisson ratio is not defined 0.3 is assumed

PCIRC,25,(25+24),90,180 !define area #1: circle inner radius R=25 mm outer radius R=49 mm

RECTNG,0,30,25,(25+24) !define area #2: rectangle 30 mm x 24 mm

AGLUE,ALL !glue both areas

LESIZE,ALL,,,8 !8 divisions per line

AMESH,ALL !mesh areas




Example 3.8. Model a wing section using solid generation and meshing tools.






/TITLE, Model extruded wing profile











MP,PRXY,1,0.3 !material #1: Poisson coefficient=0.3

MP,DENS,1,2.7e-3 !material #1: density=2.7e3 kg/m3

K,1,0,0,0 !define keypoint #1

K,2,2000,0,0

K,3,2300,200,0

K,4,1900,450,0

K,5,1000,250,0

LSTR,1,2 !define straight line between kp #1 and #2

LSTR,5,1 !defins straight line between kp #5 and #1

BSPLIN,2,3,4,5,,,-1,0,0,-1,-0.25,0 !define spline line using point #2 to #5

AL,1,2,3 !define area #1 from lines #1, #2, and #3

VOFFST,1,5000 !extrude area #1 to define volume, offset distance 5 m

ESIZE,100 !element size 0.1 m

AMESH,3,5,1 !mesh areas from #3 to #5




Example 3.9. Model the metallic plate part of a robot arm shown in 3.11.

Figure 3.11: Metallic plate as a component of a robot arm (dimensions in mm).






/TITLE, Model robot arm part (Mesh method A)








PCIRC,20,55,90,270 !generate semicircle Ri,Ro,ang1,ang2

LOCAL,11,1,320 !define new cartesian CSYS #11

WPCSYS,1,11 !put WP on CSYS #11

PCIRC,16,40,-90,90 !generate semicircle


PTXY,0,-55,320,-40,320,40,0,55 !define coordinate pairs x,y for polygon





POLY !define new polygon (area #3) with previous coordinates

PCIRC,20 !generate circle

WPCSYS,1,11 !put WP on CSYS 11


ASBA,3,4 !substract area #4 to area #3


AGLUE,ALL !glue areas (merge coincident kp and lines)

ESIZE,6 !defines size of element




ftp://amade.udg.edu/mms/PrePostFE/input_files/T3/PrePostFE_Ex309a.dat

This other input file improves the control of the mesh for the modelled metallic plate enhanc-

ing its final quality.



/TITLE, Model robot arm part (Mesh method B)








PCIRC,20,55,90,270 !generate semicircle Ri,Ro,ang1,ang2

LOCAL,11,1,320 !define new cartesian CSYS #11


PCIRC,16,40,-90,90 !generate semicircle


PTXY,0,0,320,0,320,40,0,55 !define coordinate pairs x,y for polygon


PTXY,0,-55,320,-40,320,0,0,0 !define coordinate pairs x,y for polygon



WPCSYS,1,11 !put WP on CSYS 11


WPCSYS,1,0

ASBA,3,5,,,KEEP !substract area #5 to area #3

ASBA,7,6,,,KEEP !substract area #6 to area #7



BLC4,55,-55,(320-95),110 !define rectangle (area #5) to define line #19 and #16

ASBL,4,19 !divide area #4 by line #19




ADELE,5,,,1 !delete area #5 and bellow entities

AGLUE,ALL !glue areas (merge coincident kp and lines)

LESIZE,ALL,,,6 !define number of divisions in lines

LESIZE,1,,,24,,1 !modify number of division in line

LESIZE,3,,,24,,1

LESIZE,5,,,24,,1

LESIZE,7,,,24,,1

LESIZE,12,,,24,,1

LESIZE,13,,,24,,1

LESIZE,33,,,24,,1

LESIZE,16,,,12,,1





LESIZE,19,,,12,,1

LESIZE,35,,,12,,1

LESIZE,37,,,12,,1

!LCCAT,18,28 !concatenate lines to improve mesh (uncomment it to test)

!LCCAT,14,25

!LCCAT,11,32

!LCCAT,29,36




ftp://amade.udg.edu/mms/PrePostFE/input_files/T3/PrePostFE_Ex309b.dat

Example 3.10. Model a handle shown in the Figure 3.12 using the solid generation and meshing

tools.

Figure 3.12: Handle.






/TITLE, Static model for a handle



KEYOPT,1,2,1 !element #1: keyoption 2=1 (full integration)



LOCAL,11,,,,,,-18,-22.5 !define local cartesian system to define central part

CSYS,11 !active previous defined CSYS

K,1,10,-5 !define central part

K,2,10,5

K,3,-10,5

K,4,-10,-5

K,5,7.5,-5,153

K,6,7.5,5,153

K,7,-7.5,5,153

K,8,-7.5,-5,153

V,1,2,3,4,5,6,7,8 !define volume #1 for central part

WPLANE,1,,,,1,,,1,,-1 !change WP position

CYLIND,,30,-20,20 !define cylindre volume #2

WPLANE,1,-60,,140,,,140,,,120 !change WP position

CYLIND,,15,35,50 !define cylindre volume #3





VSBV,1,2 !substract volume #2 to volume #1 (volume #1 becomes #4)

VSBV,4,3 !substract volume #3 to volume #4 (volume #4 becomes #1)

WPLANE,1,,,,1,,,1,,-1 !change WP position

CYLIND,15,30,,20 !redefine cylindre volume #2

WPLANE,1,-60,,140,,,140,,,120 !change WP position

CYLIND,7.5,15,35,50 !redefine cylindre volume #3

VGLUE,ALL !glue volumes

ESIZE,6 !define size of elements

MSHKEY,0 !not mapped mesh

MSHAPE,1,3D !mesh with tetrahedral elements

VMESH,ALL !mesh all volumes




3.4 Assessment of the mesh quality

To be confident on the results of a finite element modelization of a mechanical or structural

problem, it is necessary to assess the mesh quality, that is, to check if the mesh is fine enough to

get good simulation results. Basically, the following aspects need to be analyzed:

• Assure the right connectivity between different parts of the mesh.

• Avoid elements with high distortions, warping and inadmissible narrow or wide angles be-

tween edges.

• Error estimation of the elements, typically by means of energy hypotheses (see Chapter 6).

• Study the sensitivity of the results on different meshes of the entire FE model or of a partial

model (see Chapter 6).

3.4.1 Connectivity

The connectivity of the different parts of the mesh should be verified. The mesh edges between

different parts of the geometry must assure that the generated elements share nodes. The re-

sults of a bad connectivity definition are presented in Figure 3.13, while the correct solution is

presented in Figure 3.14.

Example 3.11. Write the input files for the simulation of the geometry shown in Figures 3.13

and 3.14 with a bad and a good connectivity of the elements.

Figure 3.13: Example of bad connectivity of different parts of the mesh. Undeformed mesh (left)

and deformed mesh (right).





Figure 3.14: Example of right connectivity of different parts of the mesh. Undeformed mesh (left)

and deformed mesh (right).






/TITLE, Connectivity (Bad connectivity)




MP,EX,1,1 !material #1: E=1 (no units)


K,1,0,0,0 !define geometry: keypoints and areas

K,2,5,0,0

K,3,5,1,0

K,4,0,1,0

K,5,10,0,0

K,6,10,5,0

K,7,5,5,0

A,1,2,3,4

A,2,5,6,7

AESIZE,2,.25

AESIZE,1,.5

AMESH,ALL



DK,1,ALL

DK,4,ALL

DK,5,UX,.1

DK,6,UX,.1




PLDISP,1





/TITLE, Connectivity (Good connectivity)











K,2,5,0,0

K,3,5,1,0

K,4,0,1,0

K,5,10,0,0

K,6,10,5,0

K,7,5,5,0

A,1,2,3,4

A,2,5,6,7

AGLUE,1,2 !to ensure right connectivity areas must be glued

AESIZE,3,.5

AESIZE,1,.25

MSHKEY,2

AMESH,ALL



DK,1,ALL

DK,4,ALL

DK,5,UX,.1

DK,6,UX,.1




PLDISP,1



3.4.2 Avoid elements with high distortions, warping and inadmissible narrow

or wide angles between edges.

Elements with high distortions, bad aspect ratio, warping and inadmissible narrow, or wide an-

gles between the element edges can lead to numerical errors. In ANSYSTM , all continuum ele-

ments (2-D and 3-D solids, 3-D shells) are tested for acceptable shape. The testing, is performed

by computing shape parameters (such as Jacobian ratio) which are functions of geometry, then

comparing them to element shape limits whose default values are functions of element type and

settings. These limits depend on the element type and formulations, therefore it is recommended

to check the guidelines given by the finite element code being used. Some example of elements

with bad aspect ratio, warping and wide angles between element edges are shown in Figure 3.15.

ANSYSTM allows to plot the elements that have some of these element geometry checks over

the limits.

Main Menu>Preprocessor>Meshing>Check Mesh>

Individual Elm>Select Warning/Error Elements

Example 3.12. Check for elements with high distortions in the geometry and mesh shown in

Figure 3.16.







Figure 3.15: Example of elements with bad aspect ratio, warping and distortion.

Figure 3.16: Example of right elements with wide corner angles. Undeformed mesh (left) and

"warning" elements of the mesh (right).




/TITLE, Distortion elements







K,2,5,0,0

K,3,5,1,0

K,4,0,1,0

K,5,6,0,0

K,6,6,5,0

K,7,5,1,0

A,1,2,3,4

A,2,5,6,7

AGLUE,1,2

AESIZE,3,.5

AESIZE,1,.25

MSHKEY,1

AMESH,ALL







3.5 Techniques to Import Models and Geometries

In order to simulate or model different geometries or problems with finite elements it is often

useful to import the geometry or the model generated by means of a CAD-system or more pow-

erful and user-friendly Pre-process utility. In this section, some techniques to import models and

geometries into a FE package are shortly described.

There are different file standards that allow us to translate geometries and finite element

entities between different packages. Here, in the following example, the IGES-format is used to

import a geometry from a CAD software.

Example 3.13. Figure 3.17 shows the schema of an arm of a grip designed to pick and carry

eggs. The grip has a total of three of these arms so the eggs are clamped in between. The

mechanism is activated by applying a contact load on the normal to the right-top circular area

of the arm. The arm can rotate around its axis and the egg is clamped by the bottom-planar

areas of the grip-arms. Model the grip-arm assuming plane-stress conditions, considering that it

is made of aluminium (E = 71000 MPa and ν = 0.33) and the rotation axis is made of steel. It is

considered that the maximum load sustained by an egg is 5 N.

25°

50

40

8°

15.95 R2

17.79 R2.5 25

R1

R30.32

Ø6

5

10

14.33

5

Figure 3.17: Geometry of the grip-arm.

Solution to Example 3.13. The geometry of the arm-grip has been already generated using a

CAD program and can be found at ftp://amade.udg.edu/mme/PrePostFE/grip_arm.igs.



dow enter /input, file, ext. Before executing the script make sure that the IGES geometry of the

model is in the working directory. You can also import the geometry using the Menu commands




ftp://amade.udg.edu/mme/PrePostFE/grip_arm.igs


and mesh the geometry avoiding the importing commands in the script.



/TITLE, Robot grip part (Model 2D plane stress)

/AUX15 !import IGES file

IOPTN,IGES,NODEFEAT !select options

IOPTN,MERGE,YES

IOPTN,SOLID,YES

IOPTN,SMALL,YES

IOPTN,GTOLER, DEFA

IGESIN,’grip_arm’,’igs’,’ ’





R,1,5 !real constant #1: thickness=5 mm





SECDATA,1e9,1e9 !section width and height (ultra rigid beam)





LGEN,,ALL,,,-1747.394,-416.3893,0,,,1 !move lines coordinates origin

LSEL,S,,,1,11 !select lines

LSEL,U,,,3,4 !unselect circle lines

AL,ALL !generate global area with all selected lines

LSEL,S,,,3,4 !select circle lines

AL,ALL !generate circle

ASBA,1,2 !substract circle to global area

N,,0,0,0 !generate node on 0,0,0 (origin)

LESIZE,3,,,10 !divisions line 3

LESIZE,4,,,10 !divisions line 4

AESIZE,ALL,2 !element size a 2 mm

MSHKEY,2 !try quadrangualar elements

AMESH,ALL !mesh all areas

TYPE,2 !set element #2

MAT,2 !set material #2

SECNUM,2 !set section #2

REAL,2 !set real #2

E,1,178 !generate element between nodes #1 and #178

*REPEAT,20,0,1 !repaeat previous command 19 times increment +1 second argument



D,1,ALL !fix articulated node

DDELE,1,ROTZ !allow rotation articulated node

D,99,UX,,,,,UY !fix horizontal and vertical displacement contact node

F,172,FX,-4.532 !apply horitzontal load node #178

F,172,FY,2.113 !apply vertical load node #178




PLNSOL,S,EQV,2 !plot Von Mises stress + undeformed edge







Problem 3.1. Reconsider the situation in Example 3.3. Using direct generation commands,

remesh the model using twice the number of divisions that were used in the example solution.

Problem 3.2. Reconsider the Example 3.3. Prepare a mesh for sensitivity analysis for the fol-

lowing meshes:

1. The mesh in the solution of the example

2. A free mesh using triangles

3. Two different quadrilateral mapped meshes with different mesh densities.

Show the obtained meshes in a report and comment on the results.

Problem 3.3. Obtain the geometry of the part presented in Example 3.9. Use a CAD program to

generate this obtained geometry and its IGES file. Export the geometry to ANSYS and mesh it.

Try to obtain a regular mesh. Show all the process in a report.

Problem 3.4. Model the U-shaped beam shown in Figure 3.18. Using a similar procedure as in

Ex. 3.6 generate the MATLAB code and the ANSYSTM command sequence to define the mesh.

The codes should define the mesh in function of the geometric parameters shown in the figure.

Figure 3.18: U-shaped beam.




Chapter 4

Boundary conditions

4.1 Introduction

The Finite Element Method may be applied to problems of very different nature which are af-

fected by different physical variables. As you may know, in the structural case, after computing

the stiffness matrix for each element, the finite element method solves a typical system of equa-

tions:

K · ~δ = ~F (4.1)

where K is the stiffness matrix, ~δ is the degree of freedom vector and ~F is the vector contain-

ing the external actions. Some of the components in ~δ and ~F will be unknown variables -and so,

the solution of our problem- and some of them will be the boundary conditions of the model. A

known variable can set a fixed value to one nodal degree of freedom (DOF) (displacements and

rotations). This type of boundary conditions are usually called Dirichlet boundary conditions or

essential boundary conditions. On the other hand, a known variable can define a nodal force or

flux, now the known variable defines a natural boundary condition (also called Newman boundary

condition). Focusing on the Engineering Mechanics realm, the typical problems may be divided

into different physics problems as is shown in the Table 4.1.

Table 4.1: Physics Problems in FEM

Problem DOFs Force or flux variables

Structural displacements

and rotations

forces, pressures, temper-

atures (for thermal strain),

gravity

Thermal temperatures heat flow rates, convection,

internal heat generation, in-

finite surface

Magnetic magnetic po-

tentials

magnetic flux, magnetic cur-

rent segments, source cur-

rent density, infinite surface

Electric electric poten-

tials (voltage

electric current, electric

charges, charge densities,

infinite surface

Fluid velocities pressures

47


For the moment, we will focus on structural problems. Thermal and structural-thermal analy-

sis will be seen further during this course. The boundary conditions in structural models include

externally applied forcing loads and constrained displacements. Loads are divided into different

categories:

• DOF constraints: constrained displacements and rotations

• Concentrated loads: forces and moments

• Distributed loads:

– loads on edges

– loads on surfaces

– body loads

– inertia loads (due to accelerations or gravity)

• Coupled-field loads: fluid-structure interaction, thermal dilatations, etc.

• Temperatures which structures are expected to work at.

Hint 9: Loads on geometry vs loads on mesh

The major commercial codes of FE allow to impose boundary conditions to geometric entities

(points, lines, surfaces, or volumes) or to mesh entities (nodes and elements).

Pros Cons

Loads on Geomet-

ric Entities

Independent of mesh. No trou-

bles in re-meshing

Possible conflicts local-global

coordinate systems

(points, lines, sur-

faces...)

Generally easier to apply (less

entities needed to select)

Less control on what you are do-

ing

Loads on Mesh En-

tities

Loads on affected nodes di-

rectly (more control)

Re-meshing involves re-

applying loads

(nodes, elements) Generally lots of

nodes/elements involved

(graphical picking may not

be convenient)

Load step

A load step or load case is simply a configuration of loads for which a solution is obtained. There-

fore, a load step in structural analysis is defined by the set of forces and moments, pressures,

body and inertia loads (as gravity), and specified displacements and rotations (constrained DOF),

all applied to the model. In a linear static or steady-state analysis, you can use different load

steps to apply different sets of loads -wind load in the first load step, gravity load in the second

load step, both loads and a different support condition in the third load step, and so on. In a

transient analysis, multiple load steps apply different segments of the load history curve.

The next sections show how different kinds of loads may be applied to a structural FE model.



Chapter 4. Boundary conditions 49

Boundary conditions in two- and three-dimensional problems

In two-dimensional problems, the following boundary conditions can be applied:

• Constrained displacements in X and Y axes, also rotations into the Z axis in flexural beam

elements.

• Concentrated forces (in X and Y directions), also moment in flexural beam elements (applied

into the Z axis).

• Distributed loads on edges and lines.

Additionally, in three-dimensional problems the following boundary conditions can be applied:

• Imposed displacements (and rotations in shell and beam elements) to all degrees of free-

dom.

• Concentrated forces (and moments in shell and beam elements) in any direction or moments

in any axis.

• Distributed loads on edges and surfaces.

• Body and inertial loads.

Hint 10: DOF notation in ANSYS

UX Displacement in X direction ROTX Rotation into X axis

UY Displacement in Y direction ROTY Rotation into Y axis

UZ Displacement in Z direction ROTZ Rotation into Z axis

Hint 11: Useful ANSYS commands to apply boundary conditions

On Mesh On Geometry

Displacements and

rotations

D on nodes DK on keypoints, DL on lines

Concentrated loads F FK on keypoints

Line loads - SFL on lines

Surface loads SF on nodes, SFE on elements SFA on areas

4.2 Load application in FEM

Loads can be applied on nodes by means of concentrated forces or moments, as it is shown in Ex-

ample 4.1. Also loads can be distributed over the elements as: surface loads, body loads, inertia

loads or other coupled-field loads (for example, thermal strains).

A Surface load is a distributed load applied over a surface, for example a pressure caused

by wind or any fluid. A body load is a volumetric load, for example expansion of material by

temperature raise in structural analysis. Inertia loads are those attributable to the inertia (mass




matrix) of a body, such as gravitational acceleration, angular velocity, and acceleration.

When a concentrated load is applied on a node in the FE model, it can be directly added to

the force vector. However, if a distributed load is applied, the element shape functions are used

by the finite element program to compute the equivalent forces vector.

Hint 12: Avoiding rigid solid modes

When modelling real life situations the boundary conditions applied to the model should guar-

antee that the model is not going to move without carrying any deformation. This may be done

by setting a zero value to all DOF in the same or different nodes. One should do this without

applying unreal boundary conditions which bring to the solution unreal stresses.

Example 4.1. Let us model a small bridge (a footbridge) intended to cross from a pier to a boat

or to a small ship (see Fig. 4.1). This footbridge is 3 m long and should bear a load of a man,

estimated in 150 kg, standing in its middle point. The section of the bridge is 300 mm × 25 mm.

The material is wood with E = 15000 MPa and ν = 0.3. Let us model this problem using beam-

type elements. At one tip of the footbridge, the tip on the ship, all displacements and rotations

are restricted and on the other one the footbridge is simply supported by the floor.

Figure 4.1: Footbridge for small ship.






/TITLE, Footbridge 2D with beam elements


!define geometric parameters (mm)

length=3000

width=300

height=25

!element definition





SECDATA,height,width !section width and height

MP,EX,1,15000 !material property #1, Young’s modulus

MP,NUXY,1,0.3 !material property #1, Poisson modulus

!define keypoints




K,1,0,0 !keypoint,#,x,y

K,2,length/2

K,3,length

!define lines

L,1,2,6 !line #1 connecting kpoint 1 and 2 in 6 parts

L,2,3,6

LMESH,all !mesh all lines

FINISH


!apply constraints

DK,1,ALL,0 !define a DOF constraint at a keypoint

DK,3,UY,0

!apply loads

FK,2,FY,-1500 !define a force load to a keypoint

SOLVE

FINISH


PRRSOL,F !list reaction forces


PLNSOL,U,SUM,0,1 !contour plot of deflection

ETABLE,TAI,SMISC,6 !Shear diagram at node i

ETABLE,TAJ,SMISC,19 !Shear diagram at node j

PLLS,TAI,TAJ,1,0 !Plot the shear diagram

PRRSOL,M !list reaction moments

ETABLE,MOI,SMISC,3 !Bending moment at node i

ETABLE,MOJ,SMISC,16 !Bending moment at node j

PLLS,MOI,MOJ,1,0 !Plot the bending moment diagram

/ESHAPE,1,1

PLNSOL,S,X



Observe that the left end of the deformed footbridge remains horizontal due to the applied

constraints while the right end rotates upwards.

Example 4.2. We want to dimension a roof. The loads to consider are the weight of the roof

and the load due to snow. The roof is 0.9m × 2.1m with a 10 mm thickness and it is settled

horizontally and supported at its four vertex. The snow load over the roof is 400 N/m2. Model the

roof and obtain the deformed shape plot. Consider carefully which are the boundary conditions

to apply. The roof is from a reinforced polymer with a Young’s modulus of 15 GPa, a Poisson’s

coefficient of 0.3, and a mass density of 40 kg/m3.






