What Is Finite Element Modeling?
Transcript of What Is Finite Element Modeling?
What Is Finite Element Modeling?I-DEAS Tutorials: Fundamental Skills
Although typically you’d use many elements, in thistutorial you’ll simulate what happens to a single thin-shellelement in the middle of a machine linkage, under atensile load.
Learn how to:
• create a finite element model• create material properties• create physical properties• create nodes• create an element• apply boundary conditions• solve the model• interpret the results
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Before you begin...
Prerequisite tutorials:
• Getting Started (I-DEAS Multimedia Training)
–or–
Quick Tips to Using I-DEAS–and–Creating Parts
• Introduction to Simulation
This tutorial assumes you have a background in theconcepts of stress as taught in a typical college coursein strength of materials.
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Setup 1 of 2
If you didn’t start I-DEAS with a new (empty) model file,open a new one now and give it a unique name.
FileOpen
Open Model File form
Model File name: any unique name
OK
Make sure you’re in the following application and task:
Simulation
Master ModelerSet your units to mm.
OptionsUnits
mm (milli newton)
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Setup 2 of 2
Save your model file. File
Save
Warning! If you are prompted by I-DEAS to save your model file,respond:
NoSave only when the tutorial instructions tell you to––notwhen I-DEAS prompts for a save.
Why:If you make a mistake at any time between saves andcan’t recover, you can reopen your model file to the lastsave and start over from that point.
Hint To reopen your model file to the previous save, pressControl-Z.
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Create a finite element model 1 of 2
Before you start an analysis, you should think about thetype of solution you want. Which solution type youchoose depends on loading and other factors. Keyfactors include:
• how loads vary with time, or the duration of theload relative to the period of natural frequency
• whether the expected result is linear
In this example, you’ll use a linear static solution.
You start by creating a finite element (FE) model. An FEmodel is always associated with a part.
In the model file, there may be more than one finiteelement model associated with the same part, withdifferent meshes.
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Create a finite element model 2 of 2
What: Switch to the Meshing task and create an FEmodel.
How:
Meshing
FE Model Create form
Part or Assembly: (any unique name)
FE Model Name: Tensile Model
OK
Things to notice Ignore the warning message if you get one. When youcreated the FE model above and entered the name of anonexisting part, a null part was created. A null partcontains no surfaces. It is displayed in Simulation as apart coordinate system.
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Create material properties 1 of 3
What: Create a material for the analysis and define themodulus of elasticity and Poisson’s ratio.
How:
AlCu
Fe
Materials form
Quick Create...
Quick Create form
Material Label & Name: Link Material(enter name only, in second field)
Property: Modulus of Elasticity
2E8
Modify Value
Continued on next page...
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Create material properties 2 of 3
Property: Poisson’s Ratio
0.3
Modify Value
OK
Do not exit from the Materials form yet.
Things to notice The value for Poisson’s ratio of 0.3 means that if thematerial is stretched one unit in the X-direction, it willshrink 30% in the Y- and Z-directions. Later in thistutorial you’ll check the displacements to see if this holdstrue.
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Create material properties 3 of 3
What: Examine the new material to make sure youentered the correct values.
How:
Materials form
Link Material
Examine...
Examine form
Review the property values in the table.
LINK MATERIALMODULUS OF ELASTICITY
2.0000E+08POISSONS RATIO
3.L0000E–01
Dismiss
Materials form
OK
Things to notice This material definition is stored in your model file andcan be referenced by any FE model you create in thesame model file.
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Create physical properties 1 of 3
The thin shell element you’ll create will be defined byfour nodes at its corners. The physical thickness of thinshell elements is defined by a separate physical propertytable.
Things to notice The link part is shown here only to indicate the thicknessof the element. Since you are creating nodes andelements manually, you won’t have a part displayed.
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Create physical properties 2 of 3
What: Create a physical property table that defines ashell thickness of 5mm.
How:
t
Thin Shell
Check I-DEAS Prompt .
physical property name: LinkThickness
No
Directory
Continued on next page...
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Create physical properties 3 of 3
TK THICKNESS [4V]
There are four values of thickness (4V)––one valueat each corner.
Check I-DEAS Prompt .
