ecc ansys workbook v2006 r1_structural analysis

ANSYS Structural Analysis Workbook V2006 R1

B6- Eureka Court, Beside BATA, Behind Chermas, Ameerpet, Hyderabad, AP, India Ph.55615442, 55625442, 55635442, 55105442, 55775442, 3755770, 23755774.

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Structural Analysis

Definition of Structural Analysis Structural analysis is probably the most common application of the finite element method. The term structural (or structure) implies not only civil engineering structures such as bridges and buildings, but also naval, aeronautical, and mechanical structures such as ship hulls, aircraft bodies, and machine housings, as well as mechanical components such as pistons, machine parts, and tools.

The seven types of structural analyses available in the ANSYS family of products are explained below. The primary unknowns (nodal degrees of freedom) calculated in a structural analysis are displacements. Other quantities, such as strains, stresses, and reaction forces, are then derived from the nodal displacements.

You can perform the following types of structural analyses.

Static Analysis--Used to determine displacements, stresses, etc. under static loading conditions. Both linear and nonlinear static analyses. Nonlinearities can include plasticity, stress stiffening, large deflection, large strain, hyper elasticity, contact surfaces, and creep.

Modal Analysis--Used to calculate the natural frequencies and mode shapes of a structure. Different mode extraction methods are available.

Harmonic Analysis--Used to determine the response of a structure to harmonically time-varying loads.

Transient Dynamic Analysis--Used to determine the response of a structure to arbitrarily time-varying loads. All nonlinearities mentioned under Static Analysis above are allowed.

Buckling Analysis--Used to calculate the buckling loads and determine the buckling mode shape. Both linear (eigenvalue) buckling and nonlinear buckling analyses are possible.

Spectrum Analysis--An extension of the modal analysis, used to calculate stresses and strains due to a response spectrum or a PSD input (random vibrations).

Explicit Dynamic Analysis--This type of structural analysis is only available in the ANSYS LS-DYNA program. ANSYS LS-DYNA provides an interface to the LS-DYNA explicit finite element program. Explicit dynamic analysis is used to calculate fast solutions for large deformation dynamics and complex contact problems.

We shall deal with Static, Modal, Harmonic, Transient Dynamic Analysis and Buckling Analysis as a part of our Structural Analysis Course.



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Static Analysis – some theory

A static analysis calculates the effects of steady loading conditions on a structure, while ignoring inertia and damping effects, such as those caused by time-varying loads. A static analysis can, however, include steady inertia loads (such as gravity and rotational velocity), and time-varying loads that can be approximated as static equivalent loads (such as the static equivalent wind and seismic loads commonly defined in many building codes).

Static analysis is used to determine the displacements, stresses, strains, and forces in structures or components caused by loads that do not induce significant inertia and damping effects. Steady loading and response conditions are assumed; that is, the loads and the structure's response are assumed to vary slowly with respect to time. The kinds of loading that can be applied in a static analysis include:

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

All of the following load types are applicable in a static analysis. Displacements (UX, UY, UZ, ROTX, ROTY, ROTZ) These are DOF constraints usually specified at model boundaries to define rigid support points. They can also indicate symmetry boundary conditions and points of known motion. The directions implied by the labels are in the nodal coordinate system. Forces (FX, FY, FZ) and Moments (MX, MY, MZ) These are concentrated loads usually specified on the model exterior. The directions implied by the labels are in the nodal coordinate system. Pressures (PRES) These are surface loads, also usually applied on the model exterior. Positive values of pressure act towards the element face (resulting in a compressive effect). Temperatures (TEMP) These are applied to study the effects of thermal expansion or contraction (that is, thermal stresses). The coefficient of thermal expansion must be defined if thermal strains are to be calculated. You can read in temperatures from a thermal analysis [LDREAD], or you can specify temperatures directly, using the BF family of commands. Fluences (FLUE) These are applied to study the effects of swelling (material enlargement due to neutron bombardment or other causes) or creep. They are used only if you input a swelling or creep equation. Gravity, Spinning, Etc. These are inertia loads that affect the entire structure. Density (or mass in some form) must be defined if inertia effects are to be included.



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Problem-1 This example was completed using ANSYS 7.0 and may slightly with the version that you are learning. The purpose of this example is to explain how to apply distributed loads and use element tables to extract data.