/TITLE, Horizontal roof


!geometric parameters (mm)





length=900

width=2100

thickness=10



R,1,thickness !real constant #1: th=10 mm

MP,EX,1,15000 !Young’s modulus for material #1 in N/mm2

MP,PRXY,1,0.3 !Poisson ratio for material #1

MP,DENS,1,40e-9 !material property #1: Density in kg/mm^3

!define keypoints

K,1,0,0 !keypoint, #, x, y

K,2,width,0

K,3,width,length

K,4,0,length

!define area

A,1,2,3,4 !area through keipoints

!element definition

AESIZE,ALL,100 !specifies the element size to be meshed onto areas


FINISH


!apply boundary conditions

DK,1,UZ,0,,,UX,UY !define a DOF constraint at a keypoint (ux=uy=uz=0)

DK,2,UZ,0,,,UY !define a DOF constraint at a keypoint (uy=uz=0)

DK,3,UZ,0

DK,4,UZ,0

!apply loads

ACEL,,,9.81 !acceleration in m/s^2, kg*m/s^2 = N

LSWRITE,1 !write the first load case

SFA,1,1,PRES,-0.4e-3 !define snow load over surface; 0.4 kN/m^2 = 0.4e-3 N/mm^2

LSWRITE,2 !write the second load case

LSSOLVE,1,2 !solve the resulting system of equations, of load cases #1 and #2

FINISH


SET,1 !first load case


SET,2 !second load case




Note that in this case the maximum displacement of the roof caused by its own weight is very

small. However, when the weight of the snow is added, the displacement of the roof is increased

considerably.

Example 4.3. Use a commercial FE code to model the effect of the distributed force q = 5 N/mm2

on the two types of short cantilever beams shown in Fig.4.2. Consider that the two beams are

made of steel with a thickness of 5 mm and take L1 = 80 mm, L2 = 40 mm, B = 150 mm and Q

= 160 mm. Use eight-noded (quadratic) quadrilateral plane elements.



then, on the command window enter /input, file, ext. Next we present two different solutions

using different approaches to implement the clamped side.





q q

(a) (b)

B

Q

Q

B

Figure 4.2: Short cantilever beams



/TITLE, Distributed load on short cantilever: direct clamped


!Geometric parameters (mm)

L1=80 !clamped length

L2=40 !short side length

B=150 !distributed load span

th=5 !thickness

ET,1,PLANE183 !element type

KEYOPT,1,3,3 !plane stress with thickness



R,1,th !real constant #1, thickness for element #1

K,1,0,0 !geometry: keypoints

K,2,0,-L1

K,3,B,0

K,4,B,-L2

L,1,2,10 !line clamped with 10 divisions

L,1,3,15 !line where load is applied, 15 divisions

L,2,4,15

L,3,4,10

AL,1,2,3,4 !area creation from lines

AMESH,ALL !mesh generation

FINISH


DL,1,,ALL !campled line #1

SFL,2,PRES,5 !load on line #2

SOLVE !solve the current load state

FINISH



PLNSOL,S,EQV !contour plot Von Mises equivalent stress

PLVECT,S !vector plot principal stress





/TITLE, Distributed load on short cantilever: not direct clamped


!Geometric parameters (mm)

L1=80 !clamped length

L2=40 !short side length

B=150 !distributed load span

Q=160 !square side

th=5 !thickness






KEYOPT,1,3,3 !plane stress with thickness



R,1,th !real constant #1, thickness for element #1


K,2,0,-L1

K,3,B,0

K,4,B,-L2

K,5,0,((Q-L1)/2)

K,6,0,-(L1+((Q-L1)/2))

K,7,-Q,((Q-L1)/2)

K,8,-Q,-(L1+((Q-L1)/2))

K,9,-Q,0

K,10,-Q,-L1

L,1,2,10

L,1,3,15 !line where load is applied, 15 divisions

L,3,4,10

L,2,4,15

L,1,5,5

L,5,7,16


L,9,1,16



L,10,2,16

L,8,6,16

L,2,6,5

!areas definition from lines

AL,1,2,3,4 !area 1

AL,5,6,7,8

AL,8,9,11,1

AL,11,10,12,13

AMESH,ALL !mesh generation

FINISH


DL,7,,ALL !campled lines

DL,9,,ALL

DL,10,,ALL

SFL,2,PRES,5 !load on line #2

SOLVE !solve the current load state

FINISH



PLNSOL,S,EQV !contour plot Von Mises equivalent stress

PLVECT,S !vector plot principal stress



After analysing the results of both short cantilever beam configurations, it can be concluded

that the inclusion of more material does not significatively affect the maximum values of the

achieved stress.

Example 4.4. Use plane strain conditions to simulate the cross-section of an hydraulic dam

shown in Fig. 4.3. The dam is made of reinforced concrete, E = 27000 MPa and ν = 0.25, and its

left side is full of water up to a height H = 30 m. The other dimensions of the dam are h = 20 m,

B = 4 m and b = 2 m. The density of the fluid contained by the dam is γ = 104 N/m3.








Figure 4.3: Cross-section of the hydraulic dam.



/TITLE, 2-D plane strain dam section


!Parameters (mm)

H=30000 !left height

H1=20000 !right height

B=4000 !bottom base

B1=2000 !top base

t=1000 !thickness

D=0.00001 !fluid density in N/mm3

p=D*H !max pressure in N/mm2


KEYOPT,1,3,2 !keyopt3=2, plane strain

!thickness is not used in plane strain

MP,EX,1,27000 !Young’s modulus

MP,NUXY,1,0.25 !Poisson


K,2,B,0

K,3,0,H1

K,4,B1,H1

K,5,0,H

K,6,B1,H

L,1,2,8 !lines by keypoints and divisions

L,1,3,20

L,2,4,20

L,3,4,8

L,3,5,10

L,4,6,10

L,5,6,8

AL,1,2,3,4 !areas by lines

AL,4,5,6,7

AMESH,ALL !mesh

FINISH

/SOLU !start SOLUTION MODULE!Boundary conditions

DL,1,,ALL !clamped nodes on base

!Apply pressure

SFGRAD,PRES,0,Y,0,-(p/H) !Gradient on surface load, type pressure, cartesian coordinate system,

! along Y direction, with slope -p/H




NSEL,S,LOC,X,0 !select left edge nodes

SF,ALL,PRES,p !apply pressure

NSEL,ALL

SOLVE

FINISH






In this case, the cross-section of the dam acts as a cantilever beam under the effect of a

distributed load following a triangular distribution.

4.3 Constrain DOF of FE model

4.3.1 Constrained displacements and rotations

In a general FEA, a node can have more than one DOF. For example, if the FE model uses beam

elements in 2-D space we have three DOF: the horizontal displacement, the vertical displacement,

and the rotation around perpendicular axis to plane. When we constraint different DOF we define

different kinds of actual boundary conditions. In the 2-D beam element case if we constraint only

the horizontal and vertical displacements we get a simple support, if we constraint all the DOF

we obtain a clamped condition.

Hint 13: ANSYS command to constrain DOF on nodes

D, nnode, dir

where nnode is the node number, dir is the degree of freedom to impose, that is uy, ux or rotz.

4.3.2 Symmetry conditions

Symmetry conditions can be applied to reduce the size of the model without loss of accuracy.

However, to use such conditions the four types of symmetry must be present: geometry, boundary

conditions, material and loads. Under this condition the solution will be also symmetric. For

example, symmetry with respect to the y-z plane involves that the nodes on the symmetry plane

have the following conditions

ui(x) = 0 θi(y) = 0 θi(z) = 0 (4.2)

Symmetric boundary conditions on nodes in the symmetry plane involve the restriction of the

out-of-plane DOF translation and the restriction of the in-plane DOF rotations.

Example 4.5. Use a commercial FE code to find the theoretical stress concentration factor in a

rectangular notched bar made of aluminium, E = 190 GPa and ν = 0.3. The dimensions and the

load state is defined in Fig. 4.4. Use eight-noded (quadratic) quadrilateral plane elements.








Figure 4.4: Rectangular notched bar, axial load 10 N/mm2.



/TITLE, Rectangular notched bar


ET,1,PLANE183 !element type #1 PLANE82 8-node 2-D

KEYOPT,1,3,3 !keyoption3=3, plane stress with thickness

R,1,4 !real constant, thickness=4 mm

MP,EX,1,190e3 !elastic modulus in N/mm2

MP,PRXY,1,0.3 !Poisson coefficient

BLC4,0,0,50,18.5 !geometry: square area 50x18.5 mm

CYL4,0,20,7.5 !circular area center (0,20) and radius 7.5 mm

ASBA,1,2 !subtract previous areas

ESIZE,1.5 !define element size

MSHKEY,0 !free mesh method

AMESH,ALL !mesh

FINISH


DL,1,,SYMM !symmetry conditions in line 1

DL,9,,SYMM !symmetry conditions in line 9

SFL,2,PRES,-10 !apply pressure on line 2 (tension)

SOLVE

FINISH


PLNSOL,S,EQV,2,1 !contour plot Von Mises equivalent stress, undeformed edge display, scale=1



The stress in the net area without stress concentration is

σo =P

A=

10 · 37 · 4

25 · 4= 14.8 MPa

The maximum horizontal stress close to the notch obtained from FE model is 28 MPa. There-

fore, the concentration factor is

k =σmax

σo= 1.89

Axysymmetric conditions can be also used when revolution geometries are simulated. In the

following example inertial loads are applied in an axisymmetric model of a flywheel.

Example 4.6. Consider the steel flywheel schematized in Figure 4.5. Its inner diameter is 50

mm and its outer diameter is 600 mm. Assume that it is rotating at 2000 rad/s. Write an input

file for ANSYSTM which models this situation and computes the total inertia of the flywheel.





rota

tio

n a

xis

Rin

Rcen1

Rr

Rout

Rcen2

B2

/2

B1

/2

B3

/2

B4

/2

Figure 4.5: Quarter section of a flywheel.






/TITLE, Flywheel


!Parameters, this is a dynamic analysis and the international

!system units are used to avoid problems

Rin=25 !mm

Rcen1=Rin*4

Rout=300

Rcen2=250

B1=300

B2=90

B3=25

B4=200

Rr=Rcen2-(25)

ET,1,PLANE182,,,1 !element type with axysymmetric condition, KEYOPTION3=1

MP,EX,1,210e3 !Young modulus, N/mm^2

MP,PRXY,1,0.3 !Poisson

MP,DENS,1,7800e-9 !Density, kg/mm^3

K,1,Rin,0 !geometry: keypoints

K,2,Rin,B1/2

K,3,Rcen1,0

K,4,Rcen1,B2/2

K,5,Rcen1,B1/2

K,6,Rr,0

K,7,Rcen2,0

K,8,Rcen2,B3/2

K,9,Rcen2,B4/2

K,10,Rout,0

K,11,Rout,B4/2

L,1,2,10 !lines defined by keypoints

L,1,3,8

L,2,5,8

L,3,4,6

L,4,5,4

L,4,6,12

L,5,8,16

L,6,7,4




L,7,8,4

L,8,9,6

L,7,10,8

L,9,11,8

L,10,11,10

AL,1,2,3,4,5 !areas by lines

AL,5,6,8,9,7

AL,10,9,11,13,12

LCCAT,4,5 !concatenate lines in one to improve meshing

LCCAT,6,8 !it helps to obtain continuity in the mesh

LCCAT,9,10 !and avoid irregularities

AMESH,ALL

FINISH


DK,1,UY

OMEGA,,2000 !rotational speed in rad/s

SOLVE

FINISH


PLDISP,1

ETABLE,evol,VOLU !compute the volume of each node, save it with "evol" label

ETABLE,ecent,CENT,X !compute the centroidal of each node, save it with "ecent" label

SMULT,mass,evol,,7800e-9 !compute the total mass in label "mass" by multiplying 7800*evol

SMULT,mx,mass,ecent,1,1 !compute the first moment of inertia in label "mx" by multiplying

! mx=1*mass+1*ecent

SMULT,einer,mx,ecent,1,1 !compute element inertia in label "einer" by multiplying

! einer=1*mx+1*ecent

SSUM !calculates the summ of every element table item

*GET,totalinertia,SSUM,,ITEM,einer !compute total inertia (kg mm^2)

*STATUS,totalinertia !List of the parameter "totalinertia"



Observe that although only a quarter section of the flywheel is modelled, the simulation takes

into account the whole flywheel and the total inertia is calculated.

4.3.3 Antisymmetry conditions

Antisymmetric conditions are similar to the symmetry conditions. They can be applied when

it exists symmetry of geometry, symmetry of boundary conditions, symmetry of material and

antisymmetry of loads. Antisymmetric boundary conditions involve the restriction of the in-plane

DOF translations of the nodes in the antisymmetry plane and the restriction of the out-of-plane

DOF rotations.

Hint 14: Symmetry conditions vs Antisymmetry conditions

The following table shows which are the DOF to restrict in function of the symmetry plane

normals.

Symmetry Antisymmetry

Normal 2D 3D 2D 3D

X UX, ROTZ UX, ROTZ, ROTY UY UY, UZ, ROTX

Y UY, ROTZ UY, ROTZ, ROTX UX UX, UZ, ROTY

Z – UZ, ROTX, ROTY – UX, UY, ROTZ





4.4 Superposition

The superposition principle states that linear problems can be decomposed as the addition of

several sub-problems. In large models involving symmetric geometry but un-symmetric loads

the superposition can be applied by decomposing the problem as the addition of a symmetric

problem and an antisymmetric one.

Example 4.7. Use a commercial FE code to find analyze the beam structure of Figure 4.6 made

of thin-walled circular tubes with an outer diameter of 20 mm and 4 mm thick. Use two-node

beam elements. Decompose the problem in a symmetric part and an antisymmetric part and

obtain the total displacements.

Figure 4.6: Geometrically symmetric unsymmetrically loaded beam structure






/TITLE, Superposition method of thin walled beam structure


!define parameters (mm)

length=3000

height=1000

Ri=6 !Inner radius of the tube

Ro=10 !Outer radius of the tube

ET,1,BEAM188 !element type #1: 3D beam (6 DoF)


SECTYPE,1,BEAM,CTUBE !section #1 = circular tube


SECDATA,Ri,Ro !inner radius, outer radius




K,2,length/4,0

K,3,length/2,0

K,7,length/2,height

K,8,length/4,height/2

L,1,2,20 !define lines by keypoints and divisions

L,2,3,20

L,3,7,20

L,3,8,20

L,2,8,20

L,1,8,20




L,8,7,20

LMESH,ALL !mesh all lines

FINISH


!Symmetric case

DK,1,ALL,0 !define a DOF constraint at a keypoint

DK,3,UX,0,,,ROTZ !constrain displacement normal to symmetric plane

DK,7,UX,0,,,ROTZ

!DSYM,SYMM,X !to apply symmetry conditions on selected nodes

FK,8,FY,-75 !apply load to a keypoint

FK,7,FY,-100

LSWRITE,1 !write first load step

!Antisymmetric case

DKDELE,ALL,ALL !delete previous constrains in keypoints

DK,1,ALL,0

DK,3,UY,0 !constrain displacent in antisymmetic plane

DK,7,UY,0

FKDELE,ALL,ALL !delete previous forces in keypoints

FK,8,FY,-25 !apply load to a keypoint

LSWRITE,2 !write second load step

LSSOLVE,1,2 !solve load steps 1 and 2

FINISH


SET,1 !activate first load step solution



SET,2 !activate second load step solution





Note that the addition of the vertical reaction force in node 1 for the two load steps results

in the total reaction force this node would have if the whole structure was simulated with the

specified loads.

4.5 Structural analysis with temperature change

Although we are focusing on structural problems we may also be interested in having some pre-

diction about our structure working at a different temperature. From the finite element software

point of view, we may understand this problem as something like:

K ·(

~δ − ~δT

)

= ~F (4.3)

where, again, K is the stiffness matrix, ~δ is the degree of freedom vector and ~F is the vector

containing the external actions, and ~δT is the variation on the degree of freedom caused by

thermal dilatation:

~δT = ~α · △T (4.4)

where ~α is the coefficient of thermal expansion (CTE) vector and△T the temperature change.

The ANSYS commands useful for these situations are summarised next.





Hint 15: ANSYS commands for temperature change in structural analysis

ANSYS command Use

MP,CTEX,NMAT,VALUE CTE through X direction

MP,CTEY,NMAT,VALUE CTE through Y direction

MP,CTEZ,NMAT,VALUE CTE through Z direction

TREF,TEMP reference temperature (at which it is measured)

TUNIF,TEMP working temperature

Example 4.8. Compute the thermal strain and the total strain of the steel plate with a hole

shown in Figure 4.7. The dimensions of the plate are measured at a temperature of 25◦C while

it is expected to work at 80◦C. Note: The coefficient of thermal expansion for steel is 12 ×10−6

◦C−1.

Figure 4.7: Steel plate with thermal expansion (Dimensions in mm)






/TITLE, Temperature change in structural analysis


!Parameters

R=10 !radius in mm

L=100 !height in mm

T=5 !thickness in mm






KEYOPT,1,1,3 !simplified enhanced strain formulation

KEYOPT,1,3,3 !element #1, keyopt 3=3, plane stress w/thickness

R,1,T !element thickness



MP,CTEX,1,12e-6 !coefficient of thermal expansion for material #1


K,2,L,0

K,3,L,L

K,4,0,L

L,1,2,20 !define lines by keypoints and divisions

L,2,3,20

L,3,4,20

L,4,1,20


CYL4,2*R,3*R,R !define circular hole

ASBA,1,2 !substract area 2 from area 1


FINISH



D,ALL,ALL !fix displacement

LSEL,S,LINE,,3 !select line 3

NSLL,S,1 !select all nodes on lines

SF,ALL,PRES,p !apply pressure on all selected nodes

NSEL,ALL

TREF,25 !reference temperature 25ºC

TUNIF,80 !uniform temperature 80ºC

SOLVE

FINISH



PLNSOL,EPTH,X !thermal strain in direction X

PLNSOL,EPEL,X !elastic strain in direction X

PLNSOL,EPTT,X !thermal+elastic strain in direction X



Observe that the value of the strain in the X direction is similar for the thermal expansion and

the elastic strain due to the mechanical load.

4.6 Loadcases

In many occasions the same structure is analyzed under different sets of boundary conditions

usually called loadcases. FE commercial codes allow to define these loadcases and to obtain the

solution separately for each of them.

Hint 16: ANSYS commands for load step definition

ANSYS command Usage

LSWRITE, LSNUM Write load steps with created loads

LSDELE, LSMIN, LSMAX, LSINC Delete load steps

LSSOLVE, LSMIN, LSMAX, LSINC Read and solve multiple load steps

SET, LSNUM Defines load step to be post-processed





Example 4.9. Reconsider the moebius tape sculpture of example 3.6 and introduce different

load cases considering: people walking on the structure, gravity load (density equal to 8 T/m3),

wind load.




FINISH !close all revious modules


/TITLE,Moebius Sculpture




SECTYPE,1,BEAM,CSOLID !section #1: preimetral beams, solid cylindrical


SECDATA,40 !section radius

SECTYPE,2,BEAM,RECT !section #2: internal beams, rectangular



SECTYPE,3,BEAM,RECT !section #3: bottom internal beams, rectangular




MP,PRXY,1,0.29 !material #1: 0.29 Poisson

MP,DENS,1,8e-6 !material #1: density=8 T/m^3

/INPUT,nodes,txt !input nodes from file: ’nodes.txt’


E,101,102 !elements to generate the perimetral circles

EGEN,47,1,1

E,148,201

E,201,202

EGEN,47,1,49

E,248,101 !define first 96 elements


E,101,1,248 !define transversal elements

E,1,201,148

E,102,2,101

E,2,202,201

EGEN,47,1,99,100

ESEL,S,,,167,172 !redefine section bottom elements

EMODIF,ALL,SECNUM,3

ALLSEL !select everything



D,236,ALL !fix sculpture to floor

D,237,ALL

D,238,ALL

ACEL,,,10 !gravity load

LSWRITE,1 !first load step: only gravity load

F,12,FY,-500 !additional loads: somebody transit over sculture

F,13,FY,-500

F,14,FY,-500

LSWRITE,2 !second load step: gravity load + people

FDEL,12,ALL

FDEL,13,ALL

FDEL,14,ALL




/INPUT,wind,txt !wind load

LSWRITE,3 !third load step: gravity load + wind

LSSOLVE,1,3 !solve load states



SET,1

PLDISP,1

SET,2

PLDISP,1

SET,3

PLDISP,1




Problem 4.1. Reconsider the situation of Example 4.1. Solve the problem again but considering

that displacements and rotations are null at both tips of the footbridge. Compare both solutions

with the analytical solution you might know.

Problem 4.2. Reconsider the situation of Example 4.2 in a different constructive solution. Con-

sider that now it is an inclined roof and that, in addition to being supported by its four corners,

it is supported by a perfectly rigid beam located under one of its shorter sides. The inclination is

such that the beam supported edge is in a height of 1 m with respect to the opposite edge. Notice

that now the normal to the roof surface is not coincident with the gravity direction. Consider how

could you apply the snow load.