1st value for thickness: 5
<Return> – for rest of the prompts
Warning! If you enter more than one value, all four values must benon-zero. To enter a uniform thickness at each corner,you need to enter only the first value.
Done
Recovery Point
FileSave
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Create nodes 1 of 2
In this example, you’ll create four nodes in the finiteelement model.
The displacements of nodes are the variables for whichthe finite element model is solved. Each node has threepotential displacements and three potential rotations, ora total of six degrees of freedom (DOF).
The four nodes in the model contain 24 DOF to besolved.
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Create nodes 2 of 2
What: Create four nodes in a 10mm square on the XYplane.
How:
Node form
OK
Check I-DEAS Prompt .
Node 1 location: 0 0
Node 2 location: 10 0
Node 3 location: 0 10
Node 4 location: 10 10
Done
Adjust the display to get a better view of the nodes. Youcould use Zoom All and F2.
Recovery Point
FileSave
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Create an element 1 of 4
Elements are displayed by lines connecting the nodes.
Why: In this example you’ll use only one element tounderstand how nodes and elements behave. Thecomputed displacements and stresses will be correctonly for simple uniform loads. In most cases, you’ll haveto create many smaller elements to converge on acorrect result.
To create elements you’ll:• specify the type of elements (1D, 2D, 3D)• define material and physical properties• pick the nodes
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Create an element 2 of 4
What: Create one thin-shell element to connect the fournodes.
How:
Element form
Make sure the following are selected on the form:
2D
Element Family: Thin Shell
Element Type:
Don’t close the Element form yet.
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Create an element 3 of 4
What: Assign material and physical properties to theelements.
How:
Element form
Material Selection...
Materials form
Link Material
OK
Element form
Physical Property:
?
Physical Property form
Link Thickness
OK
Element form
OK
Why: The default is to use the material stored with thepart that’s associated with the finite element model.Since you’re not using a meaningful part here, youmanually selected the material table you created earlier.
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Create an element 4 of 4
What: Pick the four nodes with which to create thethin-shell element. In a typical problem, you would usemore automatic methods to create elements.
How:
1
2 shift-pick
3 shift-pick
4 shift-pick
2
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1
(Done)
If the element looks wrong, you’ve picked the nodesin the wrong order. Press Control-Z and repeat thisstep before saving your model file.
Recovery Point
FileSave
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Apply boundary conditions 1 of 4
At each degree of freedom (DOF), you can either applyrestraints to the degree of freedom, or you can apply aforce.
You must apply enough restraints to hold the model as arigid body in space. Even though the linkage has equaland opposite forces on the two ends, you will restrainone end, and apply a force on the other.
1
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Apply boundary conditions 2 of 4
What: Restrain all six degrees of freedom of the nodeindicated.
How:
Boundary Conditions
1 pick node
Done
Displacement Restraint on Node form
Clamp
OK
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Apply boundary conditions 3 of 4
What: Restrain the indicated node, and leave only the Ytranslation free.
How:
1
1
Displacement Restraint on Node form
Specified
Specify Restraint...
Specified Restraint on Node form
Y Translation: Free
OK
Displacement Restraint on Node form
OK
Why: The Y-displacement is free so the element cancontract in the Y-direction as it stretches in theX-direction.
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Apply boundary conditions 4 of 4
What: Apply a force in the X-direction at two nodes.
How:
F 1
2
1
2 shift-pick
Force on Node form
X Force: 100000
OK
Recovery Point
FileSave
Things to notice The total force on the element is the sum of the twonodal forces.
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Solve the model 1 of 4
The most common type of solution is linear statics, inwhich the software simulates the response of the modelto a steady-state static load.
The software will solve the matrix equation:
{F} = [K] * {d}
At each degree of freedom, either a displacement iscalculated at the given forces, or reaction forces can becalculated where displacements are restrained.
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Solve the model 2 of 4
What: Create a solution set to define:• the type of solution to perform• the boundary condition set to use• the types of output to calculate and store• other options that affect the solution
How:
Model Solution
Manage Solution Sets form
Create...
Solution Set form
Name: Uniform Static Load
Don’t close the form; continue on the next page.
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Solve the model 3 of 4
What: Request output types of displacement and stress.List and store the output during the solve.
How:
Solution Set form
Output Selection...