A distributed load of 1000 N/m (1 N/mm) will be applied to a solid steel beam with a rectangular cross section as shown in the figure below. The cross-section of the beam is 10mm x 10mm while the modulus of elasticity of the steel is 200GPa.

Preprocessing: Defining the Problem 1. Open preprocessor menu

/PREP7 2. Give example a Title

Utility Menu > File > Change Title ... /title, Distributed Loading

3. Create Keypoints Preprocessor > Modeling > Create > Keypoints > In Active CS K,#,x,y We are going to define 2 keypoints (the beam vertices) for this structure as given in the following table:

Keypoint Coordinates (x,y)1 (0,0) 2 (1000,0)

4. Define Lines Preprocessor > Modeling > Create > Lines > Lines > Straight Line L,K#,K#



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Create a line between Keypoint 1 and Keypoint 2.

5. Define Element Types Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the BEAM3 element. This element has 3 degrees of freedom (translation along the X and Y axis's, and rotation about the Z axis). With only 3 degrees of freedom, the BEAM3 element can only be used in 2D analysis.

6. Define Real Constants Preprocessor > Real Constants... > Add... In the 'Real Constants for BEAM3' window, enter the following geometric properties:

i. Cross-sectional area AREA: 100 ii. Area Moment of Inertia IZZ: 833.333

iii. Total beam height HEIGHT: 10 This defines an element with a solid rectangular cross section 10mm x 10mm.

7. Define Element Material Properties Preprocessor > Material Props > Material Models > Structural > Linear > Elastic > Isotropic In the window that appears, enter the following geometric properties for steel:

i. Young's modulus EX: 200000 ii. Poisson's Ratio PRXY: 0.3

8. Define Mesh Size Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > All Lines... For this example we will use an element length of 100mm.

9. Mesh the frame Preprocessor > Meshing > Mesh > Lines > click 'Pick All'

10. Plot Elements Utility Menu > Plot > Elements

You may also wish to turn on element numbering and turn off keypoint numbering Utility Menu > PlotCtrls > Numbering ...

Solution Phase: Assigning Loads and Solving 1. Define Analysis Type



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Solution > Analysis Type > New Analysis > Static ANTYPE,0

2. Apply Constraints Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Pin Keypoint 1 (ie UX and UY constrained) and fix Keypoint 2 in the y direction (UY constrained).

3. Apply Loads We will apply a distributed load, of 1000 N/m or 1 N/mm, over the entire length of the beam.

o Select Solution > Define Loads > Apply > Structural > Pressure > On Beams

o Click 'Pick All' in the 'Apply F/M' window. o As shown in the following figure, enter a value of 1 in the field 'VALI Pressure

value at node I' then click 'OK'.

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



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Note:

To have the constraints and loads appear each time you select 'Replot' you must change some settings. Select Utility Menu > PlotCtrls > Symbols.... In the window that appears, select 'Pressures' in the pull down menu of the 'Surface Load Symbols' section.

4. Solve the System Solution > Solve > Current LS SOLVE

Postprocessing: Viewing the Results 1. Plot Deformed Shape

General Postproc > Plot Results > Deformed Shape PLDISP.2

2. Plot Principle stress distribution

As shown previously, we need to use element tables to obtain principle stresses for line elements.

1. Select General Postproc > Element Table > Define Table 2. Click 'Add...' 3. In the window that appears

a. enter 'SMAXI' in the 'User Label for Item' section b. In the first window in the 'Results Data Item' section scroll down and

select 'By sequence num' c. In the second window of the same section, select 'NMISC, ' d. In the third window enter '1' anywhere after the comma

4. click 'Apply' 5. Repeat steps 2 to 4 but change 'SMAXI' to 'SMAXJ' in step 3a and change '1' to

'3' in step 3d. 6. Click 'OK'. The 'Element Table Data' window should now have two variables in

it. 7. Click 'Close' in the 'Element Table Data' window. 8. Select: General Postproc > Plot Results > Line Elem Res... 9. Select 'SMAXI' from the 'LabI' pull down menu and 'SMAXJ' from the 'LabJ'

pull down menu



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Note:

o ANSYS can only calculate the stress at a single location on the element. For this example, we decided to extract the stresses from the I and J nodes of each element. These are the nodes that are at the ends of each element.

o For this problem, we wanted the principal stresses for the elements. For the BEAM3 element this is categorized as NMISC, 1 for the 'I' nodes and NMISC, 3 for the 'J' nodes. A list of available codes for each element can be found in the ANSYS help files. (ie. type help BEAM3 in the ANSYS Input window).