Figure 4.8: Inclined roof

Problem 4.3. Model the U-shaped beam shown in Figure 4.9 with ANSYSTM . The inner radius

of the beam is ri = 10 mm, the outer radius is ro = 30 mm, the thickness is t = 5 mm and the

length of the straight part is L = 30 mm. The material of the beam is steel, E = 210 GPa and

ν = 0.3, and the applied force is F = 100 N. Evaluate the value and location of the maximum

Von Mises stress. Parametrize the geometry of the model, the applied force and the material

properties. Show all work in a report.

(a) Model the whole geometry.





(b) Repeat the problem now applying symmetry conditions. Consider carefully the applied

boundary conditions to avoid unreal stresses.

(c) Compare the solutions of (a) and (b) to verify you applied the boundary conditions correctly.

(d) Assume the reference temperature is 10◦ and obtain the thermal strain and the total strain

when the temperature of the structure is 60◦. The coefficient of thermal expansion for steel

is 12×10−6.

Figure 4.9: U-shaped beam.



Chapter 5

Linear elastic material models

5.1 Introduction

Material behaviors are very varied depending on different causes: the own physical constitution

of the material (metallic materials, geomaterials, polymers, etc), the range of applied strain, or

other physic parameters such as temperature, velocity of load application, large time behavior,

etc.

The main categorization between static non-time dependent models is between linear mate-

rial models (linear elastic models) and non-linear material models (materials with hyperelasticity,

plasticity, etc). An analysis with these non-linear models requires an iterative solution. This chap-

ter is focussed on linear models.

Linear material models for structural analysis are basically destined to study the linear elastic

response of materials.

Hint 17: ANSYS command to define linear elastic properties


Define a linear material property as a

constant or a function of temperature

MP lab,mat,C0,C1,C2,C3,C4

Material property label (lab) for static structural analysis

EX - Elastic moduli (also EY, EZ) PRXY - Major Poisson’s ratios (also PRYZ, PRXZ) NUXY - Minor

Poisson’s ratios (also NUYZ, NUXZ) GXY - Shear moduli (also GYZ, GXZ) DENS - Mass density

These properties (which may be functions of temperature) are called linear properties be-

cause typical non-thermal analyses with these properties require only a single iteration.

X, Y, and Z refer to the element coordinate system. In general, if a material is isotropic, only

the EX, the PRXY and possibly the DENS terms are used.

The models should be defined in the different element types according to their formulation:

uniaxial models for one-dimensional elements (truss, pipes, etc.), plane models for bi-dimensional

elements (plane solid elements, beams, etc.), or three-axial approach for three-dimensional ele-

ments (shells, solids).

67


5.2 One-dimensional elastic properties

The constitutive equation for a one-dimensional lineal material model is

σx = Eǫx (5.1)

The model only needs a single property, the Young’s modulus (E) (in ANSYS denoted by label

EX). If inertial loads are employed in any loadcase, the density should be also introduced (in

ANSYS denoted by label DENS). Examples 2.1, 2.2, 4.1, and so on use this material model.

5.3 Bi-dimensional and Three-dimensional elastic properties

Now, the material constitutive equation is a tensorial expression,

σij = Cijklǫkl (5.2)

Structural material properties must be input as an isotropic, orthotropic, or anisotropic ma-

terial.

5.3.1 Isotropic material

When the material is isotropic the stiffness tensor in plane linear elasticity is

C =E

1− ν2

1 ν 0

ν 1 0

0 01− ν

2

(5.3)

For a plane-stress case, E = E and ν = ν, while for a plane-strain or axisymmetrical model

E = E/(1− ν2) and ν = ν/(1− ν).

The stiffness tensor in isotropic solid linear elasticity is

C =

1

E

−ν

E

−ν

E0 0 0

−ν

E

1

E

−ν

E0 0 0

−ν

E

−ν

E

1

E0 0 0

0 0 01

G0 0

0 0 0 01

G0

0 0 0 0 01

G

−1

(5.4)

where G = E/2(1 + ν).

Only two properties are necessary to define this model: Young’s modulus (E) and the Pois-

son’s ratio (ν).



Chapter 5. Linear elastic material models 69

Hint 18: ANSYS properties for isotropic linear elastic materials

In ANSYS, Young’s modulus must be input using the label EX. Poisson’s ratio (PRXY or NUXY)

should not be equal to or greater than 0.5, the value assumed by default is 0.3, if a zero value

is desired input PRXY or NUXY with a zero or blank value. The shear modulus (GXY) defaults to

EX/(2(1+NUXY)) (ANSYS internally converts PRXY to NUXY when solving the analysis). If GXY is

input, it must match EX/(2(1+NUXY)). Hence, the only reason for inputting the shear modulus

is to ensure consistency with the other two properties.

Isotropic Elastic Example: High Carbon Steel

MP,EX,1,210e3 !MPa

MP,PRXY,1,.29 !no units

MP,DENS,1,7850e-9 !kg/mm3

Examples 2.4, 2.6, 4.5 and so on use isotropic plane stress materials model. Examples 2.5, 2.2

and 4.4 use isotropic plane strain model. Example 4.6 uses isotropic plane strain with axysym-

metry. Example 3.10 uses solid isotropic material model.

5.3.2 Orthotropic material

The stiffness tensor in orthotropic solid linear elasticity is

C =

1

Ex

−νyxEy

−νzxEz

0 0 0

−νxyEx

1

Ey

−νzyEz

0 0 0

−νxzEx

−νyzEy

1

Ez0 0 0

0 0 01

Gyz0 0

0 0 0 01

Gxz0

0 0 0 0 01

Gxy

−1

(5.5)

where (νyx/Ey) = (νxy/Ex), (νzx/Ez) = (νxz/Ex), and (νzy/Ez) = (νyz/Ey). Depending on Ex, Ey,

and Ez values, the pairs related Poisson’s ratios become major o minor coefficients.




Hint 19: ANSYS properties for orthotropic linear elastic material

In ANSYS, the properties required for an orthotropic material are:

• Young’s moduli, EX, EY, EZ,

• Poisson’s ratios, may be input in either major (PRXY, PRYZ, PRXZ) or minor (NUXY, NUYZ,

NUXZ) form, but not both for a particular material, and

• shear moludi GXY, GYZ, and GXZ.

Note that, for example, if only EX and EY are input (with different values) to a plane stress

element, an error will result indicating that the material is orthotropic and that GXY and NUXY

are also needed.

For axisymmetric analyses, the X, Y, and Z labels refer to the radial (R), axial (Z), and hoop

(θ) directions, respectively.

Orthotropic Elastic Example: Aluminum Oxide

MP,EX,1,307e3 !MPa

MP,EY,1,358.1e3 !MPa

MP,EZ,1,358.1e3 !MPa

MP,GXY,126.9e3 !MPa

MP,GXZ,126.9e3 !MPa

MP,GYZ,126.9e3 !MPa

MP,NUXY,1,.20 !no units

MP,NUXZ,1,.20 !no units

MP,NUYZ,1,.20 !no units


Example 5.1. Use ANSYS to model a rectangular plate with dimensions ax =4000 mm, ay =2000

mm, thickness t =10 mm subjected to a uniform tensile load q0 = 1200 N/mm applied in the 2000

mm edge. The material is an unidirectional laminate AS4D/9310 which elastic properties are

summarised in Table 5.1, with the fibers oriented in the x-axis. Obtain the elongation of the plate

under the applied load and compare it with the analytical solution obtained using Hooke’s Law.

Table 5.1: Material properties of unidirectional AS4D/9310 carbon/epoxy composite.

Property AS4D/9310

E1 133.86 GPa

E2 = E3 7.71 GPa

G12 = G13 4.36 GPa

G23 2.76 GPa

ν12 = ν13 0.301

ν23 0.396

Solution to Example 5.1. As the problem can be considered as a 2D situation, one of the

coordinates, in the thickness direction, can be eliminated from the governing equations so that

the 3D problem simplifies to 2D. In the process, the thickness becomes a parameter, which is

known and supplied to the modeling software. Typically, we differentiate material properties and




parameters even though both are supplied as known input data. Most software packages make

this difference as well. For example, the solid thickness is supplied to ANSYS as a real constant

set (R command), while material properties are entered separately (MP command).

In the FE model we use PLANE183, a 8-node bi-dimensional solid element. Symmetry with

respect to the x − z and y − z planes is used to model 1/4 of the plate. The ANSYSTM command

sequence for this example is listed below. You can either type these commands on the command

window, or you can type them on a file, then, on the command window enter /input, file, ext.



/TITLE,Rectangular plate under distributed load


t=10 !thickness

ET,1,PLANE183,,,3 !element type #1, 2D solid, plane stress w/thickness

R,1,t !real constant #1, thickness

!material properties FOR AS4D/9110 orthotropic laminate

MP,EX,1,133.86e3 !MPa

MP,EY,1,7.706e3

MP,EZ,1,7.706e3

MP,GXY,1,4.36e3

MP,GYZ,1,2.76e3

MP,GXZ,1,4.36e3

MP,PRXY,1,0.301

MP,PRYZ,1,0.396

MP,PRXZ,1,0.301

RECTNG,0,2000,0,1000 !geometry: square area

ESIZE,250 !element size 250 mm

AMESH,all !mesh the area

FINISH


ANTYPE,STATIC !static analysis

DL,1,1,SYMM !symmetry conditions in line 1

DL,4,1,SYMM

SFL,2,PRES,-1200/t !apply uniform pressure N/mm in line #2

SOLVE

FINISH



PLESOL,S,X !contour plot of x direction stress



The obtained maximum displacement of the model is 1.813 mm, due to used symmetry condi-

tions, the total elongation of the plate is the double, ∆L = 3.626 mm. Applying the Hooke’s Law

the total elongation is

∆L = FL

E A= q0

axEx t

= 3.586mm

Consequently, both solutions are consistent with each other.





5.3.3 Anisotropic material

The stiffness tensor for an general anisotropic material with linear elasticity is

C =

C11 C12 C13 C14 C15 C16

C22 C23 C24 C25 C26

C33 C34 C35 C36

Sym C44 C45 C46

C55 C56

C66

(5.6)

which is defined by 21 constants.

Hint 20: ANSYS properties for anisotropic linear elastic material

In ANSYS the order of the vector is expected as {x, y, z, xy, yz, xz}, whereas in the Voigt nota-

tion the order is given as {x, y, z, yz, xz, xy}.

The anisotropic values are defined in ANSYS using TB,ANEL,1 and one TBDATA command for

each value (see example next). The sequence number in TBDATA command is obtained following

the rows showed in the above stiffness tensor (C11, C12, C13, C14, C15, C16, C22, C23...).

Anisotropic Elastic Example: Cadmium


TB,ANEL,1

TBDATA,1,121e3 !C11 (MPa)

TBDATA,2,48.1e3 !C12 (MPa)


TBDATA,7,121e3 !C22 (MPa)



TBDATA,16,18.5 !C44 (Pa)

TBDATA,19,18.5 !C55 (Pa)

TBDATA,21,24.2 !C66 (Pa)

Only some elements in ANSYS have the anisotropic elasticity capability (SOLID5, PLANE13,

SOLID64, SOLID98, SHELL181, PLANE182, PLANE183, SOLID185, SOLID186, SOLID187,

among others).

5.4 Element coordinate system

During the definition of the model when we use orthotropic or anisotropic material, we need

to be sure about that the material properties are applied in the correct orientation. In most of

FEA commercial codes, the material coordinate system is associated to the element coordinate

system, one right-handed orthogonal system associated to each element. Also, the element co-

ordinate system can be used to obtain the derived results (strains and stress) in these material

directions.




The orientation of the element coordinate system is associated with the element typology. For

bar or beam elements the orientation of the x-axis is generally along the element line defined

by two end nodes. For solid elements (in two and three dimension), generally the orientation is

defined parallel to the global or geometric coordinate system. However, for shell elements this is

not useful as the x and y directions must be defined on the element surface and the z-axis always

normal to it. The default orientation of x and y axes depends of the commercial code and the

element type.

There are various ways to define the default orientation of x and y in shell elements. Two

of them are shown in Fig. 5.1. In Fig. 5.1(a) the x-axis is aligned with the edge defined by the

first and second nodes of each element, the z-axis normal to the shell surface (with the outward

direction determined by the right-hand rule), and the y-axis perpendicular to the x- and z-axis

(ANSYS uses this rule as default). Other packages, such as MSC.MARC, calculate the orientation

of x-axis from the lines defined by the middles points of the edges as shown in Fig. 5.1(b).

Figure 5.1: Default orientations of element (or material) coordinate systems in shells elements:

(a) ANSYS, (b) MSC.MARC.

In the example 5.1, we have defined only a rectangular plate with rectangular elements and all

of them have the first and the second nodes aligned with the global x-axis. The material axes have

been chosen parallel as the global axis. But this is not the general case. Most commercial codes

have utilities to control or change the element coordinate system. In Example 5.2, it is shown

how it can be done in a shell with curvature, and in Example 5.3, how different orientations in

different locations of the structure can be defined.

Example 5.2. Define a 3D curved shell in ANSYS and mesh it aligning the coordinate system of

the elements with the global y-axis.

Solution to Example 5.2. For shells defined in 3D, the ESYS orientation uses the projection of

the local system on the shell surface. The element x-axis is determined from the projection of the

local x-axis on the shell surface. The z-axis is determined normal to the shell surface (with the

outward direction determined by the right-hand rule), and the y-axis perpendicular to the x- an

z-axis. For elements without midside nodes (linear shape functions), the projection is evaluated

at the element centroid and it is assumed constant in direction throughout the element. For

elements with midside nodes (quadratic shape functions), the projection is evaluated at each

integration point and may vary in direction throughout the element.

In ANSYS a local coordinate system, which can be cartesian, cylindrical or spherical, must

defined using the LOCAL command. Then, each element is associated to a previously defined local

coordinate system using the element property ESYS. The result is the orientation of the x-axis

element coordinate system parallel to x-axis local coordinate system. Also, it is possible to define

element coordinate system orientations by user written subroutines.

The ANSYSTM command sequence to align the element x-axis with the global y-axis for this

example is listed below. You can either type these commands on the command window, or you

can type them on a file, then, on the command window enter /input, file, ext.






/TITLE, Element orientation in a curved shell


ET,1,SHELL281 !element type #1: SHELL281 (6 DOF)



K,1,30,0,2 !define geometry by keypoints and lines

K,2,0,5,5

K,3,-30,0,2

K,4,20,35,0

K,5,0,35,2

K,6,-20,35,0

L,1,4

L,3,6

BSPLIN,1,2,3 !create spline between nodes 1, 2 and 3

BSPLIN,4,5,6 !create spline between nodes 4, 5 and 6

AL,ALL !create a area with all lines

LOCAL,11,0,8,,,90,0,0 !define local coord. system, with label 11,according to cartesian

!coordinate system, with origin in point (8,0,0)

!and rotation about local Z,X,Y equal to (90º,0º,0º)

ESYS,11 !set coord. system for elements meshed

ESIZE,7 !define element size

AMESH,1 !mesh the area #1

CSYS,0 !go back to default coord. system

/PSYMB,ESYS,1 !set on display laminate orientation

/TYPE,1,0 !not hidden surfaces

/VSCALE,1,1.5,0 !change arrow scaling

EPLOT !display elements



The results of the orientation of the elements coordinate systems can be observed in Figure

5.3. The element coordinate systems may be displayed as a triad with the /PSYMB command with

the label ESYS.

Figure 5.2: Element coordinate system orientation, not using "ESYS" command (left) and using

"ESYS" command (right).

Example 5.3. Model in ANSYS a flange pipe with axial and radial material orientation. In the

cylindrical part the reference axis (x-axis) will be in longitudinal direction. In the flange the

reference axis will be radial (see Fig. 5.3).

Solution to Example 5.3. In the model different orientation systems in different model locations

must be defined. Two local reference axes are defined and activated using the ESYS command.

Then, the elements on the cylindrical part are aligned to the axial direction and the elements on





Figure 5.3: Reference x-axis in a pipe flange.

the flange are aligned according to the radial direction. The ANSYSTM command sequence for

this example is listed below. You can either type these commands on the command window, or

you can type them on a file, then, on the command window enter /input, file, ext.



/TITLE, Flange pipe with axial and radial material orientations


ET,1,SHELL281 !element type #1 SHELL281 (6 DOF)



CYL4,0,0,350,,,,300 !create geometry by solid modeling

CYL4,0,0,350,,550

AGLUE,3,4,5 !glue areas, area 5 becomes area 6

LOCAL,11,0,,,,0,0,90 !define rotation=90 deg around Y (cylinder)

LOCAL,12,1,,,,0,0,0 !define polar coordinate system (flange)

ESIZE,50 !define element size


AMESH,3,4 !mesh the cylindrical areas (areas 3 and 4)


AMESH,6 !mesh the flange area (area 6)

CSYS,0 !go back to default coord. system

/PSYMB,ESYS,1 !set on display local orientation

/TYPE,1,0 !not hidden surfaces

EPLOT !display elements



Observe that independently of the geometry of the element, all the element coordinate systems

in the cylindrical part are oriented in the longitudinal direction and those of the flange part are

oriented in the radial direction.


Problem 5.1. Reconsider the rectangular plate made of an orthotropic material described in

Ex.5.1 but introducing the same material properties by using the anisotropic behavior option

command instead of the orthotropic one. Check that the obtained results are identical to those

of Ex. 5.1.

Problem 5.2. Using PLANE182 element, generate a mesh from the spring geometry shown in

Figures 5.4, where L =100 mm, ri =10 mm ,ro =30 mm and thickness equal to 10mm. The spring

is made of the same orthotropic material as in Ex. 5.1. The dashed line on middle of the strip

shows the y-axis of the material orientation. Report graphically the obtained mesh and show the

material axis orientation.





Figure 5.4: Spring geometry.



Chapter 6

Postprocessing

6.1 Introduction

The gathering, interpretation and understanding of the calculated results in a Finite Element

problem is as important as a good definition of the boundary conditions, type of element, etc. In

this chapter, different useful methods to obtain and analyze the calculated results are presented.

All Finite Element programs have the possibility to analyze the obtained results by using two

different methods: general postprocessing and time-history postprocessing. The first method is

intended to review the results for the whole model at specific steps and sub-steps (specific time-

increments of frequencies). With this method, for instance, the stress distribution in the model

can be plotted for a structural analysis. The second method allows to obtain the variation of a

certain variable or result item at particular locations in the model with respect to time, frequency

or other result variables. In a nonlinear structural analysis, for example, the deflection of a

certain node can be plot versus time or the applied load.

Hint 21: ANSYSTM methods for post-processing

• The general postprocessing in ANSYSTM is carried out by using the General Postproc

option in the Main Menu or invoking the /POST1 module in a command script.

• The time-history postprocessing in ANSYSTM is carried out by using the TimeHist Postpro

option in the Main Menu or invoking the /POST26 module in a command script.

A brief description of both methods in ANSYSTM is included in the following sections.

6.2 General postprocessing

The general postprocessor is used to review the results for the entire model or parts of the

model for a specific time, frequency or step. In this module, simple graphic representations

of the results can be obtained as well as listing the results in a table and more complex data

manipulation. By default, different results are directly obtained after a simulation. These results

depend on the type of analysis and the type of element used.

77


Reading results data

The first step to analyze the obtained results is to handle the results for the specific step, sub-step

or time increment desired. By default, most of the commercial Finite Element programs read the

results data for the last calculated increment.

Hint 22: ANSYSTM methods for reading results data

• SET reads all the results data from the results file for the whole model at a particular

time.

• SUBSET reads all the results data from the results file for specific parts of the model at

a particular time. Other commands such as LAYER, TOP, MID, BOTTOM can be used to

specify particular parts of the model where results are to be considered.

• APPEND reads all the results data from the results file for the whole model at a particular

time and adds them to the previously stored results data.

Reading selected results data

Most of the results data, such as stresses, strains, reaction forces and displacements in a struc-

tural analysis, are available during postprocessing in commercial Finite Element codes. In some,

these data results need to be specified beforehand (this is the case of ABAQUSTM , for instance).

In others, not all the data results are directly available in the general postprocessing but have to

be obtained afterwards (this is the case of ANSYSTM ).

Hint 23: ANSYSTM methods for reading selected results data

• In ANSYSTM some results (depending on the analysis and element type) must be accessed

by creating an Element Table, ETABLE.

• ETABLE allows accessing, storing and manipulating data results for afterwards represen-

tation.

• The filling of the element table and the use of the command ETABLE must be specifically

done in accordance to the available results for the considered element type and analysis

type. It is strongly recommended to check which results are available for the element and

analysis type and identify the variable names or numbers that correspond to the desired

results.

Coordinate system for the results data

Commercial Finite Elements codes allow reviewing the results data in different coordinate sys-

tems. Usually, results are calculated and stored in the nodal or the element coordinate systems

and rotated into the active coordinate system, which, by default, is the global cartesian system.



Chapter 6. Postprocessing 79

These results, can be afterwards rotated into any previously defined coordinate system for dis-

play, list and storage.

Hint 24: ANSYSTM coordinate systems for results data

• Use RSYS,0 to obtain the results in the global cartesian coordinate system (default).

• Use RSYS,1 to obtain the results in the global cylindrical coordinate system.

• Use RSYS,2 to obtain the results in the global spherical coordinate system.

• Use RSYS,N to obtain the results in the local coordinate system N .

• Use RSYS,SOLU to obtain the results in the nodal or element coordinate systems.

Manipulation of the results data

Most commercial Finite Elements programs allow the manipulation of the data results, including

arithmetic operations, comparisons, failure criteria definition, etc.

Hint 25: ANSYSTM manipulation of results data

• Use SADD to add two different specified stored results.

• Use SMULT to multiply two different specified stored results.