Output Selection form
Output Type: Displacements
Store/List: Store and List
Output Type: Stress
Store/List: Store and List
OK
Solution Set form
OK
Manage Solution Sets form
Dismiss
Recovery Point
FileSave
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Solve the model 4 of 4
What: Define the location where temporary files andoutput files are stored. Then solve the model.
How:
Solve form
Output file: results1.lis
Solve
Things to notice The windows will temporarily disappear when the solverstarts. As the solve progresses, status messages aredisplayed in the I-DEAS List window. After the solve, anyerrors or warnings are listed in the I-DEAS List window.
During the solve, the file results1.lis is created to storedetails of the analysis. You may want to examine this filelater, then delete it.
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Interpret the results 1 of 8
When reviewing results you must ask:• Are the results from the model correct?• What do the results mean?
Some simple hand calculations can increase yourconfidence that the results are correct.
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Interpret the results 2 of 8
What: Select the deformation results to examine.
How:
Post Processing
Results Selection form
Display Results: Clear
Displacement_1 (select)
Deformation Results
OK
Why: Deformation results will be shown as displacednodes. Display results are shown in color. By clearingthe display results, you can see the displacements moreclearly.
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Interpret the results 3 of 8
What: Display the deformation.
How:
(to select all elements)
It’s good practice to look at displacements firstbecause they are more intuitive than stresses. Makesure the displacements look reasonable before youtry to interpret stresses.
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Interpret the results 4 of 8
What: Perform a hand calculation to check thedisplacements.
How:
The displacement in the X-direction should be:
X = PL/AE
P = 100,000 * 2L = 10A = 10 * 5E = 2E8
X = 0.0002
Things to notice This displacement value is small, verifying that theassumption of using linear static analysis is valid.
Most workstations have a desktop calculator you canuse to try this calculation.
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Interpret the results 5 of 8
What: Generate a report of the displacements to checkthe values.
How:
Report Writer form
Output Similar To: Deformed Display
Generate Report
Check I-DEAS List . Scroll back in the I-DEAS List window.X-displacements at nodes 2 and 4 should matchyour hand calculation. If they don’t match, you mayhave made a mistake in entering boundaryconditions, material properties, or physicalproperties.
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Interpret the results 6 of 8
What: Check Poisson’s ratio.
How:Using the values shown in the I-DEAS List window,divide the Y-translation by the X-translation at node 2:
Y-translation / X-translation = 0.3
Things to notice In the plot of deflection, the element shrinks in theY-direction as it stretches in the X-direction. Under auniaxial load, this value should match the value youentered for Poisson’s ratio (0.3).
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Interpret the results 7 of 8
What: Display the stress results.
How:
Results Selection form
Stress_2 (select)
Display Results
Component: Maximum Principal
OK
Things to notice Due to uniform loading, all values for the element are thesame. The fact that the stresses are less than the yieldstress verifies that using linear static analysis is a goodassumption.
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Interpret the results 8 of 8
What: Perform a hand calculation to check the stress.
How:
Stress = P/A
P = 100,000 * 2A = 10 * 5
Stress = 4000
Things to notice If 4000 is not the value on your plot, you may haveincorrectly entered the nodal coordinates, physicalproperties, or forces.
In this simple case, one element gives the correctanswer since the stress is uniform. In most modelswhere the stress is not uniform, the elements can onlyapproximate the true values, so you’ll require moreelements to obtain a correct answer.
You may want to examine the output.lis file on theoperating system. It will list the same results youdisplayed.
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Tutorial wrap-up
You have completed the What Is Finite ElementModeling tutorial.
Delete the FE model, then the part. This part is not usedin any other tutorial.
Hint
See also...For additional information on the concepts covered inthis tutorial, see the following:
Help, Manuals, Table of Contents
Simulation: Finite Element Modeling User’s GuideSimulation OverviewUsing Simulation Tools
Simulation: Model Solution/Optimization User’s GuideUsing the Solvers
Using Linear Statics Analysis
What’s next?You know that one element is not enough to model mostproblems, but how many should you use? Here we usedthin-shell elements, but when should you use other typesof elements? These questions and others will bediscussed in other tutorials.
To exit this tutorial, select: File
Exit
Warning! Do not use the menu in the I-DEAS Icons window to exit.Use the menu in the Acrobat Reader window.
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