As shown in the plot below, the maximum stress occurs in the middle of the beam with a value of 750 MPa.



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Problem-2 This example was created using ANSYS 7.0 to solve a simple 2D Truss problem. Problem Description Determine the nodal deflections, reaction forces, and stress for the truss system shown below (E = 200GPa, A = 3250mm2).

(Modified from Chandrupatla & Belegunda, Introduction to Finite Elements in Engineering, p.123)

Preprocessing: Defining the Problem 1. Give the Simplified Version a Title

In the Utility menu bar select File > Change Title:

The following window will appear:

Enter the title and click 'OK'. This title will appear in the bottom left corner of the 'Graphics' Window once you begin. Note: to get the title to appear immediately, select Utility Menu > Plot > Replot

2. Enter Keypoints The overall geometry is defined in ANSYS using keypoints which specify various principal coordinates to define the body. For this example, these keypoints are the ends of each truss.

o We are going to define 7 keypoints for the simplified structure as given in the following table

coordinate keypoint x y

1 0 0 2 1800 3118 3 3600 0 4 5400 3118 5 7200 0 6 9000 3118 7 10800 0

o (these keypoints are depicted by numbers in the above figure)



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o From the 'ANSYS Main Menu' select:

Preprocessor > Modeling > Create > Keypoints > In Active CS

The following window will then appear:

o To define the first keypoint which has the coordinates x = 0 and y = 0:

Enter keypoint number 1 in the appropriate box, and enter the x,y coordinates: 0, 0 in their appropriate boxes (as shown above). Click 'Apply' to accept what you have typed.

o Enter the remaining keypoints using the same method. Note: When entering the final data point, click on 'OK' to indicate that you are finished entering keypoints. If you first press 'Apply' and then 'OK' for the final keypoint, you will have defined it twice! If you did press 'Apply' for the final point, simply press 'Cancel' to close this dialog box.

Units Note the units of measure (ie mm) were not specified. It is the responsibility of the user to ensure that a consistent set of units are used for the problem; thus making any conversions where necessary. Correcting Mistakes When defining keypoints, lines, areas, volumes, elements, constraints and loads you are bound to make mistakes. Fortunately these are easily corrected so that you don't need to begin from scratch every time an error is made! Every 'Create' menu for generating these various entities also has a corresponding 'Delete' menu for fixing things up.

3. Form Lines The keypoints must now be connected



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We will use the mouse to select the keypoints to form the lines.

o In the main menu select: Preprocessor > Modeling > Create > Lines > Lines > In Active Coord. The following window will then appear:

o Use the mouse to pick keypoint #1 (i.e. click on it). It will now be marked by a

small yellow box. o Now move the mouse toward keypoint #2. A line will now show on the screen

joining these two points. Left click and a permanent line will appear. o Connect the remaining keypoints using the same method. o When you're done, click on 'OK' in the 'Lines in Active Coord' window,

minimize the 'Lines' menu and the 'Create' menu. Your ANSYS Graphics window should look similar to the following figure.



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Disappearing Lines Please note that any lines you have created may 'disappear' throughout your analysis. However, they have most likely NOT been deleted. If this occurs at any time from the Utility Menu select:

Plot > Lines 4. Define the Type of Element

It is now necessary to create elements. This is called 'meshing'. ANSYS first needs to know what kind of elements to use for our problem:

o From the Preprocessor Menu, select: Element Type > Add/Edit/Delete. The following window will then appear:

o Click on the 'Add...' button. The following window will appear:



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o For this example, we will use the 2D spar element as selected in the above

figure. Select the element shown and click 'OK'. You should see 'Type 1 LINK1' in the 'Element Types' window.

o Click on 'Close' in the 'Element Types' dialog box. 5. Define Geometric Properties

We now need to specify geometric properties for our elements: o In the Preprocessor menu, select Real Constants > Add/Edit/Delete

o Click Add... and select 'Type 1 LINK1' (actually it is already selected). Click

on 'OK'. The following window will appear:

o As shown in the window above, enter the cross-sectional area (3250mm):



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o Click on 'OK'. o 'Set 1' now appears in the dialog box. Click on 'Close' in the 'Real Constants'

window. 6. Element Material Properties

You then need to specify material properties: o In the 'Preprocessor' menu select Material Props > Material Models

o Double click on Structural > Linear > Elastic > Isotropic

We are going to give the properties of Steel. Enter the following field:

EX 200000

o Set these properties and click on 'OK'. Note: You may obtain the note 'PRXY will be set to 0.0'. This is poisson's ratio and is not required for this element type. Click 'OK' on the window to continue. Close the "Define Material Model Behavior" by clicking on the 'X' box in the upper right hand corner.