• SEXP can be used to exponentiate and multiply two different specified stored results.

• SMIN and SMAX allow to compare and obtain the minimum and maximum, respectively, of

two different specified stored results.

• SALLOW can be used to define the allowable stress for safety factor calculations.

• SFCALC calculates safety factors for stored results.

Reviewing results data

All the commercial Finite Elements codes allow reviewing the results data both in graphical plots

and listings.

Graphical plots include different options such as contour displays, deformed shapes, vector

plots, path plots, etc. Contour plots display the variation of the results variable over the model.

Deformed shapes represent the final shape of the system in equilibrium after taking into account

all the boundary conditions and loads. Vector plots show the variation and direction of a vector

quantity result using arrows. Path plots are useful to observe the variation of a results variable

along a predefined path.




Hint 26: ANSYSTM methods for graphically reviewing results data

• PLNSOL displays continuous contour lines over the entire model.

• Use PLESOL to display contour lines that are discontinuous across the elements.

• PLETAB displays contour lines of stored data in element tables.

• Use PLDISP to display the deformed shape of the model.

• PLVECT generates vector displays using arrows.

• PLPATH plots the results data on the predefined path.

Listings are an efficient tools to report resulting nodal, element solution and reaction data.

Hint 27: ANSYSTM methods for listing results data

• Use PRNSOL to list specified nodal solution data.

• Use PRESOL to list specified results for the selected elements.

• PRRSOL lists the node reactions for the selected nodes.

• PRNLD lists the summed element nodal loads for the selected nodes.

Example 6.1. Consider the clamped beam simulated in Chapter 1 and the input file generated

in Ex. 1.2. After simulating the beam, postprocess the results and plot the deformed shape of the

beam, the contour plots for the distribution of longidtudinal stress, equivalent (Von Mises) stress

and principal elastic strain and list the reaction forces for 5th and the last substeps.

Solution to Example 6.1. As it was already explained in Chapter 1, the requested plots can be

obtained by means of the GUI menus, the Results Viewer module or commands. In this case, an

input file with all the required commands is used. The ANSYSTM command sequence for this

example is listed below.


SET,1,5 !set step 5






SET,1,LAST !set last step (default)












Observe that as the analysis is linear elastic, the results for the 5th substep are half of those

for the last.


in Ex. 1.2. After simulating the beam, postprocess the results and obtain the variation of the

equivalent stress along the vertical line defined by the nodes on the centre of the clamped section.

Solution to Example 6.2. The ANSYSTM commands to define the required pathplot, obtain the

results along this pathplot and represent it are summarised in the following input file.


PATH,vertax,2 !definition of PATH "vertax" by 2 points

PPATH,1,,0,0,100 !start node located on X,Y,Z

PPATH,2,,0,100,100 !end node located on X,Y,Z

/PBC,PATH,1 !show path

/REPLOT

PDEF,Svertax,S,X,AVG !project x-stress on path, name it ’Svertax’

PLPATH,Svertax !plot defined ’Svertax’



Observe that as expected for a linear elastic analysis, the variation of the equivalent stress

along the path is linear.

Example 6.3. A clamped plate 100 mm long, 25 mm wide and 5 mm thick is made of AS4D/9310

unidirectional composite material (see Ex. 5.1 for material properties). The material is oriented

at 45◦ with respect to the longitudinal axis of the plate. The plate is bended 10 mm at the free

end. Simulate the plate using shell elements and postprocess the results to obtain the contour

plot of the resulting stresses in the material. As a simple failure criterion, it is assumed that the

material fails when the absolute value of the stress in the fibre direction is equal or greater than

500 MPa. Compute the failure criterion and obtain its contour plot.

Solution to Example 6.3. The ANSYSTM commands to generate, simulate and postprocess the

previous model are listed in the summarised input file.



/TITLE, Clamped plate

L=100 !plate length

B=25 !plate width

h=5 !plate thickness

fi=45 !orientation angle

FA=500 !allowable fibre stress




SECTYPE,1,SHELL !shell section type for material#1

SECDATA,h !section data, with thickness h=5mm

MP,EX,1,133.86E3 !material is UD AS4D/3100 Carbon/Epoxy

MP,EY,1,7.706E3

MP,EZ,1,7.706E3

MP,GXY,1,4.36E3

MP,GYZ,1,2.76E3

MP,GXZ,1,4.36E3





MP,PRXY,1,0.301

MP,PRYZ,1,0.396

MP,PRXZ,1,0.301

RECTNG,0,L,0,B !creates a rectangle with x=L and y=B

ESIZE,5 !element size 5 mm

LOCAL,11,0,,,,fi !define material orientation

ESYS,11 !use material orientation

AMESH,ALL !mesh the area

CSYS,0 !return to global system

FINISH

/SOLU

NSEL,S,LOC,X,0 !select encastred nodes

D,ALL,ALL !fix encastred nodes

NSEL,S,LOC,X,L !select end nodes

D,ALL,UZ,-10 !apply displacement end nodes

NSEL,ALL !select all


FINISH

/POST1

/VIEW,1,1,1,1 !iso-view

PLNSOL,S,X,2,1 !stress in direction 1, global

RSYS,11 !results in material orientation

PLNSOL,S,X,2,1 !stress in direction 1, local

ETABLE,S1,S,X !store fibre stress as S1

SABS,1 !use absolute values

SMULT,FF,S1,,1/FA !store fibre failure as FF

PLETAB,FF,AVG !plot fibre failure, averaged



Observe that the stress contour plots depend on the coordinate system employed. If the global

coordinate system is used, the x-stress is in the longitudinal direction of the plate while when the

element coordinate system is used, the x-stress corresponds to the fibre stress.

The contour plot of the longitudinal stress in the material direction shows a maximum of

789.67 MPa, which is higher than the material allowable stress in the fibre direction. However,

the contour plot for the assumed failure criterion shows a maximum of 0.75, indicating that there

is no material failure. This incoherence is due to the fact that the failure criterion employed

has been calculated using the ETABLE command and element results. Actually, ETABLE assumes a

unique interpoled value for the whole element from the nodal values. In this way, part of the in-

formation is lost and some differences appear. These differences can be reduced by reducing the

size of the element or performing a mesh quality assessment (see Section 6.5). Try to recalculate

the same example using smaller elements and compare the results.

Another possibility to obtain more accurate results, especially for failure prediction, is to use

the nodal results. In ANSYSTM the FC command can be used for this purpose. The following

input file shows an example of the use of the maximum stress failure criterion in the previous

example.

/POST1

!Failure criterion definition and calculation

FC,1,S,XTEN,500 !mat1,stress,max stress 11t

FC,1,S,XCMP,-500 !mat1,stress,max stress 11c

FC,1,S,YTEN,1e6 !mat1,stress,max stress 22t (large value, don’t compute)

!max stress 22c=-22t

FC,1,S,ZTEN,1e6 !mat1,stress,max stress 33t (large value, don’t compute)





!max stress 33c=-33t

FC,1,S,XY,1e6 !mat1,stress,max stress 12 (positive, large value, don’t compute)

FC,1,S,YZ,1e6 !mat1,stress,max stress 23 (positive, large value, don’t compute)

FC,1,S,XZ,1e6 !mat1,stress,max stress 13 (positive, large value, don’t compute)

PRNSOL,S,FAIL !print table with FAIL index

PLNSOL,FAIL,SMAX !maximum stress failure criterion


ftp://amade.udg.edu/mms/PrePostFE/input_files/T6/PrePostFE_Ex603_fail.dat

Observe that in this case the predicted failure index almost doubles that of the initial calcula-

tion.

6.3 Time-history postprocessing

The time-history postprocessor is used to review the results for specific parts of the model as

a function of time or frequency. As in the case of the general postprocessor, it is also possible

to obtain graphic plots and listings as well as perform data manipulation. General uses of the

time-history postprocessing include the variation plot of a certain result as a function of time in

a transient analysis or obtain the force-deflection plot in a nonlinear structural analysis.

Obtaining results data

The first step to analyse the obtained results is to handle the results for the specific part of the

model and assign them to a results variable.

Hint 28: ANSYSTM methods for reading time-history results data

In the time-history postprocessor of ANSYSTM different data results for specific locations are

retreived from the results file by using different commands and assigning them a numerical

label.

• Use NSOL to specify the nodal data from the results file to be stored.

• ESOL specifies the element data from the results file to be stored.

• Use ANSOL to specify the averaged nodal data from the results file to be stored in the

solution coordinate system.

• RFORCE specifies the total reaction force to be stored.

Manipulating results data

Once the time-history result variables are defined, they can be manipulated and combined to

obtain new variables.

Hint 29: ANSYSTM methods for manipulating time-history results data

In the time-history postprocessor of ANSYSTM there are different commands to manipulate

and combine result variables in order to obtain new ones. Some of the more useful are:

ABS, ADD, ATAN, EXP, PROD, SQRT. Check their use in the HELP module.



ftp://amade.udg.edu/mms/PrePostFE/input_files/T6/PrePostFE_Ex603_fail.dat


Reviewing results data

As in the case of the general postprocessing, time-history variables can be both graphically dis-

played and listed.

Hint 30: ANSYSTM methods for reviewing time-history results data

• Use PLVAR to graph the result variables. Up to ten variables can be plot.

• XVAR specifies the result variable to be used for the X-axis in the graphs. By default, is

time in static and transient analyses and frequency in harmonic analyses.

• PRVAR lists up to six specified result variables. Times or frequencies for which results are

to be listed can be controlled with NPRINT and PRTIME.


in Ex. 1.2. After simulating the beam, postprocess the results and obtain the evolution of the

longitudinal stress and strain at the top node of the vertical line on the centre of the clamped

section. Plot the evolution of both variables in a combined plot versus time taking into account a

scale factor if necessary. List the results and plot the variation of the longitudinal stress versus

the longitudinal strain.

Solution to Example 6.4. The ANSYSTM command sequence for obtaining the time-history

postprocessing of this example is listed below.

/POST26 !history plot module

ANSOL,2,708,EPEL,X,EPELX !create evolution of x-strain at node 708, name it as EPELX

!and save it with reference number 2

ANSOL,3,708,S,X,SX !create evolution of x-stress at node 708, name it as SX

!and save it with reference number 3

PROD,4,2,,,EPELX5,,,1e5 !create a new variable, with 4 as number reference and EPELX5

!as name, that takes the variable with reference number 2 and applies

!a scaling factor of 1e5

PLVAR,EPELX5,SX !plot x-strain (scaled) and stress evolution

PRVAR,EPELX,SX !list x-strain and stress evolution

XVAR,2 !set the variable with reference number 2 to be on the x-axis

PLVAR,SX !plot the variable SX versus EPELX (x-stress versus x-strain)



Observe that the longitudinal stress is proportional to the longitudinal strain, as expected for

a linear elastic model. A scale factor of 1× 105 has been considered to plot in the same graph the

evolution of the longitudinal stress and strain.

6.4 Result analysis of combined stresses or strains

When performing a structural simulation, the comparison between the resulting strains or stresses

with an allowable value in a uniaxial model is straightforward. However, when there is more than





one component of strain or stress, these components are normally combined into one resulting

strain or stress and then compared with the allowable value. Especific techniques for strain and

stress combination are presented next.

6.4.1 Combined strain

Principal strains can be calculated from the strain components by the cubic equation:

∣

∣

∣

∣

∣

∣

εx − ε0 γxy/2 γxz/2

γxy/2 εy − ε0 γyz/2

γxz/2 γyz/2 ǫz − ε0

∣

∣

∣

∣

∣

∣

= 0 (6.1)

where ε0 are the three solutions and principal strains. The three principal strains are labeled ǫ1,

ǫ2, and ǫ3 (in ANSYSTM output as 1, 2 and 3 with strain items such as EPEL). The principal strains

are ordered so that ǫ1 is the most positive and ǫ3 is the most negative.

The strain intensity ǫI (output as INT with strain items such as EPEL) is the largest of the

absolute values of ǫ1 - ǫ2, ǫ2 - ǫ3, or ǫ3 - ǫ1. That is:

ǫI = max (|ǫ1 − ǫ2|, |ǫ2 − ǫ3|, |ǫ3 − ǫ1|) (6.2)

The von Mises or equivalent strain ǫe (in ANSYSTM output as EQV with strain items such as

EPEL) is computed as

ǫe =1

1 + ν

(

1

2

(

(ǫ1 − ǫ2)2 + (ǫ2 − ǫ3)

2 + (ǫ3 − ǫ1)2)

)1

2

(6.3)

6.4.2 Combined stress

Principal stresses are calculated from the stress components by the cubic equation:

∣

∣

∣

∣

∣

∣

σx − σ0 σxy σxzσxy σy − σ0 σyzσxz σyz σz − σ0

∣

∣

∣

∣

∣

∣

= 0 (6.4)

where σ0 are the three solutions and principal stresses. The three principal stresses are labeled

σ1, σ2 and σ3 (in ANSYSTM output as S1, S2 and S3). The principal stresses are ordered so that

σ1 is the most positive and σ3 is the most negative.

The stress intensity σI (in ANSYSTM output as SINT) is the largest of the absolute values of

σ1 - σ2, σ2 - σ3, or σ3 - σ1. That is:

σI = max (|σ1 − σ2|, |σ2 − σ3|, |σ3 − σ1|) (6.5)

The von Mises or equivalent stress σe (in ANSYSTM output as SEQV) is computed as

σe =

(

1

2

(

(σ1 − σ2)2 + (σ2 − σ3)

2 + (σ3 − σ1)2)

)1

2

(6.6)

Example 6.5. Graph the contour plots of the principal and equivalent (Von Mises) strains and

stresses for the cantilever beam simulated in Ex. 1.2. Also graph the contour plot of the stress

and strain intensity and a vector plot of the principal stress and strain.




Solution to Example 6.5. The following ANSYSTM input file can be used to obtain the above

strains and stresses contour plots and vector plots.


PLNSOL,S,1 !contour plot of 1st principal stress

PLNSOL,S,2 !contour plot of 2nd principal stress

PLNSOL,S,3 !contour plot of 3th principal stress

PLVECT,S !vector plot of principal stress

PLNSOL, S,INT, !contour plot of stress intensity

PLNSOL, S,EQV, !contour plot of Von Mises stress

PLNSOL,EPEL,1 !contour plot of 1st principal strain

PLNSOL,EPEL,2 !contour plot of 2nd principal strain

PLNSOL,EPEL,3 !contour plot of 3th principal strain

PLVECT,EPEL !vector plot of principal strain

PLNSOL,EPEL,INT !contour plot of strain intensity

PLNSOL,EPEL,EQV !contour plot of Von Mises strain



Observe that in the vector plot of the principal strains and stresses, the maximum principal

stress and strain are located on the top part of the beam, while the minimum principal stress

and strain are located on the bottom part of the beam and represented by an inverted arrow

(negative). In both cases, the direction of the arrows is longitudinal with the beam, indicating

the direction of the principal stress and strain directions. Also observe that the size of the arrow

is proportional to the value of the represented stress or strain.

The equivalent (Von Mises) stress and strain can be also calculated using the ETABLE com-

mand. However, as in the case of Ex. 6.3, there is a certain difference between the values

calculated in this way and the values directly obtained with ANSYSTM . The following input file

can be used to calculate the Von Mises stress for the current example.


ETABLE,Sp1,S,1 !store max principal stress in Sp1

ETABLE,Sp2,S,2

ETABLE,Sp3,S,3

SADD,Si1,Sp1,Sp2,,-1 !store Si1 as Sp1-Sp2

SADD,Si2,Sp2,Sp3,,-1

SADD,Si3,Sp3,Sp1,,-1

SEXP,Si11,Si1,,2 !store Si11 as (Sp1-Sp2)^2

SEXP,Si22,Si2,,2

SEXP,Si33,Si3,,2

SADD,Si12,Si11,Si22 !store Si12 as (Sp1-Sp2)^2+(Sp2-Sp3)^2

SADD,Si123,Si12,Si33 !store Si123 as (Sp1-Sp2)^2+(Sp2-Sp3)^2+(Sp3-Sp1)^2

SMULT,S123,Si123,,0.5 !store S123 as Si123/2

SEXP,Seq,S123,,0.5 !store Seq1 as S123^0.5

PLETAB,Seq,avg


ftp://amade.udg.edu/mms/PrePostFE/input_files/T6/PrePostFE_Ex605_eqv.dat




ftp://amade.udg.edu/mms/PrePostFE/input_files/T6/PrePostFE_Ex605_eqv.dat


6.5 Assessment of the mesh and results quality

As mentioned in Section 3.4, it is necessary to assess the mesh quality to be confident on the

results of a mechanical or structural simulation. That is, to check if the mesh is fine enough to

get good simulations results. Basically, the following aspects need to be analysed:

• Assure the right connectivity between different parts of the mesh.

• Avoid elements with high distortions, warping and inadmissible narrow or wide angles of

the element edges.

• Error estimation of the elements, typically by means of energy hypotheses.

• Study the sensitivity of the results on different meshes of the entire FE model or of a partial

model.

From the above, the error estimation of the elements by means of energy hypotheses is carried

out during the postprocessing. The mesh quality assessment in this way is explained next.

6.5.1 Energy norm to estimate the error of the elements

The finite element solution is an approximation to the true solution of a mathematical prob-

lem. For the analyst is important to know the magnitude of error involved in the solution. The

ANSYSTM program offers a method for a posteriori estimation of the solution error due to mesh

discretisation. The method involves calculating the energy error within each finite element and

expressing this error in terms of a global error energy norm.

The primary unknown in conventional finite element analysis is the displacement at every

node in the model. The finite elements used in the model have shape functions associated with

them that characterize the resulting displacement field of the physical model. The principal of

minimization of total potential energy is used to obtain the equation in the form of:

[K] {x} = {F} (6.7)

where [K] is the global stiffness matrix, {x} is the displacement vector, and {F} is the force vec-

tor.

This equation is solved for the unknown displacement vector {x}. The stresses and strains are

then obtained as the first derivative of these displacements. So while the displacement field in

the finite element model is continuous, the stress field in the model is dicontinuous. The stress at

a node, as printed from ANSYSTM , is the average of the stresses from all the elements attached

to that node. This introduces an error in the magnitude of stress at a node and is referred to as

mesh discretization error. The phenomenon of mesh discretization error is graphically illustrated

in Figure 6.1. The coarser the mesh, the greater the potential for this error. Similarly, for a very

fine mesh, the contribution to stress at a node from all elements attached to it will be the same,

exact value. However, rarely do we have the luxury of repeated mesh refinement.

This way, the nodal stress error vector is defined as:

{∆σin} = {σa

n} − {σin} (6.8)

where {∆σin} is the stress error vector of node n of element i, {σa

n} is the average stress vector

at node n, and {σin} is the stress vector of node n of element i. As an example, the nodal stress

error vector of node 13 of element 6, {∆σ613}, in Figure 6.1 can be computed as:




1 52 43

6 107 98

11 1512 14

16 2017 1918

21 2522 2423

1 2 3 4

5 6 7 8

14 15 1613

9 10 11 12

σx=50

τxy=10

σy=15

σx=100

τxy=20

σy=30

σx=150

τxy=30

σy=45

σx=200

τxy=40

σy=60

Figure 6.1: Mesh discretization error.

σx13 =50 + 100 + 150 + 200

4= 125; σy13 = 37.5; τxy13 = 25 (6.9)

{σa13}

T = {125 37.5 25}T (6.10)

{σ6

13}T = {50 15 10}T (6.11)

{∆σ6

13}T = {σa

13}T − {σ6

13}T = {75 22.5 15}T (6.12)

Once all the nodal stress error vectors are computed, the energy error for each element can

be defined as:

Ei =1

2

∫

V(∆σi

n)jkDkl(∆σin)lidV (6.13)

where Ei is the energy error of element i and Dkl is the constitutive tensor.

By adding all element energy errors Ei, the global energy error in the model, E, can be

determined:

E =

Ne∑

i=1

Ei (6.14)

The energy error can be normalized against the strain energy U and expressed as a percent

error in energy norm, E:

ERR =

(

E

U + E

)1

2

% (6.15)

The percent error in energy norm E is a good overall global estimate of the discretization

or mesh accuracy. According to this, in ANSYSTM , the structural energy error (labeled SERR)

is a measure of the discontinuity of the stress field from element to element, and the thermal

energy error (TERR) is a measure of the discontinuity of the heat flux from element to element.

With these energy errors (SERR and TERR) ANSYSTM program calculates a percent error in energy




norm (SEPC for structural percent error and TEPC for thermal percent error). PRERR command

lists SEPC and TEPC for all selected elements, and PLESOL displays the contours of SERR or TERR

to find out where to refine the mesh.

Example 6.6. Evaluate the accuracy of the mesh and the solution of the FE model used to solve

problem 4.5. Plot nodal stresses, element stresses, and the percentage of energy error. Evaluate

two different meshes.

Solution to Example 6.6. Two different element types and two different mesh refinements are

considered in order to establish a comparison. The first element type considered is a 8-noded

2D solid element, PLANE183, while the second is the 4-noded version of the same, PLANE182. The

first mesh corresponds to an element size of about 5.5 mm, while in the second the element size

is of 1.5 mm.

Before getting the energy error quantities the PowerGraphics option of ANSYSTM has to be

disabled, /GRAPHICS,FULL. To plot the percentage of structural energy error of each element

type, PLESOL,SERR,,0,1.0. To list the total pencentage of energy error of each element type

PRERR.

The first model to be analyzed corresponds to a mesh with 8-noded 2D solid elements and an

element size of 5.5mm. The mesh obtained and the energy error of each element are shown in

Figure 6.2. The total percentage error in energy norm returned by ANSYSTM is STRUCTURAL

PERCENTAGE ERROR IN ENERGY NORM (SEPC) = 5.0734. This difference is appreciated if the

nodal and the element quantities for the stress in the direction perpendicular to the load axis

are compared. These stresses are shown in Figure 6.3.

Figure 6.2: Mesh and energy error with elements of nominal size of 5.5 mm and 8-noded plane

elements.

The results obtained when 4-noded plane elements of nominal size 5.5 mm are used are shown

in Figure 6.4. The total percentage error in energy norm returned by ANSYS in this case is

STRUCTURAL PERCENTAGE ERROR IN ENERGY NORM (SEPC) = 11.389. The difference between the

nodal and the element quantities for the stress in the direction perpendicular to the load axis is

higher than when 8-noded elements are used, as it is shown in Figure 6.5.




Figure 6.3: Nodal (left) and element (right) stress in the direction perpendicular to the load axis

with elements of nominal size of 5.5 mm and 8-noded plane elements.


elements.



Using plane stress elements with 8-noded (PLANE183) and a mesh with elements with a nom-




inal size of 1.5mm, the energy error is smaller as it is shown in Figure 6.6. Now, the total percent-

age error in energy norm returned by ANSYS is STRUCTURAL PERCENTAGE ERROR IN ENERGY NORM

(SEPC) = 0.28475. The difference between the nodal and the element quantities for the stress

in the direction perpendicular to the load axis is smaller than in previous analysis as it is shown

in Figure 6.7.


elements.



For comparison purpouses, the last simulation considers plane stress elements with 4-noded

(PLANE182) and a mesh with elements with a nominal size of 1.5mm (see Figures 6.8 and 6.9).

In this last case, the total percentage error in energy norm returned by ANSYS is STRUCTURAL

PERCENTAGE ERROR IN ENERGY NORM (SEPC) = 0.4.9002. This energy error is larger than the

one of the 8-noded and 1.5mm size case, but lower than the one of the 8-noded and 5.5mm size

case (see summary in Table 6.1).





elements.



plane183 PLANE182

8-noded 4-noded

Element size1.5 mm 0.28475 4.9002

5.5 mm 5.0734 11.389

Table 6.1: Variaton on the percent error energy norm (SEPC) as a function of the type of element

and the element size.




Reminder: Reporting results data

Appart from reviewing the results and assess their quality, reporting the obtained results is

also an important task. This can be done in a simple way by capturing an image of the display

or capturing a list. In ANSYSTM this can be done in two ways:

1. PlotCtrls>Write Metafile and select Standard Color or Invert White/Black.

2. PlotCtrls>Capture Image and print the captured image, either to a file or in paper.

3. PlotCtrls>Hard Copy and either send the captured image to a printer or save it with a

image file extension (i.e. BMP, Postscript, TIFF, JPEG, PNG).

Moreover, in the case of ANSYSTM , there is also the option to elaborate a report including

images, tables, etc. (File>Report Generator).

For the specific case of ANSYSTM , but also for other commercial Finite Element codes,

the reviewing of the different results data can be more convenient by using the Graphical

User Interface (GUI) than using commands. For instance, you can use the Results Viewer tool

»General Postproc>Results Viewer to plot contour plots, list nodal or element quantities, easily

change substeps and animate the results among many other utilities. Try to repeat the previous

examples using the GUI or Results Viewer and decide which method is more convenient for you.


Problem 6.1. Reconsider the problem in Ex. 2.4. Recalculate it applying symmetry conditions

on the meshes listed below. Evaluate the energy error norm obtained for the three different

meshes used. Analyse the difference on the nodal stresses and element stresses.

1. A mesh using 10 quadrilateral elements in the vertical direction.

2. A mesh using 80 quadrilateral elements in the vertical direction.

3. A free mesh using 10 triangle elements in the vertical direction.

Problem 6.2. Reconsider the problem in Ex. 2.4. Calculate the principal stresses using ETABLE

commands and considering plane stress, see equation 6.16. Compare these results whith the

principal stresses obtained directly from the program. Calculate the relative error between them.

σ1 =σx + σy

2+

√

√

√

√

[

(

σx − σy2

)2

+ τ2xy

]

σ2 =σx + σy

2−

√

√

√

√

[

(

σx − σy2

)2

+ τ2xy

]

(6.16)



Chapter 7

Coupling and constrain equations.

Submodelling

7.1 Coupling and constrain equations. Introductory concepts

In the Finite Element method the relationships between the different degrees of freedom (DOF)

of the nodes in the mesh are defined by the elements. This is represented by the stiffness matrix

as expressed in the next equation:

~F = K · ~δ (7.1)

where K is the stiffness matrix, ~δ is the DOF vector and ~F is the vector containing the equivalent

external loads.

However, some degree-of-freedom related features such as rigid regions, structural joints,

sliding symmetry boundaries, or periodic conditions, cannot be described by means of elements.

A general option which most FE commercial codes offer is the possibility of defining special asso-

ciations, or internodal connections between nodal DOF in ways that elements are not able. One

possibility is to use equalities between a DOF of a set of nodes (coupling or linking DOF). Another

option is to define equations between different DOF (constraint equations). Both techniques add

equations to equation (7.1).

7.1.1 Coupled DOF sets

Coupling a degree of freedom of a set of nodes imposes that the DOF results calculated for one

member of the set must be the same for all the members in the set for this particular coupled

DOF. Coupling can be used to model various joint and hinge effects. A more general form of

coupling can be done with constraint equations (see section 7.1.2).

For structural analyses, the DOF available for a node depend upon the DOF associated to the

element type. For example, degrees of freedom available with two-dimensional beam elements

are UX, UY, and ROTZ only. A set of coupled nodes in a nodal coordinate direction are forced

to take the same displacement (or rotation) in the specified direction. Coupling operates in the

nodal coordinate system of each node coupled. Therefore, you should usually keep the nodal

coordinate systems consistent.

In commercial codes, one of the DOF of the coupled set is the reference (or prime) DOF (in

ANSYS is the first DOF on the coupled set). All other degrees of freedom in the coupled sets are

95


eliminated from the solution matrices by their relationship to the reference DOF. For this reason,

DOF with specified constraints (grounded DOF) should not be included in a coupled set and the

coupled DOF must not appear in more than one coupled set.

Hint 31: Useful ANSYS commands to define or modify Coupled DOF sets


Defines (or adds to) a set of coupled DOF CP nset,label,nd1,nd2,nd3,...

Defines (or adds to) a set of coupled DOF CPNGEN nset,label,n1,n2,ninc

List coupled set DOF CPLIST cp1,cp2,cpinc

Delete coupled set DOF CPDELE cp1,cp2,cpinc

Example 7.1. Model a structure that consists of two pipes of 10 mm in diameter and 2 mm thick

and a steel bar of cross-sectional area 20 mm2. One of the beams has a distributed load of 5

N/mm2 on its edge. The material is steel with E = 210 GPa and ν = 0.3. The two beams are

joined by a pin. Model this problem using beam type elements for beams and link bar elements

for the bar.

200

5 N/m

m2

Steel pipe: Φ10x2

Figure 7.1: Pin joint structure (Dimensions in mm).

Solution to Example 7.1. The joint between the two beam-pipes can be solved using a common

node because the beam element have rotational z-axis DOF. For this reason, the coupling DOF

of UX and UY is used there, defining a cylindrical joint in this point. The ANSYSTM command

sequence for this example is listed below. You can either type these commands on the command

window, or you can type them on a file, then, on the command window enter /input, file, ext.

FINISH

/CLEAR

/TITLE, Pinned structure




ET,2,LINK180 !element type #2: 3D truss (3 DoF)

SECTYPE,1,BEAM,CTUBE !section #1=hollow tube


SECDATA,3,5 !section internal and external radius

R,2,20 !geometric properties bar: cross-section

MP,EX,1,210e3 !material #1: Young’s modulus




Chapter 7. Coupling and constrain equations. Submodelling 97

N,1,-100 !define node #1 on coordinates

N,11,0,50 !pinned node of left steel pipe

N,12,0,50 !pinned node of right steel pipe

N,22,100

FILL,1,11 !define nodes between #1 and #11

FILL,12,22

CP,1,UX,11,12 !define coupling DOF #1: UX direction node set #11 and #12

CP,2,UY,11,12 !define coupling DOF #2: UY direction node set #11 and #12

TYPE,1 !select BEAM element type (#1) and real constant set #1

REAL,1

E,1,2 !define BEAM elements

EGEN,10,1,1

E,12,13

EGEN,10,1,11

TYPE,2 !select TRUSS element type (#2) and real constant set #2

REAL,2

E,6,17 !define TRUSS elements for rope

FINISH


ANTYPE,STATIC

D,1,UX,,,,,UY,UZ,ROTX,ROTY !displacement X and Y for node #1

D,22,UX,,,,,UY,UZ,ROTX,ROTY !displacement X and Y for node #22

SFBEAM,1,2,PRES,5 !apply distributed load on element #2

*REPEAT,10,1 !repeats 9 times more the previous command incrementing +1 number element

/PBC,ALL !show BC’s when solve


FINISH



/ESHAPE,1,1




Observe that the structure deforms in such a way that the pinned-node remains part of the

two pipes, imposing common displacements at this point, but allowing them to rotate indepen-

dently from each other.

Forces applied to coupled nodes (in the coupled DOF direction) will be summed up and ap-

plied to the reference DOF. Output forces are also summed at the reference DOF. A set of coupled

nodes which are not coincident, or which are not along the line of the coupled displacement di-

rection, may produce an applied moment which will not appear in the reaction forces. If the

structure rotates, a moment may be produced in the coupled set in the form of a force couple.

This moment is in addition to the real reaction forces and may make it appear that moment equi-

librium is not satisfied by just the applied forces and the reaction forces.

Example 7.2. A cantilever T-beam is loaded with a vertical 1 kN force distributed on the end

section (see Fig. 7.2). The T-beam is made of aluminum (E =69 GPa and ν = 0.3) with plates 2.2

mm thick. Obtain the deformed shape and the stress in the Z direction.

Solution to Example 7.2. A coupling of vertical DOF set is applied on the nodes of the end sec-

tion. In this way, applying the total vertical force on one of the nodes of the section is equivalent

to apply a distributed force. The ANSYSTM command sequence for this example is listed below.







Figure 7.2: Cantilever T-beam (Dimensions in mm).

FINISH

/CLEAR

/TITLE, Cantilever T-beam




MP,EX,1,69e3 !material #1: Young’s modulus


R,1,2.2 !real constant set #1: 2.2 mm thickness

K,1,0,0,0 !defines coordinates keypoint #1

K,2,0,20,0

K,3,-15,0,0

K,4,15,0,0

KGEN,3,1,4,1,,,80,4 !generates keypoint #5 to #12

K,10,0,5,160 !redefines keypoint #10

A,1,2,6,5 !defines area #1






ESIZE,5 !set 5 mm elements size

MSHKEY,1


NSEL,S,LOC,Z,160 !select nodes on section z=160

CP,1,UY,ALL !coupling UY on all selected nodes

FINISH


ANTYPE,STATIC

NSEL,S,LOC,Z,0 !BC on z=0 mm

D,ALL,ALL

NSEL,ALL

NODE1=NODE(0,0,160) !set reference node NODE1 on coordinates 0,0,160

F,NODE1,FY,-1e3 !apply total force of 1 kN on z=160


FINISH



/ESHAPE,1,1

PLNSOL,S,Z !contour plot of stress Z distribution







7.1.2 Constraint equations of DOF

Linear constraint equations provide a more general way of relating degree of freedom values

than it is possible with simple coupling. Constraint equations must have the form:

A =n∑

i=1

αiUi (7.2)

where A is a constant, Ui is the degree of freedom of term i, αi is the coefficient of term i, and n

is the number of terms in the equation.

The first unique degree of freedom in the equation is eliminated in terms of all other degrees

of freedom in the equation. A unique degree of freedom is one which is not specified in any other

constraint equation, coupled node set, specified displacement set, or master degree of freedom

set. It is recommended that the first term of the equation be the degree of freedom to be elim-

inated. The first term of the equation cannot contain a master degree of freedom, and no term

can contain coupled degrees of freedom. The same degree of freedom may be specified in more

than one equation but care must be taken to avoid over-specification (over-constraint).

The degrees of freedom specified in the equation (i.e., UX, UY, ROTZ, etc.) must also be in-

cluded in the model (as determined from the element types [ET]). Also, each node in the equation

must be defined on an element (any element type containing that degree of freedom will do).

Hint 32: Useful ANSYS commands to define constraint equations


Defines (or adds to) a set of coupled DOF CE neq,A,nd1,lab1,coeff1,...

List constraint equation CELIST ce1,ce2,ceinc

Delete constraint equation CEDELE ce1,ce2,ceinc

Example 7.3. A composite material has two different components distributed in a regular array

(see Figure 7.3). The matrix material has a Young’s modulus of 5 GPa, while for the rectangular

particles is 150 GPa. Model one RVE (Representative Volume Element) of this geometry with

appropriate periodic conditions. Note, take a = b = 5 mm.




Figure 7.3: RVE example.

Hint 33: Two-dimensional periodic conditions for a rectangular RVE

In an rectangular RVE, such as shown in Figure 7.3, four faces (x1 = ±a and x2 = ±b) and the

four edges defined by these faces need to be considered to define the constraint equations to

apply the periodic conditions.

On the periodic pair of faces x1 = ±a and x2 = ±b, the constrain equations are

u1(a, x2)− u1(−a, x2)− 2aε011

= 0

u2(a, x2)− u2(−a, x2)− 2bε021

= 0∀x2 ∈ (−b,+b) (7.3)

u1(x1, b)− u1(x1,−b)− 2aε012

= 0

u2(x1, b)− u2(x1,−b)− 2bε022

= 0∀x1 ∈ (−a,+a) (7.4)

The intersection of planes x1 = ±a and x2 = ±b define two pairs of periodic edges, for

which the following equations apply,

u1(+a,+b)− u1(−a,−b)− 2aε011

− 2bε012

= 0

u2(+a,+b)− u2(−a,−b)− 2aε021

− 2bε022

= 0(7.5)

u1(+a,−b)− u1(−a,+b)− 2aε021

+ 2bε012

= 0

u2(+a,−b)− u2(−a,+b)− 2aε021

+ 2bε022

= 0(7.6)




FINISH

/CLEAR

/TITLE, Square RVE with two materials


a=5 !geometry of the cell

b=5




a1=3*(a/5) !geometry of the material #2

b1=2*(b/5)

a2=1*(a/5) !position of center of the material

b2=1*(b/5)

epxx=0.001 !horizontal strain

epyy=0.000 !vertical strain

epxy=0.001 !shear strain







BLC5,0,0,(2*a),(2*b) !rectangle, centered on 0,0, size 2a x 2b

LESIZE,ALL,,,10 !10 divisions per edge

TYPE,1

MAT,1

AMESH,1 !mesh area #1

NSEL,S,LOC,X,(-(a1+(a/100))+a2),((a1+(a/100))+a2) !select nodes material #2

NSEL,R,LOC,Y,(-(b1+(b/100))+b2),((b1+(b/100))+b2)

ESLN,S,1 !select elements material #2

EMODIF,ALL,MAT,2 !set material #2

ALLSEL,ALL !reselect all items

/PNUM,MAT,1

EPLOT !plot elements

!periodic conditions of vertex (2 couple of vertex)

CE,1,(epxx*(a*2))+(epxy*(b*2)),1,UX,-1,12,UX,1

CE,2,(epxx*(a*2))-(epxy*(b*2)),22,UX,-1,2,UX,1

CE,3,(epyy*(b*2))+(epxy*(a*2)),1,UY,-1,12,UY,1

CE,4,(epyy*(b*2))-(epxy*(a*2)),2,UY,-1,22,UY,1

!periodic conditions of vertical sides (9 couple of nodes)

CE,11,(epxx*(a*2)),32,UX,-1,21,UX,1 !uniaxial strain XX

CE,12,(epxx*(a*2)),33,UX,-1,20,UX,1

CE,13,(epxx*(a*2)),34,UX,-1,19,UX,1

CE,14,(epxx*(a*2)),35,UX,-1,18,UX,1

CE,15,(epxx*(a*2)),36,UX,-1,17,UX,1

CE,16,(epxx*(a*2)),37,UX,-1,16,UX,1

CE,17,(epxx*(a*2)),38,UX,-1,15,UX,1

CE,18,(epxx*(a*2)),39,UX,-1,14,UX,1

CE,19,(epxx*(a*2)),40,UX,-1,13,UX,1

CE,21,(epxy*(a*2)),32,UY,-1,21,UY,1 !shear strain XY

CE,22,(epxy*(a*2)),33,UY,-1,20,UY,1

CE,23,(epxy*(a*2)),34,UY,-1,19,UY,1

CE,24,(epxy*(a*2)),35,UY,-1,18,UY,1

CE,25,(epxy*(a*2)),36,UY,-1,17,UY,1

CE,26,(epxy*(a*2)),37,UY,-1,16,UY,1

CE,27,(epxy*(a*2)),38,UY,-1,15,UY,1

CE,28,(epxy*(a*2)),39,UY,-1,14,UY,1

CE,29,(epxy*(a*2)),40,UY,-1,13,UY,1

!periodic conditions of horizontal sides (9 couple of nodes)

CE,32,(epyy*(b*2)),3,UY,-1,31,UY,1 !uniaxial strain yy

CE,33,(epyy*(b*2)),4,UY,-1,30,UY,1

CE,34,(epyy*(b*2)),5,UY,-1,29,UY,1

CE,35,(epyy*(b*2)),6,UY,-1,28,UY,1

CE,36,(epyy*(b*2)),7,UY,-1,27,UY,1

CE,37,(epyy*(b*2)),8,UY,-1,26,UY,1

CE,38,(epyy*(b*2)),9,UY,-1,25,UY,1

CE,39,(epyy*(b*2)),10,UY,-1,24,UY,1

CE,40,(epyy*(b*2)),11,UY,-1,23,UY,1

CE,43,(epxy*(b*2)),3,UX,-1,31,UX,1 !shear strain XY

CE,44,(epxy*(b*2)),4,UX,-1,30,UX,1

CE,45,(epxy*(b*2)),5,UX,-1,29,UX,1




CE,46,(epxy*(b*2)),6,UX,-1,28,UX,1

CE,47,(epxy*(b*2)),7,UX,-1,27,UX,1

CE,48,(epxy*(b*2)),8,UX,-1,26,UX,1

CE,49,(epxy*(b*2)),9,UX,-1,25,UX,1

CE,50,(epxy*(b*2)),10,UX,-1,24,UX,1

CE,51,(epxy*(b*2)),11,UX,-1,23,UX,1

FINISH


NSEL,S,LOC,X,0 !select central node

NSEL,R,LOC,Y,0

D,ALL,ALL !central node constrained

ALLSEL


FINISH



PLNSOL,U,SUM !total displacement contour

PRNSOL,S !list stress state of nodes



See how the obtained solution is symmetric in different directions for different parameters.

To simplify the definition of the periodic conditions, they can be defined by a loop, see the

following command sequence.

FINISH

/CLEAR

/TITLE, Square RVE with two materials


a=5 !geometry of the cell

b=5

a1=3*(a/5) !geometry of the material #2

b1=2*(b/5)

a2=1*(a/5) !position of center of the material

b2=1*(b/5)

epxx=0.001 !horizontal strain

epyy=0.000 !vertical strain

epxy=0.001 !shear strain







BLC5,0,0,(2*a),(2*b) !rectangle, centered on 0,0, size 2a x 2b

LESIZE,ALL,,,10 !10 divisions for edge

TYPE,1

MAT,1

AMESH,1 !mesh area #1

NSEL,S,LOC,X,(-(a1+(a/100))+a2),((a1+(a/100))+a2) !select nodes material #2

NSEL,R,LOC,Y,(-(b1+(b/100))+b2),((b1+(b/100))+b2)

ESLN,S,1 !select elements material #2

EMODIF,ALL,MAT,2 !set material #2

ALLSEL,ALL !reselect all items

/PNUM,MAT,1


!periodic conditions of vertex (2 couple of vertex)





CE,1,(epxx*(a*2))+(epxy*(b*2)),1,UX,-1,12,UX,1

CE,2,(epxx*(a*2))-(epxy*(b*2)),22,UX,-1,2,UX,1

CE,3,(epyy*(b*2))+(epxy*(a*2)),1,UY,-1,12,UY,1

CE,4,(epyy*(b*2))-(epxy*(a*2)),2,UY,-1,22,UY,1

!periodic conditions

j=0

*DO,i,1,9,1

CE,i+10,(epxx*(a*2)),i+31,UX,-1,21-j,UX,1 !uniaxial strain XX

CE,i+20,(epxy*(a*2)),i+31,UY,-1,21-j,UY,1 !shear strain XY

CE,i+31,(epyy*(b*2)),i+2,UY,-1,31-j,UY,1 !uniaxial strain YY

CE,i+42,(epxy*(b*2)),i+2,UX,-1,31-j,UX,1 !shear strain XY

j=j+1

*ENDDO

FINISH


NSEL,S,LOC,X,0 !select central node

NSEL,R,LOC,Y,0

D,ALL,ALL !central node constrained

ALLSEL


FINISH



PLNSOL,U,SUM !total displacement contour

PRNSOL,S !list stress state of nodes



7.2 Submodelling

As was introduced in Chapter 3, finite-element software users generally establish in the meshing

process some equilibrium between the available computational resources (where coarser meshes

are preferred) and the needs of the solved problem (where finer meshes may be needed). More-

over, some structural components may exhibit nearly constant stress through its dominium but a

small area showing large stress concentration. Submodelling is a useful tool to analyze parts of

a model which exhibit stress concentrations or which need a detailed geometrical modelling.