7. Mesh Size The last step before meshing is to tell ANSYS what size the elements should be. There are a variety of ways to do this but we will just deal with one method for now.

o In the Preprocessor menu select Meshing > Size Cntrls > ManualSize > Lines > All Lines



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o In the size 'NDIV' field, enter the desired number of divisions per line. For this

example we want only 1 division per line, therefore, enter '1' and then click 'OK'. Note that we have not yet meshed the geometry, we have simply defined the element sizes.

8. Mesh Now the frame can be meshed.

o In the 'Preprocessor' menu select Meshing > Mesh > Lines and click 'Pick All' in the 'Mesh Lines' Window

Your model should now appear as shown in the following window

Plot Numbering To show the line numbers, keypoint numbers, node numbers...

• From the Utility Menu (top of screen) select PlotCtrls > Numbering... • Fill in the Window as shown below and click 'OK'



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Now you can turn numbering on or off at your discretion

Saving Your Work Save the model at this time, so if you make some mistakes later on, you will at least be able to come back to this point. To do this, on the Utility Menu select File > Save as.... Select the name and location where you want to save your file. It is a good idea to save your job at different times throughout the building and analysis of the model to backup your work in case of a system crash or what have you.

Solution Phase: Assigning Loads and Solving You have now defined your model. It is now time to apply the load(s) and constraint(s) and solve the the resulting system of equations. Open up the 'Solution' menu (from the same 'ANSYS Main Menu').

1. Define Analysis Type First you must tell ANSYS how you want it to solve this problem:

o From the Solution Menu, select Analysis Type > New Analysis.

o Ensure that 'Static' is selected; i.e. you are going to do a static analysis on the

truss as opposed to a dynamic analysis, for example. o Click 'OK'.



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2. Apply Constraints

It is necessary to apply constraints to the model otherwise the model is not tied down or grounded and a singular solution will result. In mechanical structures, these constraints will typically be fixed, pinned and roller-type connections. As shown above, the left end of the truss bridge is pinned while the right end has a roller connection.

o In the Solution menu, select Define Loads > Apply > Structural > Displacement > On Keypoints

o Select the left end of the bridge (Keypoint 1) by clicking on it in the Graphics

Window and click on 'OK' in the 'Apply U,ROT on KPs' window.

o This location is fixed which means that all translational and rotational degrees

of freedom (DOFs) are constrained. Therefore, select 'All DOF' by clicking on it and enter '0' in the Value field and click 'OK'. You will see some blue triangles in the graphics window indicating the displacement contraints.

o Using the same method, apply the roller connection to the right end (UY constrained). Note that more than one DOF constraint can be selected at a time in the "Apply U,ROT on KPs" window. Therefore, you may need to 'deselect' the 'All DOF' option to select just the 'UY' option.

3. Apply Loads



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As shown in the diagram, there are four downward loads of 280kN, 210kN, 280kN, and 360kN at keypoints 1, 3, 5, and 7 respectively.

o Select Define Loads > Apply > Structural > Force/Moment > on Keypoints. o Select the first Keypoint (left end of the truss) and click 'OK' in the 'Apply F/M

on KPs' window.

o Select FY in the 'Direction of force/mom'. This indicate that we will be

applying the load in the 'y' direction o Enter a value of -280000 in the 'Force/moment value' box and click 'OK'. Note

that we are using units of N here, this is consistent with the previous values input.

o The force will appear in the graphics window as a red arrow. o Apply the remaining loads in the same manner.

The applied loads and constraints should now appear as shown below.

4. Solving the System

We now tell ANSYS to find the solution: o In the 'Solution' menu select Solve > Current LS. This indicates that we desire

the solution under the current Load Step (LS).



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o The above windows will appear. Ensure that your solution options are the same

as shown above and click 'OK'. o Once the solution is done the following window will pop up. Click 'Close' and

close the /STATUS Command Window..

Postprocessing: Viewing the Results 1. Hand Calculations

We will first calculate the forces and stress in element 1 (as labeled in the problem description).