Submodelling is also known as the cut boundary displacement method or the specified bound-

ary displacement method. The cut boundary is the boundary of the submodel which represents a

cut through the global model. Displacements calculated on the cut boundary of the coarse model

are specified as boundary conditions for the submodel.

Submodelling is based on St. Venant’s principle, which states that if an actual distribution

of forces is replaced by a statically equivalent system, the distribution of stress and strain is al-

tered only near the regions of load application. The principle implies that stress concentration

effects are localized around the concentration; therefore, if the boundaries of the submodel are

far enough away from the stress concentration, reasonably accurate results can be calculated in

the submodel.

Aside from the obvious benefit of giving more accurate results in a region of your model, the

submodelling technique has other advantages:

• It reduces, or even eliminates, the need for complicated transition regions in solid finite

element models.





• It enables you to experiment with different designs for the region of interest (different fillet

radii, for example).

Some restrictions for the use of submodelling are:

• It is valid only for solid elements and shell elements.

• The principle behind submodelling assumes that the cut boundaries are enough far away

from the stress concentration region. You must verify that this assumption is adequately

satisfied.

In the following sections you can find a summary of Chapter 9 of ANSYS 13 Help, regarding

submodelling.

7.2.1 The global model

In many cases, the coarse model does not need to include local geometric details. However, the

finite element mesh must be fine enough to produce a reasonably accurate degree of freedom

solution. This is important because the results of the submodel are almost entirely based on

interpolated degree of freedom results at the cut boundary.

7.2.2 The submodel

In the submodel the same element type (solid or shell) that was used in the global model should

be employed1. Also, the same element real constants (such as shell thickness) and material prop-

erties should be introduced.

Specify appropriate node rotations. Node rotation angles on cut boundary nodes should not

be changed after they have been written to the node file in interpolation step 1. Command:

NROTAT 2.

The presence or absence of node rotation angles in the coarse model has no effect upon the

submodel.

7.2.3 Submodelling procedure

The basic steps for any submodelling procedure are:

1. Perform Cut Boundary Interpolation. This is the key step in submodelling. You identify

the nodes along the cut boundaries, and the finite element program calculates the DOF val-

ues (displacements) at those nodes by interpolating results from the global model. For each

node of the submodel along the cut boundary, the program uses the appropriate element

from the coarse mesh to determine the DOF values. These values are then interpolated

onto the cut boundary nodes using the element shape functions.

1Shell-to-solid may also be used but will not be covered in this chapter. See ANSYS help.2Be aware that node rotation angles might be changed by application of nodal constraints [DSYM], by transfer of

line constraints [SFL], or by transfer of area constraints [SFA], as well as by more obvious methods [NROTAT and

NMODIF]




Hint 34: Cut boundary interpolation in ANSYS. Global model

• Identify and write the cut boundary nodes of the submodel to a file (Jobname.NODE by

default). You can do this in PREP7 by selecting nodes along the cut boundaries and then

using one of these methods to write the nodes to a file with command NWRITE (you perform

temperature interpolation see ANSYS help).

• To restore the full set of nodes, use command ALLSEL.

• To write the database to Jobname.DB, use command SAVE.

• To do the cut boundary interpolation (and the temperature interpolation), the database

must contain the geometry for the coarse model. Therefore, you must resume the database

use command RESUME, making sure to identify the name of the coarse model database file.

• Enter POST1 with command /POST1. Interpolation can only be performed in POST1.

• Point to the coarse results file with command FILE.

• Read in the desired set of data from the results file with command SET.

• Initiate cut boundary interpolation with command CBDOF. By default, the CBDOF command

assumes that the cut boundary nodes are on file Jobname.NODE. The ANSYS program will

then calculate the cut boundary DOF values and write them in the form of D commands

to the file Jobname.CBDO.

• All interpolation work is now done, so leave POST1 (FINISH) and restore the submodel

database (RESUME).

2. Analyze the Submodel. In this step, you define the analysis type and analysis options,

apply the interpolated DOF values (and temperatures), define other loads and boundary

conditions, specify load step options, and obtain the submodel solution. It is important that

on the submodel any other loads and boundary conditions that existed on the coarse model

are introduced as well. Examples are symmetry boundary conditions, surface loads, inertia

forces (such as gravity), concentrated force loads, etc.




Hint 35: ANSYS commands for the Submodel

• The first step is to enter SOLUTION (/SOLU).

• Define the appropriate analysis type (usually static) and analysis options.

• To apply the cut boundary DOF constraints, simply read in the file of D commands (created

by CBDOF) with command /INPUT,,CBDO.

• Specify what DOF constraint values and nodal body force loads are to be accumulated

with commands DCUM,ADD and BFCUM,,ADD. Be sure to reset the DCUM and BFCUM com-

mands to their default status before proceeding.

• Then specify load step options (such as output controls) and initiate solution calculations

using Command(s): SOLVE.

• After the solution is obtained, leave SOLUTION (FINISH).

3. Verify the Distance Between the Cut Boundaries and the Stress Concentration. The

final step is to verify that the cut boundaries of the submodel are far enough away from the

concentration. You can do this by comparing results (stresses, magnetic flux density, etc.)

along the cut boundaries with those along the corresponding locations of the coarse model.

If the results are in good agreement, it indicates that proper cut boundaries have been cho-

sen. Otherwise, you will need to recreate and reanalyze the submodel with different cut

boundaries further away from the region of interest.

Example 7.4. Model a variation of the hydraulic dam considered in Ex. 2.5 using the sub-

modelling procedure according to what is shown in Figure 7.4. Start with a simple model with

geometric parameters h1 = 10 m, h2 = 5 m, L = 2 m and W = 1 m and consider that the dam is

subjected to a uniform 10 kg/cm2 pressure at its deepest end.




FINISH

/CLEAR

/TITLE, Dam submodeling - Global Model

/FILNAME,Dam_Global


h1=10000 !total wall height

h2=5000 !height

L=8000 !length

W=2000 !width




KEYOPT,1,3,2 !element type #1: keyoption 3=2 (plane strain)

MP,EX,1,27000 !Young Modulus

MP,PRXY,1,0.25 !Poisson’s ratio

K,1,0,0 !define geometry by keypoints




h1

h2

L

W

Cut Boundary

Submodel

Figure 7.4: Submodelling of a dam.

K,2,L,0

K,3,L,(h1-h2)/2

K,4,L,h1-2*h2/3

K,5,L,h1

K,6,L-W,h1

K,7,L-W,h1-2*h2/3

K,8,L-W,h1-h2

K,9,(L-W)/2,(h1-h2)/2

K,10,L-W/2,(h1-h2)/2

K,11,L-W/2,h1-2*h2/3

L,1,2,40 !define lines by keypoints

L,2,3,20

L,3,4,20

L,4,5,20

L,5,6,10

L,6,7,20

L,7,8,10

L,8,9,20

L,9,1,20

L,9,10,15

L,10,11,20

L,11,7,5

AL,1,2,3,4,5,6,12,11,10,9 !create areas from lines

AL,10,11,12,7,8

MSHKEY,2 !use mapped meshing if possible


SAVE !save the file GLOBAL.db

FINISH


DL,1,1,ALL !fix the line 1 of area 1

LSEL,S,LINE,,2,4,1 !select lines 2,4,1

SFL,ALL,PRES,p !apply the pressure

ALLSEL !select all

SOLVE

FINISH



PLNSOL,S,EQV !show Von Mises stress

SAVE,ALL !save solution






Observe that a stress concentration is produced in the corner where the straight part from

the dam and its basis unite.

Example 7.5. Introduce a round corner, that is, a more detailed geometry, in the cross-section

of the previous hydraulic dam and analyze the results. Set the fillet radius to 2 m.




FINISH

/CLEAR

/TITLE, Dam submodeling - Submodel

/FILNAME,Dam_Submodel


h1=10000 !total wall height

h2=5000 !height

L=8000 !length

W=2000 !width

divl=100 !element size

RAD=2000 !fillet radius






K,7,L-W,h1-2*h2/3 !define geometry by keypoints

K,8,L-W,h1-h2

K,9,(L-W)/2,(h1-h2)/2

K,10,L-W/2,(h1-h2)/2

K,11,L-W/2,h1-2*h2/3

L,7,8 !define lines by keypoints

L,8,9

L,9,10

L,10,11

L,11,7

LFILLT,1,2,RAD !define fillet between intersection lines #1 and #2

LESIZE,ALL,RAD*2*3.14159/96 !the divisions of the fillet radius

AL,1,2,3,4,5,6 !define area by lines



NROTAT,ALL !rotate nodal coordinate sys into local

LSEL,S,LINE,,3,5,1 !select cut boundary lines

NSLL,S !select nodes on lines

NWRITE !write node list and coordinates into file SUBMODEL.node

ALLSEL !select all

SAVE !save the file SUBMODEL.db

RESUME,Dam_Global,DB !resume Global Model previously saved


FILE,Dam_Global !point to the global model results file

SET !defines the data set to be read from the results file GLOBAL.db

CBDOF !perform cut boundary interpolation (file SUBMODEL.cbdo)

FINISH

RESUME,Dam_Submodel,DB !resume the file SUBMODEL.db


DCUM,ADD !cumulative loads

/INPUT,,CBDO !input BC from global model (file SUBMODEL.cbdo)





DCUM,REPL !cumulative DOF, replace

SOLVE

FINISH






Note that the submodel allows for a more detailed analysis of the corner or fillet area. In fact,

the maximum Von Mises stress in the submodel is about 23 MPa, while in the global model it is

approximated to about 31 MPa.

Example 7.6. Use ANSYSTM to model the steel trapezoidal cantilever beam with holes shown in

Figure 7.5. The beam is subjected to a uniform pressure of 20 MPa applied on its top edge. The

diameter of the holes is 7 mm. Use the submodel technique to obtain the Von Mises stress in the

area of the holes. Consider a plane strain model.

35

35

50

55

55

350

20 MPa

510

10

750

80

Figure 7.5: Trapezoidal cantilever beam with holes.

Solution to Example 7.6. First, only the general beam, without the small holes, is analyzed to

obtain the global model. The ANSYSTM command sequence for this example is listed below. You

can either type these commands on the command window, or you can type them on a file, then,

on the command window enter /input, file, ext.

FINISH

/CLEAR

/TITLE, Cantilever beam submodeling - Global Model

/FILNAME,CANTILEVER_GLOBAL


h1=350 !cantilevered tip height

h2=50 !free tip height

L=750 !length

d=7 !hole diameter

top_off=10 !offset to top holes

p=20 !pressure

holes1x=510 !x coord left holes

holes2x=590 !x coord right holes

sep1=55 !vertical separation left holes

sep2=35 !vertical separation right holes

xmin=490 !x coordinates of submodel





xmax=610






K,1,0,0 !define geometry by keypoints

K,2,L,0

K,3,L,-h2

K,4,0,-h1

L,1,2,40 !define lines by keypoints

L,2,3,20

L,3,4,40

L,4,1,20

AL,1,2,3,4 !create areas from lines



SAVE !save the file as CANTILEVER_GLOBAL.db

FINISH


DL,4,1,ALL !cantilever boundary conditions

LSEL,S,LINE,,1 !select line 1

SFL,ALL,PRES,p !apply pressure to line 1

ALLSEL !select all

SOLVE

FINISH




SAVE,ALL !save the solution



Afterwards, the hole region is analysed by using the submodelling technique incorporating

the results of the global model. The ANSYSTM command sequence for this example is listed

below. You can either type these commands on the command window, or you can type them on a

file, then, on the command window enter /input, file, ext.

FINISH

/CLEAR

/TITLE, Cantilever beam submodeling - Submodel

/FILNAME,CANTILEVER_SUBMODEL


h1=350 !cantilevered tip height

h2=50 !free tip height

L=750 !length

d=7 !hole diameter

top_off=10 !offset to top holes

p=20 !pressure

holes1x=510 !x coord left holes

holes2x=590 !x coord right holes

sep1=55 !vertical separation left holes

sep2=35 !vertical separation right holes

xmin=490 !x coordinates of submodel

xmax=610










K,1,xmin,0 !define geometry by keypoints

K,2,xmax,0

K,3,xmax,-h1+xmax*(h1-h2)/L

K,4,xmin,-h1+xmin*(h1-h2)/L

L,1,2 !define lines by keypoints

L,2,3

L,3,4

L,4,1

AL,1,2,3,4 !create areas from lines

CYL4,holes1x,-top_off,d/2 !generate holes

CYL4,holes1x,-(top_off+sep1),d/2

CYL4,holes1x,-(top_off+2*sep1),d/2

CYL4,holes2x,-top_off,d/2

CYL4,holes2x,-(top_off+sep2),d/2

CYL4,holes2x,-(top_off+2*sep2),d/2

ASEL,S,AREA,,1,7 !select areas 1 to 7

ASBA,1,ALL !substract areas from areas

LESIZE,ALL,d/2 !define the element size



NROTAT,ALL !rotate nodal coordinate sys into local

LSEL,S,LINE,,2,4,2 !select cut boundary lines

NSLL,S !select nodes on lines

NWRITE !write node list into file

ALLSEL !select all

SAVE

RESUME,CANTILEVER_GLOBAL,DB !resume Global Model


FILE,CANTILEVER_GLOBAL !point to the global model results file

SET !data set to be read from the results file

CBDOF !perform cut boundary interpolation

FINISH

RESUME,CANTILEVER_SUBMODEL,DB !resume Submodel


LSEL,S,LINE,,1

SFL,ALL,PRES,p

DCUM,ADD !cumulative loads

/INPUT,,CBDO !input BC from global model

DCUM,REPL !cumulative DOF, replace

SOLVE

FINISH






Observe that although the general tendency of the Von Mises stress is the same for the results

of the global and submodel, in the latter stress concentrations can be observed around the holes,

which cannot be captured by the global model.






Problem 7.1. Using the same geometry and model of Example 3.9, apply a force of 100 N in

one of the holes producing a bending moment on the plate while the other hole is clamped. Use

coupled DOF set to apply the force.

Problem 7.2. Using similar code of Ex. 7.3 model a RVE representative of a masonry wall (see

Figure 7.6). The bricks are 240 mm long and 45 mm thick. The vertical gap between bricks is 30

mm while the horizontal one is 60 mm. Use the results of the model to compute the homogenized

Young’s Modulus of the masonry wall.

Figure 7.6: Masonry RVE.

Problem 7.3. Reconsider the situation of Ex. 7.4. Solve the problem again with h1 = 30 m, h2 =

18 m, L = 6 m and W = 2.5 m. Notice that a new stress concentration appears. Construct a new

submodel for this area and try to introduce a fillet radius to avoid stress concentrations. Comment

the obtained results for different values of the fillet radius and the validity of the analysis.

Problem 7.4. Model the simple structure of Figure 7.7. The material is steel and the parameters

A = 11 m, B = 2.5 m and C = 3.5 m. The applied load is F = 10 N/m. Construct submodel for

the area showing stress concentrations and try to introduce a fillet radius to avoid large stress

concentration.

A

B

F

C

C

Figure 7.7: Submodel for a T-structure.



Chapter 8

Nonlinear analysis. Geometric

nonlinearities

8.1 Introduction

Many structural problems present nonlinear behaviours that have to be taken into account for a

correct simulation. For example, in a cantilever beam subjected to an end load (see Figure 8.1),

as the beams deflects the distance between the load application point and the encastred area

is reduced, consequently reducing the corresponding bending moment. Moreover, as the beam

deflects and rotates the effect of the load on the beam can change due to the relative orientation

between beam and load. For instance, the bending behaviour of the system varies if the load

has a predefined direction, e.g. gravitational load, or the applied force follows the surface of

the beam, e.g. pressure. Obviously, this effect is only important when large displacements are

involved. Another example of structural nonlinear behaviour is the case of a wood shelf loaded

with books. Even if the load is kept constant, as time passes, the deflection of the shelf increases.

The interaction of different parts of a model also corresponds to a nonlinear structural problem.

Figure 8.1: Geometric nonlinear behaviour in a cantilever beam.

Actually, the three previous examples correspond to the three different types or categories of

nonlinear structural behaviour:

• Geometric nonlinearities

• Material nonlinearities

113


• Change of state

In the first case, the nonlinear structural behaviour is due to a change in the geometric con-

figuration of the structure. This change is usually characterised by large displacement and/or ro-

tations. Material nonlinearities are caused by nonlinear stress-strain relationships. These can be

influenced by different factors such as plasticity, temperature or creep. Finally, structures involv-

ing change of state can also results in a nonlinear structural behaviour: a cable can only transmit

load under tension, a roller support is either in contact or not, etc. In all the cases, though, the

load-deflection curve shows a variation in structural stiffness, which is the basic characteristic

of the nonlinear structural behaviour. Figure 8.2 shows characteristic load-deflection curves for

the three categories of nonlinear structural behaviour. In general, this nonlinear behaviour im-

plies more convergence problems complicating the analysis and requiring more iterations and

different methods and strategies to ensure a converged solution.

Figure 8.2: Characteristic nonlinear load-deflections curves for: (a) geometric nonlinear be-

haviour, (b) material nonlinear behaviour and (c) change of state.

In this chapter, the nonlinear structural behaviour caused by geometric nonlinearities are

considered. Two different types of geometric nonlinearities are taken into account:

• Geometric nonlinearities due to large displacements and/or rotations

• Geometric nonlinearities due to buckling

8.2 Geometric nonlinear behaviour due to large displacements

Most of the structural analyses involve small displacements, rotations and deflections resulting

in null or small stiffness variations. However, some problems involving large displacements,

rotations and deflections result in stiffness variations. In these cases, a nonlinear analysis has to

be carried out activating large displacements or geometric nonlinearities. In order to facilitate

the convergence of the solution during the analysis, it is convenient to divide the application

of load or displacement in different steps. The smaller the steps, the better the options for a

converged solution. However, many load or displacements increments might result in excessive

calculation times.

Hint 36: Activating the geometric nonlinear effects

In the majority of the commercial finite element codes, the geometric nonlinearities are taken

into account by activating a nonlinear geometric effects option. In the case of ANSYSTM this

option is activated by using NLGEOM,ON.



Chapter 8. Nonlinear analysis. Geometric nonlinearities 115

Example 8.1. Consider the cantilever beam shown in Figure 8.3. The end-load P = 100 kN is

distributed along 50 mm in the longitudinal direction. The beam is made of steel (E = 207 GPa

and ν = 0.3), is 2 m long, its base is 20 mm and its thickness is 100 mm. Compare the results

obtained considering a linear behaviour of the system, a nonlinear behaviour with a constant

direction of the applied load (gravity) and a nonlinear behaviour with the load always applied

perpendicular to the surface of the beam.

Figure 8.3: Cantilever beam subjected to a distributed end-load.

Solution to Example 8.1. The cantilever beam in Figure 8.3 has been simulated with and with-

out considering the geometric nonlinearities. The two different cases for the nonlinear system

have been taken into account. In the first case, the direction of the load remains the same during

the whole simulation, gravity load. In the second case, the direction of the load varies accord-

ingly to the deflection of the beam in order to remain perpendicular to the load application area

on the beam.

Hint 37: Defining a Follower Force

In ANSYSTM an applied force always perpendicular to the surface of the element in a nonlinear

problem has to be simulated defining the force as pressure. In other commercial finite element

codes, such as ABAQUSTM and MSC-MarcTM there is an option to indicate that the direction

of the force varies according to the deflection of the element, FOLLOWER FORCE.

As expected, the results of the simulation show a linear relationship between the applied load

and the deflection of the beam for the linear case. However, when the nonlinear geometric effects

such as large displacements are taken into account, the resulting relationship between applied

load and deflection is not linear. Figure 8.4 summarises the load-deflection curves for the three

cases considered. Note that the applied load is the same for the three cases but the resulting

deflection varies depending on the assumptions. If the axial stress is considered, the linear model

predicts an axial stress of 758.5 MPa, while the nonlinear model with gravity load 685 MPa and

the nonlinear model with pressure 742 MPa. Note that in the linear case the results are exactly

the same if the load is applied as a gravity load or as pressure.



window enter /input, file, ext.



/TITLE, Nonlinear cantilever beam


!Parameters

H=100 !beam height in mm

L=2000 !beam span in mm




0.E+00

2.E+04

4.E+04

6.E+04

8.E+04

1.E+05

0 200 400 600 800

Deflection (mm)

Lo

ad

(N

)Linear

Nonlinear-gravity

Nonlinear-pressure

1 × 10�

8 × 10�

6 × 10�

4 × 10�

2 × 10�

0

Figure 8.4: Resulting load-deflection curves for the cantilever beam.

B=20 !beam base in mm

p=1e5 !end-load





SECDATA,H,B !section width and height



K,1,0,0 !generate key-points

K,2,L,0

L,1,2 !define line

LESIZE,1,50 !define element size

LMESH,1

FINISH



NLGEOM,ON !activate geometric nonlinearities




NSEL,S,LOC,X,0 !encastred node

D,ALL,ALL

NSEL,S,LOC,X,L-50,L !apply end-load nodes

ESLN,S,1 !select end elements

!SFBEAM,ALL,1,PRES,p/50 !distributed end-load on beam element

F,ALL,FY,-p/2 !concentrated end-load in 2 nodes (gravity load)

ALLSEL !select all the entities


FINISH


PLNSOL,U,Y !plot vertical displacement

/POST26 !start HISTORY-PLOT MODULE

NSOL,2,41,U,Y,uy !evolution of Y-displacement at node 41

ABS,3,2,,,U_Y !absolute value

XVAR,3 !use as X-axis

RFORCE,4,1,F,Y,F_Y !evolution of reaction force

PLVAR,4 !plot reaction force vs. displacement

PRVAR,3,4 !list displacement and reaction force







8.3 Buckling analysis

Buckling and geometric instabilities in structural parts are typical problems that exhibit geomet-

ric nonlinear effects. Actually, a structure subjected to compressive loads can present no lateral

displacements until a critical load is reached, Pcr. Beyond this critical load, or bifurcation point,

the structure deforms in such a way that lateral displacements occur. Depending on the config-

uration of the structure, the occurrence lateral displacements can imply the global instability of

the structure. In other cases, the occurrence of these lateral displacements only imply a change

of behaviour and the structure can still sustain load under the post-buckled configuration. Fig-

ure 8.5 schematically shows the possible theoretical load versus lateral displacement curves for

a structure under buckling loads. The curve on the left corresponds to a stable structure that can

sustain load after reaching the critical load as the structure tends to stiffen. On the contrary, the

curve on the right is for a unstable structure where after reaching Pcr the structure can deform in

such a way that the lateral displacement increases without increasing the applied load. Although

in this case this behaviour occurs in one direction for the lateral displacement, it is not possible

to ensure that the structure will deform on the other way.

Figure 8.5: Theoretical load versus lateral displacement curves for a structure under buckling

loads.

In general, two different techniques are usually available in all FEM codes for predicting the

buckling load and buckling mode shape of a structure: nonlinear buckling analysis and eigen-

value (or linear) buckling analysis. Because the two methods can yield different results, it is

necessary to understand the differences between them.

8.3.1 Eigenvalue buckling analysis

A real structure sustaining load can suffer buckling or become unstable in many ways and adopt-

ing different deformed shapes that depend on the applied load. For every buckling mode or

deformed shape there is a critical load or bifurcation point. Eigenvalue buckling analysis can

be used to predict the theoretical buckling load (or critical load or bifurcation point) of an ideal

linear elastic structure. For instance, an eigenvalue buckling analysis of a column will match the

classical Euler solution. However, real structures usually contain imperfections and nonlineari-

ties that might cause the instability or buckling at a force lower than the theoretical critical load.

Thus, eigenvalue buckling analysis often yields unconservative results and must be used carefully

and just as a tentative analysis. The following equation summarises the calculation required to

determine the eigenvalues of a structure:




([K] + λi[S])φi = 0 (8.1)

where [K] is the stiffness matrix, [S] is the stress stiffness matrix, λi is the i-th eigenvalue and φi is

the i-th eigenvector of displacements. As the stress stiffness matrix is required for the calculation

of the eigenvalues, this matrix has to be previously obtained. In the case of ANSYSTM , this can

be achieved by performing first a static analysis activating the PSTRES option. The resulting

critical loads are obtained multiplying the applied load per the resulting eigenvalues. It is a

common practice to simulate the problem applying unit loads so the resulting eigenvalues directly

correspond to the critical loads. The eigenvectors are normalized so that the largest component

is 1.0. Thus, the stresses (when output) may only be interpreted as a relative distribution of

stresses.

Hint 38: Negative eigenvalues.

In general, negative eigenvalues indicate that the applied load should be applied in the oppo-

site direction to obtain the represented eigenvalue. However, in ANSYSTM , if the first eigen-

value closest to the shift point is negative (indicating that the loads applied in a reverse sense

will cause buckling), the program will terminate.

Example 8.2. Compute the first ten bifurcation loads of the prismatic bi-supported column

shown in Figure 8.6 using the eigenvalue buckling analysis. The column is made of steel (E

= 207 GPa and ν =0.3), its base is b = 200 mm, its thickness is h = 100 mm and the length is L

= 5 m.

Figure 8.6: Prismatic column under compressive load.

Solution to Example 8.2. Although the geometry of the column is prismatic with a thickness

half of the base and the first buckling modes will be in the same deformation plane (X-Y), it is

possible that one or more of the ten first eigenvalues correspond to a deformed shape in the

other deformation plane (X-Z). Thus, a plane analysis could not take into account some buckling

modes and a 3D analysis is required.






/TITLE, Prismatic column


!Parameters

H=100 !column thickness in mm

L=5000 !column span in mm

B=200 !column base in mm




p=1 !applied load





SECDATA,H,B !section width and height




K,2,L,0

L,1,2,100 !define line and divisions

LMESH,1 !mesh the line

FINISH



PSTRES,ON !calculate stress stiffness matrix

D,1,UX,,,,,UY,UZ !fixed displacements node 1

D,2,UY,,,,,UZ !fixed displacements node 2

F,2,FX,-p !apply load node 2

SOLVE

FINISH


ANTYPE,BUCKLE !buckling analysis

BUCOPT,SUBSP,10 !buckling options: subspace iteration method, 10 eigenvalues

MXPAND !expand all buckling modes

SOLVE


SET,LIST !list eigenvalues

SET,FIRST !set first eigenvalue

PLDISP,1 !plot buckling mode

SET,NEXT !set next eigenvalue

PLDISP,1 !plot buckling mode



As expected, the buckling mode that corresponds to a deformed shape in the X-Z plane is one

of the first modes which would not be considered in a 2-D analysis. Actually, this buckling mode

is the third found, although the critical load coincides with that of the second buckling mode in

the X-Y plane. Table 8.1 summarises the critical loads for the first ten calculated eigenvalues.

Table 8.1: Resulting eigenvalues for the bi-supported prismatic column.

Eigenvalue 1 2 3 4 5 6 7 8 9 10

Critical load (×106 N) 1.36 5.43 5.43 12.1 21.4 21.4 33.2 47.3 47.3 63.6

Example 8.3. The cantilever right angle frame shown in Figure 8.7 is subjected to an in-plane

end load F . Determine the first five bifurcation loads and deformed shapes for the structure. The

frame is made of steel with the following geometry: L = 250 mm, b = 30 mm and t = 1 mm.

Solution to Example 8.3. As in the previous example, the deformed shape of the structure can

produce out-of-plane displacements. Thus, a 3-D analysis has to be carried out. In this case, shell

elements will be used to this end.





t

F

b

b

L

L

Figure 8.7: Cantilever structural frame.






/TITLE, L-Frame: eigenvalues


!Parameters


L=250 !span in mm

b=30 !column base in mm

p=1 !applied load

ET,1,SHELL181 !element type #1: SHELL181 (6DoF)

KEYOPT,1,3,2 !element #1: keyoption 3=2 (full integration imcompatible modes)





K,2,0,-b

K,3,L-b,-b

K,4,L,-b

K,5,L,0

K,6,L-b,-L

K,7,L,-L


L,2,3,44

L,3,4,6

L,4,5,6

L,5,1,50

L,3,6,44

L,6,7,6

L,7,4,44

AL,1,2,3,4,5 !define areas through lines

AL,8,7,6,3

AGLUE,1,2 !bond areas

AMESH,ALL

FINISH







NSEL,S,LOC,X,0

D,ALL,ALL !fixed encastred nodes

NSEL,S,LOC,X,L

NSEL,R,LOC,Y,-L

F,ALL,FX,p !apply load

NSEL,ALL

SOLVE

FINISH





SOLVE

FINISH


SET,LIST !list eigenvalues

SET,FIRST !set first eigenvalue

PLNSOL,U,Z,2,10 !plot buckling mode

SET,NEXT !set next eigenvalue

PLNSOL,U,Z,2,10 !plot buckling mode



Note that the first eigenvalue corresponds to a load F = 17.53 N.

8.3.2 Nonlinear buckling analysis

Nonlinear buckling analysis is usually the more accurate approach and the recommended for

design or evaluation of real structures. This technique uses a nonlinear static analysis with

gradually increasing loads to determine the load level at which the structure becomes unstable.

Using the nonlinear technique, the model can include features such as initial imperfections,

plastic behaviour, gaps, and large-deflection response. In addition, using deflection-controlled

loading, the post-buckled performance of the structure that buckles into a stable configuration

can be evaluated.

Example 8.4. Consider the structural frame in Ex. 8.3 and use a nonlinear buckling analysis to

determine the critical load and buckling curve for the structure. Use an out-of-plane perturbation

load applied at the free end of the frame to drive to the structure to buckling. This perturbation

is removed close to the buckling load.

Solution to Example 8.4. The out-of-plane perturbation load is only used to cause a small de-

formation and force the structure to buckle as a consequence of an assumed imperfection. In this

case, this assumed imperfection causes a small out-of-plane deflection. This perturbation is set

to Fz = 1 ×10−2 N.

As the eigenvalues of the structure have been evaluated in the previous example and it has

been determined that the first buckling mode occurs when F is 17.66 N, during the simulation

a load F = 24 N will be considered in order to obtain the post-buckling behaviour of the structure.

As expected, the critical load or bifurcation point predicted by using the nonlinear buckling

method is close to the first eigenvalue found in Ex. 8.3 although slightly lower. Figure 8.8 shows

the resulting buckling curve where it can be seen that the bifurcation point is about 16 N while

the structure starts to behave in a nonlinear way at about 10 N. It can be also seen in the figure

that the behaviour of the structure is stable after the bifurcation point as the structure tends to

stiffen and larger displacements requires larger loads.





1

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

VALU

0

8

16

24

32

40

48

56

64

72

80

TIME

L-Frame: nonlinear buckling - load perturbation

NOV 27 2010

01:10:17

POST26

F_X

Displacement (mm)

Lo

ad (

N)

Figure 8.8: Buckling curve for the cantilever structural frame - load perturbation.






/TITLE, L-Frame: nonlinear buckling - load perturbation


!Parameters


L=250 !span in mm








K,2,0,-b

K,3,L-b,-b

K,4,L,-b

K,5,L,0

K,6,L-b,-L

K,7,L,-L


L,2,3,44

L,3,4,6

L,4,5,6

L,5,1,50

L,3,6,44

L,6,7,6

L,7,4,44


AL,8,7,6,3


AMESH,ALL




nNODE=node(L-b/2,-L,0) !store node number

FINISH




ARCLEN,ON !use Arc Length method (for improved convergence)

NSUBST,2 !use 2 substeps

NSEL,S,LOC,X,0


ALLSEL

F,nNODE,FZ,1e-2 !apply perturbation load

SOLVE

FINISH


OUTRES,ALL,ALL !write results for every substep


F,nNODE,FX,24 !apply load

SOLVE

FINISH


SET,LAST !set last eigenvalue

PLNSOL,U,Z !plot buckling mode


NSOL,2,nNODE,U,Z,uz !evolution of Z-displacement at nNODE

ABS,3,2,,,U_Z !absolute value


ESOL,4,389,nNODE,F,X,F_1 !evolution of applied force (element 389, nNODE)


ADD,6,4,5,,F_X,,,-1,-1 !evolution of applied force (change sign)





Note that in this case the calculation requires more time as it is a nonlinear calculation. Thus,

it is important to adjust correctly all the control parameters in order to reduce the calculation

time and ensure the convergence of the solution.

Example 8.5. Consider the structural frame in Ex. 8.3 and 8.4 and use a nonlinear buckling

analysis to determine the critical load and buckling curve for the structure. In this case, use the

deformed shape of the first buckling mode multiplied by 1/10 of the thickness of the frame as the

initial perturbation.

Solution to Example 8.5. In this case, the eigenvalue analysis must be run first in order to

upload the deformed geometry of the structure for the first buckling mode as the initial pertur-

bation in the structure. As in the previous case, the eigenvalue analysis of the structure results

in the first buckling mode taking place when F is 17.66 N. Thus, during the simulation a load F

= 24 N will be considered in order to obtain the post-buckling behaviour of the structure.

Similarly to the previous example, the critical load or bifurcation point predicted is close to

the first eigenvalue but slightly lower and the structure is stable after the bifurcation point. As it

can be seen in Figure 8.9, the bifurcation point in the resulting buckling curve is about 16.5 N,

while the nonlinear behaviour of the structure starts at about 10 N. Observe that the resulting

load-displacement curve shown in Figure 8.9 is very similar to the one obtained in Ex. 8.4.








1

0

2.5

5

7.5

10

12.5

15

17.5

20

22.5

25

VALU

0

8

16

24

32

40

48

56

64

72

80

TIME

L-Frame: nonlinear buckling - deformed shape perturbation

NOV 27 2010

01:05:26

POST26

F_X

Displacement (mm)

Lo

ad (

N)

Figure 8.9: Buckling curve for the cantilever structural frame - deformed shape perturbation.



/TITLE, L-Frame: nonlinear buckling - deformed shape perturbation


!Parameters


L=250 !span in mm


p=1 !applied load

delta=t/10 !shape factor multiplier (t/10)







K,2,0,-b

K,3,L-b,-b

K,4,L,-b

K,5,L,0

K,6,L-b,-L

K,7,L,-L


L,2,3,44

L,3,4,6

L,4,5,6

L,5,1,50

L,3,6,44

L,6,7,6

L,7,4,44


AL,8,7,6,3


AMESH,ALL

nNODE=node(L-b/2,-L,0) !store node number

FINISH







NSEL,S,LOC,X,0


ALLSEL

F,nNODE,FX,p !apply load

SOLVE

FINISH





SOLVE

FINISH

!Add displacement from previous analysis

!multiplier for added displacements,

!load step, substep (equivalent to mode=1),

!results file containing displacements and extension


UPGEOM,delta,1,1,file,rst

FINISH

/SOLU !start SOLUTION MODULE - continuation loads

ANTYPE,STATIC !static analysis


ARCLEN,ON !use Arc Length method (for improved convergence)

OUTRES,ALL,ALL !write results of every substep


F,nNODE,FX,24 !apply load

SOLVE

FINISH


SET,LAST !set last eigenvalue

PLNSOL,U,Z !plot buckling mode


NSOL,2,nNODE,U,Z,uz !evolution of Z-displacement at nNODE

ABS,3,2,,,U_Z !absolute value




ADD,6,4,5,,F_X,,,-1,-1 !evolution of applied force (change sign)






Problem 8.1. The bi-supported column in Example 8.2 can be considered as a symmetric struc-

ture and therefore could be simulated using symmetry boundary conditions (as schematically

shown in Fig. 8.10). Determine if the resulting buckling eigenvalues are the same for the com-

plete structure and the symmetric half column or not. Justify the results.

Problem 8.2. Compute the bifurcation load and the post-buckling behaviour of the column in

Example 8.2 by means of a nonlinear analysis.

Problem 8.3. Extract the first five bifurcation loads and deformed shapes of the structure pre-

sented in Figure 8.11. The thickness of the beam is 1.5 mm and it is made of steel.





P

L/2

x

y

z

S

S

Figure 8.10: Buckling curve for the cantilever structural frame.

Figure 8.11: Cantilever T-beam.



Chapter 9

Material nonlinearities

9.1 Introduction

As has been introduced in Chapter 8, a nonlinear structural behaviour can be induced by a geo-

metric nonlinearity, a contact or boundary nonlinearity, or by a material nonlinearity. This chapter

deals with the nonlinear structural behaviour casued by material nonlinearities. It must be taken

into account that an analysis including nonlinear models requires an iterative solution that might

compromise its convergence and can be very expensive in terms of computational time.

Nonlinear material models arise from the presence of a time-independent behaviour, such as

hyperelasticity, plasticity or damage, or to take into account other time-dependent behaviours,

such as creep or viscoplasticity.

A number of material-related factors can cause changes in the stiffness of the structure dur-

ing the course of an analysis. Nonlinear stress-strain relationships of plastic, multilinear elastic,

and hyperelastic materials will cause a stiffness change in the structure at different load levels

(and, typically, at different temperatures). Creep, viscoplasticity, and viscoelasticity will give rise

to nonlinearities that can be time-, rate-, temperature-, or stress-related. Swelling will induce

strains that can be a function of temperature, time, neutron flux level (or some analogous quan-

tity), and stress. Some nonlinear material models are presented here as an example involving

different physical responses.

9.2 Plasticity

Most common engineering materials exhibit a linear stress-strain relationship up to a stress

level known as the proportional or elastic limit. Beyond this limit, the stress-strain relationship

becomes nonlinear, although not necessarily inelastic. Plastic behaviour, characterized by non-

recoverable strain, begins when the stress level exceeds the material yield point. Because there

is usually little difference between the yield point and the proportional limit, the most common

plasticity approach assumes that these two points are coincident in plasticity analyses (see Fig-

ure 9.1).

Plasticity is a nonconservative, path-dependent phenomenon. In other words, the sequence

in which loads are applied and in which plastic responses occur affects the final solution results.

If plastic response is anticipated for the analysis, loads should be applied as a series of small

incremental load steps or time steps.

127


9.2.1 Bilinear Kinematic Plasticity

As shown in Figure 9.1, bilinear plasticity assumes an elastic linear behaviour up to the yield

point, and another linear behaviour after this point. The unload path is parallel to the elastic

response. This model is known as the bilinear kinematic hardening model and is recommended

for small-strain analyses involving materials that obey von Mises yield criteria (most metals).

Figure 9.1: a) Uniaxial plasticity behaviour, and b) Bilinear Plasticity approach

Hint 39: ANSYS properties for Bilinear Kinematic Hardening model

The bilinear kinematic hardening model is activated in ANSYS with the command TB,BKIN

(see the following example). The sequence number in TBDATA command is the yield stress in

the first position and the tangent modulus d.

Bilinear Kinematic Plasticity Example: Nickel Alloy

FINISH

/CLEAR

/PREP7

MP,EX,1,180e9 !Pa


MP,DENS,1,8490 !kg/m3

TB,BKIN,1,1 !bilinear kinematic hardening, material #1, 1 temperature

TBDATA,1,900e6,445e6 !Yield stress (Pa) and Tangent modulus (Pa)

TBPLOT,BKIN,1 !display the data table

In the case of an isotropic hardening, the hardening of the material is independent of the di-

rection and it is the same for tension and compression. The Bilinear Isotropic Hardening model

uses the von Mises yield criteria coupled with an isotropic work hardening assumption. This

model is recommended for large strain analyses and it is activated in ANSYS with the command

TB,BISO. The definition parameters are the same as for the case of the bilinear kinematic model.

Example 9.1. Use a bilinear plasticity model to simulate in ANSYS an hexagonal Allen wrench

10 mm wide across flats made of high carbon steel (HCS) with a Young’s modulus of 210 ×103

MPa, a Poisson’s ratio of 0.3, a Yield stress of 500 MPa, and a Tangent modulus of 20 ×103 MPa.

Solution to Example 9.1. Small strains are expected for this analysis and the Bilinear Kine-

matic Hardening model is used. Two different load cases are considered to show the effect of



Chapter 9. Material nonlinearities 129

the bilinear model and the generated plastic strain. Observe that once the first applied load is

removed there is no remaining deformation or plastic strain. However, when the second applied

load is removed, there is a remaining deformation and plastic strain. The ANSYS command list

is presented next.

FINISH

/CLEAR

/TITLE, Allen wrench with plastic strain

/UNITS,MPA !units are in mm, MPa, and Newtons


W_HEX=10 !distance across flats (10mm)

*AFUN,DEG !specifies units for angular functions in parameter expressions

W_FLAT=W_HEX/2/TAN(60)*2 !wrench edge length

L_SHANK=60 !shank length

L_HANDLE=112.5 !handle length

BENRAD=10 !bending radius between the shank and the handle

L_ELEM=7.5 !element length

NO_D_HEX=2 !number of element divisions in the hexagon


KEYOPT,1,2,3 !element #1: keyoption 2=3 (simplified enhanced strain integration)



MP,EX,1,2.1E5 !define material, plastic steel, Young’s modulus

MP,PRXY,1,0.3 !Poisson coeficient

TB,BKIN,1,1 !bilinear isotropic hardening, material #1, 1 temperature

TBDATA,1,500,0.2E5 !yield stress, tangent modulus

TBLIST,BKIN,1 !list plastic model properties

/XRANGE,0,0.01 !x-axis range on TBPLOT

TBPLOT,BKIN,1 !plot plastic strain-stress model

RPOLY,6,W_FLAT !define the geometry of the Allen wrench section (hexagon)

K,7 !define the shank and handle lengths

K,8,,,-L_SHANK

K,9,,L_HANDLE,-L_SHANK

L,4,1

L,7,8

L,8,9

LFILLT,8,9,BENRAD !generates a fillet line between intersecting lines 8 and 9

ASBL,1,7 !subtracts lines from areas

CM,BOTAREA,AREA !groups geometry entities in the variable BOTAREA

LESIZE,2,,,NO_D_HEX !define control mesh and meshing of the hexagonal section

LESIZE,5,,,NO_D_HEX

LESIZE,1,,,1

LESIZE,3,,,1

LESIZE,4,,,1

LESIZE,6,,,1

TYPE,2 !change the element type to mesh the section

ESHAPE,2

AMESH,ALL

TYPE,1 !change the element type to mesh the volume

ESIZE,L_ELEM !define control mesh

VDRAG,2,3,,,,,8,10,9 !volume by dragging an area along a path, area, line path

CMSEL,,BOTAREA !select the group BOTAREA and assemblies

ACLEAR,ALL !clear the area and associate nodes

ASEL,ALL !select all areas

/VIEW,1,1,1,1




ANTYPE,STATIC

AUTOTS,ON !automatic stepping on

NSUBST,3,5,1 !number of substeps, max, min

CNVTOL,F,,,,1E-30 !set convergence tolerance on force, minimum

CMSEL,,BOTAREA !apply boundary conditions: clamp bot area

LSEL,,EXT !select the external lines of the BOTAREA

NSLL,,1 !select nodes associated with selected lines

D,ALL,ALL !clamp




NSEL,S,LOC,X,W_FLAT/2-0.1,W_FLAT+0.1 !select nodes to apply load

NSEL,R,LOC,Y,L_HANDLE-L_ELEM-0.1,L_HANDLE+0.1

CM,NLOAD,NODE !define group of nodes

ALLSEL

T=60 !apply a load of 60 N to nodes

F,NLOAD,FX,-T

SOLVE !solve 1st load set

FDELE,ALL,ALL !delete the loads

T=1e-10 !apply a load of 0 N to nodes

F,NLOAD,FX,-T

SOLVE !solve 2nd load set: no loads


F,NLOAD,FX,-T

SOLVE !solve 3rd load set

FDELE,ALL,ALL !delete the loads

T=1e-10 !apply a load of 0 N to nodes

F,NLOAD,FX,-T

SOLVE !solve 4th load set: only remain plastic strain



/DSCALE,,1 !do not scale displacements

SET,1

PLNSOL,EPEL,EQV !contour plot of von Mises equivalent elastic strains

PLNSOL,EPPL,EQV !contour plot of von Mises equivalent plastic strains

SET,2



SET,3



SET,4





The same problem can be simulated by defining an initial deformed state for a final load step

corresponding to the deformation-state resulting from an initial load step. Once this final load

step is solved with zero loads applied, the remaining deformations are the plastic deformations

of the initial step. In ANSYS this can be achieved by defining this initial deformed state by using

the INISTATE command. The ANSYS command list for the generation of the initial load step and

writing of the initial state file for the considered the example is presented next.

FINISH

/CLEAR

/TITLE, Allen wrench with plastic strain

/UNITS,MPA !units are in mm, MPa, and Newtons


W_HEX=10 !distance across flats (10mm)

*AFUN,DEG !specifies units for angular functions in parameter expressions

W_FLAT=W_HEX/2/TAN(60)*2 !wrench edge length

L_SHANK=60 !shank length

L_HANDLE=112.5 !handle length

BENRAD=10 !bending radius between the shank and the handle

L_ELEM=7.5 !element length

NO_D_HEX=2 !number of element divisions in the hexagon


KEYOPT,1,2,3 !element #1: keyoption 2=3 (simplified enhanced strain integration)







MP,EX,1,2.1E5 !define material, plastic steel, Young’s modulus

MP,PRXY,1,0.3 !Poisson coeficient

TB,BKIN,1,1 !bilinear isotropic hardening, material #1, 1 temperature

TBDATA,1,500,0.2E5 !yield stress, tangent modulus

TBLIST,BKIN,1 !list plastic model properties

/XRANGE,0,0.01 !x-axis range on TBPLOT

TBPLOT,BKIN,1 !plot plastic strain-stress model

RPOLY,6,W_FLAT !define the geometry of the Allen wrench section (hexagon)

K,7 !define the shank and handle lengths

K,8,,,-L_SHANK

K,9,,L_HANDLE,-L_SHANK

L,4,1

L,7,8

L,8,9

LFILLT,8,9,BENRAD !generates a fillet line between intersecting lines 8 and 9

ASBL,1,7 !subtracts lines from areas

CM,BOTAREA,AREA !groups geometry entities in the variable BOTAREA

LESIZE,2,,,NO_D_HEX !define control mesh and meshing of the hexagonal section

LESIZE,5,,,NO_D_HEX

LESIZE,1,,,1

LESIZE,3,,,1

LESIZE,4,,,1

LESIZE,6,,,1

TYPE,2 !change the element type to mesh the section

ESHAPE,2

AMESH,ALL

TYPE,1 !change the element type to mesh the volume

ESIZE,L_ELEM !define control mesh

VDRAG,2,3,,,,,8,10,9 !volume by dragging an area along a path, area, line path

CMSEL,,BOTAREA !select the group BOTAREA and assemblies

ACLEAR,ALL !clear the area and associate nodes

ASEL,ALL !select all areas

/VIEW,1,1,1,1




ANTYPE,STATIC

AUTOTS,ON !automatic stepping on

NSUBST,3,5,1 !number of substeps, max, min

CNVTOL,F,,,,1E-30 !set convergence tolerance on force, minimum

CMSEL,,BOTAREA !apply boundary conditions: clamp bot area

LSEL,,EXT !select the external lines of the BOTAREA

NSLL,,1 !select nodes associated with selected lines

D,ALL,ALL !clamp

NSEL,S,LOC,X,W_FLAT/2-0.1,W_FLAT+0.1 !select nodes to apply load

NSEL,R,LOC,Y,L_HANDLE-L_ELEM-0.1,L_HANDLE+0.1

CM,NLOAD,NODE !define group of nodes

ALLSEL


!T=120 !uncomment to apply load of 120 N to nodes

F,NLOAD,FX,-T

INISTATE,WRITE,1,,,,0,EPPL !write plastic deformation state in file

SOLVE !solve initial load set








The ANSYS command list for the generation of final load step and reading of the initial state

file for the considered example is presented next.





FINISH


INISTATE,READ,file.ist !read plastic deformation state in file

FDELE,ALL,ALL !delete all loads

SOLVE !solve final load set: no loads







ftp://amade.udg.edu/mms/PrePostFE/input_files/T9/PrePostFE_Ex901c.dat

Observe that additionally to the expected plastic deformation there is a remaining elastic

deformation and stress. What is the cause of this remaining elastic deformation and stress?

9.2.2 Multiple-point Isotropic Plasticity

Multiple-point plasticity models can be used to define multilinear plasticity behaviours.

Hint 40: ANSYS properties for Multiple-point Isotropic Plasticity model

The multiple-point isotropic plasticity model is activated in ANSYS with the command TB,MISO

(see example next), the sequence number in TBPT command are the different stress-strain

points that define the material response.

Multiple-point Isotropic Plasticity Example: Aluminium

FINISH

/CLEAR

/PREP7

MP,EX,1,69000 !material #1: Elastic modulus

MP,PRXY,1,0.3 !material #1: Poisson coeficient

TB,MISO,1,1,11 !material #1, Multi-point isotropic plasticity, n- points

TBPT,,0.003,207 !nonlinear curve point,, x coordinate, y coordinate

TBPT,,0.004,230.7885

TBPT,,0.005,242.7903

TBPT,,0.006,252.0527

TBPT,,0.007,259.8827

TBPT,,0.008,266.7906

TBPT,,0.009,273.0415

TBPT,,0.01,278.7934

TBPT,,0.024,335.648

TBPT,,0.04,379.7314

TBPT,,0.063,428.6213

TBPLOT,MISO,1

Example 9.2. Considerer a thin-walled aluminum sphere with a radius r = 250 mm and a thick-

ness t = 5 mm. The sphere is subjected to an internal pressure p = 12 MPa. The elastic modulus



ftp://amade.udg.edu/mms/PrePostFE/input_files/T9/PrePostFE_Ex901c.dat


and Poisson ratio of the material are E = 69 GPa and ν = 0.3. The plastic behaviour of aluminium

is governed by

σe = 207 + 930(ǫp)1/2

where σe is the effective stress and ǫp is the plastic strain. The aim is to obtain the radial dis-

placement with the internal pressure and the unrecoverable radial displacement after unloading.

Solution to Example 9.2. The following MATLAB file is used to obtain the stress vs. total strain

curve shown in Figure 9.2 and write the command sequence to define the material behaviour in

the ANSYS input file.

%% PrePostFE. Solution of Example 9.2.

% J.A. Mayugo, 2008

clear all;close all

E=69000 % MPa

nu=0.3

sigma_eff_0=207 % MPa

sigma_eff=930 % MPa

%% Elastic response

epsilon_elastic=[0:0.0001:0.005]; %elastic strain

sigma_elastic=E.*epsilon_elastic; %elastic stress

%% Plastic response

epsilon_p=[0:0.0001:0.065]; %plastic strain

sigma_eff=sigma_eff_0+sigma_eff.*(epsilon_p).^0.5; %plastic stress

%% Total response

epsilon_e=sigma_eff./E;

epsilon=epsilon_e+epsilon_p;

%% Multi-point plasticity model

epsilon_MISO=[[0.003:0.001:0.01],0.024,0.040,0.063];

sigma_MISO=interp1(epsilon,sigma_eff,epsilon_MISO,’linear’);

[m,n_MISO]=size(sigma_MISO)

figure1=figure

hold on

plot(epsilon,sigma_eff,’LineWidth’,1.8);

plot(epsilon_elastic,sigma_elastic,’r--’,’LineWidth’,1.8);

plot(epsilon_MISO,sigma_MISO,’kd’,’LineWidth’,1.8,’MarkerFaceColor’,’w’,’MarkerSize’,6);

hold off

xlabel(’Total strain’,’Fontsize’,16);ylabel(’Stress’,’Fontsize’,16);

legend(’Plastic response’,’Elastic response’,’Discrete response to MISO’,’Location’,’Best’);legend(’boxoff’);

set(gca,’Fontsize’,16);

saveas(figure1,[’PrePostFE_Ex902’],’eps’);

n_file = ’PrePostFE_Ex902’;

tline{1} = [’MP,EX,1,’,num2str(E),’ !material #1: Elastic modulus ’];

tline{2} = [’MP,NUXY,1,’,num2str(nu),’ !material #1: Poisson coeficient’];

tline{3} = [’TB,MISO,1,1,’,num2str(n_MISO),’ !material #1, Multi-point isotropic plasticity, n- points’];

for ii=1:n_MISO

tline{3+ii} = [’TBPT,,’,num2str(epsilon_MISO(ii)),’,’,num2str(sigma_MISO(ii))];

end

fid0 = fopen([n_file,’.txt’],’w’);

fprintf(fid0,’%10s\n\n’,’!INPUT FILE TO ANSYS, Multi-point isotropic plasticity’);

for i=1:n_MISO+3

fprintf(fid0,’%10s\n’,tline{i});

end

fclose(fid0);







0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080

50

100

150

200

250

300

350

400

450

Total strain

Str

ess

Plastic responseElastic responseDiscrete response to MISO

Figure 9.2: Plasticity uniaxial behaviour

See input file for ANSYS below.



/TITLE, Aluminium sphere subjected to internal presure


ET,1,SHELL208 !element #1: 2-d axisymmetric SHELL

SECTYPE,1,SHELL !section #1: type shell

SECOFFSET,MID !section #1: offset=0

SECDATA,5,1,,5 !section #1: thickness, material and integration points

MP,EX,1,69000 !material #1: Elastic modulus

MP,PRXY,1,0.3 !material #1: Poisson coeficient

TB,MISO,1,1,11 !material #1, Multi-point isotropic plasticity, n- points

TBPT,,0.003,207 !non-linear curve point,, x coordinates, y coordinates

TBPT,,0.004,230.7885

TBPT,,0.005,242.7903

TBPT,,0.006,252.0527

TBPT,,0.007,259.8827

TBPT,,0.008,266.7906

TBPT,,0.009,273.0415

TBPT,,0.01,278.7934

TBPT,,0.024,335.648

TBPT,,0.04,379.7314

TBPT,,0.063,428.6213

K,1, !create keypoints

K,2,250

K,3,,250

LARC,2,3,1,250 !create arc

LESIZE,ALL,,,20 !specify 20 elements on line

LMESH,ALL !mesh

FINISH !exit PRE-PROCESSOR MODULE


!OUTRES,ALL,1 !write results for each sub-step

NSUBST,,100,1 !number of substeps, max, min

NSEL,S,LOC,Y,0 !select nodes at Y=0

D,ALL,UY,0 !fix the node displacement in Y direction

SFE,ALL,1,PRES,,12 !apply internal pressure 12 MPa

NSEL,ALL !select all

SOLVE !solve the first load-step




SFEDELE,ALL,1,PRES !delete all pressure loads

SOLVE !solve a second load-step

FINISH !exit SOLUTION MODULE


SET,1 !set the first load-step

PLDISP,1

SET,2 !set the second load-step

PLDISP,1

FINISH !exit POST-PROCESSOR MODULE



9.3 Two parameter Mooney-Rivlin Hyperelastic

A material can be considered as hyperelastic if there is an elastic potential function (or strain

energy density function), which is a scalar function of one of the strain or deformation tensors,

whose derivative with respect to a strain component determines the corresponding stress com-

ponent. Hyperelasticity can be used to analyse elastomers or rubber-like materials subjected to

large strains and displacements with small volume changes (nearly incompressible materials).

Therefore, it is necessary to activate the large strain theory during the analysis. One of the hy-

perelasticity models that can be used in ANSYS is the Mooney-Rivlin model.

In the two parameter Mooney-Rivlin model the form of the strain energy potential is:

u = c10(

I1 − 3)

+ c01(

I2 − 3)

+1

d(J − 1)2 (9.1)

where u is the strain energy potential, I1 the first deviatoric strain invariant, I2 the second devia-

toric strain invariant, c01 and c10 are material constants characterising the deviatoric deformation

of the material, J determinant of the elastic deformation gradient and d is the material incom-

pressibility parameter.

The initial shear modulus is defined as:

µ = G = 2 (c10 + c01) (9.2)

and the initial bulk modulus is defined as:

K =2

d(9.3)

where if d is not defined explicity but as

d =1 + 2ν

c10 + c01(9.4)





Hint 41: ANSYS properties for Mooney-Rivlin model

Mooney-Rivlin model for elements SHELL181, PLANE182, PLANE183, SOLID185, SOLID186,

SOLID187, SOLSH190, SHELL208, and SHELL209.

The Mooney-Rivlin model is activated in ANSYS with the command TB,HYPER,,,,MOONEY or

TB,MOONEY (see example next), the sequence number in TBDATA command is c10, c01, and d.

Mooney-Rivlin Example: Rubber

MP,DENS,1,.0018 !lb/in3


TB,MOONEY,1

TBDATA,1,80 !C10 (psi)

TBDATA,2,20 !C01 (psi)

Example 9.3. Use ANSYS to model a circular plate made of rubber with applied pressure in one

surface. Use an hyperelastic model.

Solution to Example 9.3. See input file for ANSYS below.



/TITLE, Balloon / circular plate problem


MP,EX,1,1.E6 !Young’s modulus

MP,PRXY,1,0.5 !Poison coefficient

MP,DENS,1,0.1 !density

TB,HYPER,1,,,MOONEY !data table for Mooney-Rivlin hyperelastic model

TBDATA,1,80.0,20.0 !constants c_10 and c_01

ET,1,SHELL181 !element type SHELL181

R,1,0.5 !real constant set #1, thickness of 0.5 mm

N,1,0.,0. !geometry and mesh

N,2,0.17143,0.

N,3,0.47143,0.

N,4,0.90000,0.

N,5,1.4571,0.

N,6,2.1429,0.

N,7,2.9571,0.

N,8,3.9000,0.

N,9,4.9714,0.

N,10,6.1714,0.

N,11,7.5000,0.

N,101,0.,0.

N,102,0.16996,2.23759E-02

N,103,0.46740,6.15338E-02

N,104,0.89230,0.11747

N,105,1.4447,0.19020

N,106,2.1245,0.27970

N,107,2.9318,0.38598

N,108,3.8666,0.50905

N,109,4.9289,0.64890

N,110,6.1186,0.80553

N,111,7.4358,0.97895

E,1,2,102,102

E,2,3,103,102




E,3,4,104,103

E,4,5,105,104

E,5,6,106,105

E,6,7,107,106

E,7,8,108,107

E,8,9,109,108

E,9,10,110,109

E,10,11,111,110

LOCAL,11,0,0.0,0.0,0.0,7.5,0.0,0.0 !new local coordinate system

NROTAT,102,111,1 !rotates nodes: 102 to 111



ANTYPE,STATIC !set static analysis

AUTOTS,ON !use automatic time stepping

NSUBST,400,1200,25 !specify the number of substeps, max, min

NLGEOM,ON !non-linear analysis

NROPT,FULL,,OFF !option of convergence: Newton-Raphson

OUTRES,ALL,ALL !solution data to be written

NEQIT,20 !max number of equilibrium iterations

D,1,UY,0.0,,11,1,ROTX,ROTZ !boundary conditions

D,102,UY,0.0,,111,1,ROTX,ROTZ

D,11,UX,0.0,,111,100,UY,UZ

D,1,UX,0.0,,,,UY,ROTX,ROTY,ROTZ

SF,ALL,PRES,50.0 !apply uniform pressure in psi




/NOPR !suppress graphing data

/VIEW,1,,-1 !change the view to plane x-z

!/ANG,1

!/USER

/FOCUS,1,4,,8,0 !setup center of graphics screen for disp. plot

/DIST,,12 !set distance to zoom out

/TRIAD,OFF !hide the coordinate system

SET,FIRST !set displacement data for first substep

PLDISP,0 !plot displacement data

/NOERASE !set display to overlay plot

SET,,10

PLDISP,0

SET,,20

PLDISP,0

SET,,25

PLDISP,0

SET,LAST

PLDISP,1 !plot final displacement with original position

/ERASE

/TRIAD,ON

/GOPR !reactivate supressed printout (\NOPR)

/ESHAPE,0

FINISH !finish POST-PROCESSOR MODULE



The following file allows to obtain the thickness evolution an the displacement evolution (in

this case with an experimental comparison).

/POST26 !start TIME-HISTORY MODULE

/XRANGE,0,3.0 !x-axis scale range

/YRANGE,0,1 !y-axis scale range

/AXLAB,X,UZ OF CENTER/R-INITIAL !x-axis label

/AXLAB,Y,THICKNESS/ORIGINAL THICKNESS !y-axis label

NSOL,2,1,U,Z,UZ_1 !nodal data to be stored





ESOL,3,1,,SMIS,17,TH_1 !element data bo be stored

ADD,4,2,,,UZRATIO,,,0.13333333,0,0, !adds variables

ADD,5,3,,,SH.181,,,2,0,0,

/COLOR,CURVE,MRED !curve color

XVAR,4 !x-variable to be displayed

PLVAR,5 !y-variable to be displayed

/ERASE

/NOPR !supress the input data

*DIM,X,TABLE,20,1 !add a table: experimental data

*DIM,Y,TABLE,20,1

X(1,1)= 1.25 !table x data point

Y(1,1)= 1.25 !table y data pointX(2,1)= 1.8

Y(2,1)= 2.5

X(3,1)= 2.25

Y(3,1)= 4.0

X(4,1)= 2.6

Y(4,1)= 5.9

X(5,1)= 2.9

Y(5,1)= 7.8

X(6,1)= 3.2

Y(6,1)= 9.8

X(7,1)= 3.5

Y(7,1)= 11.6

X(8,1)= 3.62

Y(8,1)= 12.6

X(9,1)= 4.1

Y(9,1)= 15.3

X(10,1)= 4.9

Y(10,1)= 18.8

X(11,1)= 5.7

Y(11,1)= 22.1

X(12,1)= 6.2

Y(12,1)= 24.0

X(13,1)= 7.2

Y(13,1)= 27.9

X(14,1)= 8.3

Y(14,1)= 31.2

X(15,1)= 8.9

Y(15,1)= 32.9

X(16,1)= 9.9

Y(16,1)= 35.8

X(17,1)= 10.9

Y(17,1)= 38.0

X(18,1)= 13.1

Y(18,1)= 42.9

X(19,1)= 14.4

Y(19,1)= 45

X(20,1)= 15.2

Y(20,1)= 46

/GOPR !reactivates the supressed printout

/XRANGE,0,20 !x-axis scale range

/YRANGE,0,60 !y-axis scale range

/AXLAB,X,UZ OF CENTER (IN) !x-axis label

/AXLAB,Y,PRESSURE (LB/SQ IN) !y-axis label

/COLOR,CURVE,YGRE !curve color

*VPLOT,X(1,1),Y(1,1) !plot a curve from an array

/NOERASE

NSOL,2,1,U,Z,UZ_1 !nodal data to be stored

PROD,7,1,,,SH.181,,,50,0,0, !multiply solution by 50

/COLOR,CURVE,MRED !curve color

XVAR,2 !specify x variable to be displayed

PLVAR,7 !display solution in graph file

/ERASE

FINISH !finish TIME-HISTORY MODULE


ftp://amade.udg.edu/mms/PrePostFE/input_files/T9/PrePostFE_Ex903_post.dat



ftp://amade.udg.edu/mms/PrePostFE/input_files/T9/PrePostFE_Ex903_post.dat



Problem 9.1. Reconsider the situation in Example 3.3 and introduce a bilinear plasticity model.

The yield stress is 270 MPa and ET = E/180. Increase the load until some plastic deformation

appears. Obtain the Von Misses contour plot of the plastic strain and comment the results. Show

your results in a report.

Problem 9.2. Generate an ANSYS model to simulate the behavior of a 5 mm thick rectangular

plate (see figure 9.3) with material nonlinearities defined by a bilinear kinematic plasticity law.

Use PLANE182 elements and consider the following material properties to define the non-linear

behaviour: an elastic modulus of 210 GPa, Poisson’s coefficient ν = 0.3, and a yield stress of 500

MPa. Define an appropriate value for the applied pressure p and obtain the stress-strain curve of

the bilinear plasticity approach when: a) the tangent modulus is 1% of the elastic modulus and

b) the tangent modulus is 0.1% of the elastic modulus.

Figure 9.3: Rectangular plate

Use the same applied pressure as before to simulate the same problem when the tangent

modulus is 0.01% of the initial modulus. Comment the obtained results.



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