2. Results Using ANSYS

Reaction Forces A list of the resulting reaction forces can be obtained for this element

o from the Main Menu select General Postproc > List Results > Reaction Solu.



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o Select 'All struc forc F' as shown above and click 'OK'

These values agree with the reaction forces claculated by hand above.

Deformation o In the General Postproc menu, select Plot Results > Deformed Shape. The

following window will appear.

o Select 'Def + undef edge' and click 'OK' to view both the deformed and the

undeformed object.



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o Observe the value of the maximum deflection in the upper left hand corner

(DMX=7.409). One should also observe that the constrained degrees of freedom appear to have a deflection of 0 (as expected!)

Deflection For a more detailed version of the deflection of the beam,

o From the 'General Postproc' menu select Plot results > Contour Plot > Nodal Solution. The following window will appear.

o Select 'DOF solution' and 'USUM' as shown in the above window. Leave the

other selections as the default values. Click 'OK'.



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o Looking at the scale, you may want to use more useful intervals. From the

Utility Menu select Plot Controls > Style > Contours > Uniform Contours... o Fill in the following window as shown and click 'OK'.

You should obtain the following.



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o The deflection can also be obtained as a list as shown below. General Postproc

> List Results > Nodal Solution select 'DOF Solution' and 'ALL DOFs' from the lists in the 'List Nodal Solution' window and click 'OK'. This means that we want to see a listing of all degrees of freedom from the solution.

o Are these results what you expected? Note that all the degrees of freedom were

constrained to zero at node 1, while UY was constrained to zero at node 7. o If you wanted to save these results to a file, select 'File' within the results

window (at the upper left-hand corner of this list window) and select 'Save as'. Axial Stress For line elements (ie links, beams, spars, and pipes) you will often need to use the Element Table to gain access to derived data (ie stresses, strains). For this example we should obtain axial stress to compare with the hand calculations. The Element Table is different for each element, therefore, we need to look at the help file for LINK1 (Type help link1 into the Input Line). From Table 1.2 in the Help file, we can see that SAXL can be obtained through the ETABLE, using the item 'LS,1'



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o From the General Postprocessor menu select Element Table > Define Table o Click on 'Add...'

o As shown above, enter 'SAXL' in the 'Lab' box. This specifies the name of the

item you are defining. Next, in the 'Item,Comp' boxes, select 'By sequence number' and 'LS,'. Then enter 1 after LS, in the selection box

o Click on 'OK' and close the 'Element Table Data' window. o Plot the Stresses by selecting Element Table > Plot Elem Table o The following window will appear. Ensure that 'SAXL' is selected and click

'OK'

o Because you changed the contour intervals for the Displacement plot to "User

Specified" - you need to switch this back to "Auto calculated" to obtain new values for VMIN/VMAX.

Utility Menu > PlotCtrls > Style > Contours > Uniform Contours ...



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Again, you may wish to select more appropriate intervals for the contour plot

o List the Stresses From the 'Element Table' menu, select 'List Elem Table' From the 'List Element Table Data' window which appears ensure

'SAXL' is highlighted Click 'OK'

Note that the axial stress in Element 1 is 82.9MPa as predicted analytically.



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Problem-3 Analysis of a power transmission tower Introduction: In this example you will learn to use the 2-D Truss element in ANSYS. Physical Problem: A power transmission tower is a common example of a structure that is made up of only truss members. These towers are actually 3-D structures, but for the sake of simplicity we will take a cross-sectional face of the tower. The tower is mainly subjected to loading in the vertical direction due to the weight of the cables. Also it is subjected to forces due to wind. In this example we will consider only loading due to the weight of the cables, which is in the vertical direction. Problem Description:

The tower is made up of trusses. You may recall that a truss is a structural element that experiences loading only in the axial direction. Units: Use S.I. units ONLY Geometry: the cross sections of each of the truss members is 6.25e-3 sq. meter. Material: Assume the structure is made of steel with modulus of elasticity E=200 GPa. Boundary conditions: The tower is constrained along X and Y directions at the bottom left corner, and along Y direction at the bottom right corner. Loading: The tower is loaded at the top. The load is in horizontal direction only, and its magnitude is 5000 N. Objective:

To determine deflection at each joint. To determine stress in each member. To determine reaction forces at the base.

You are required to hand in print outs for the above. Figure:

The five trusses at the top are each 3m in length.



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IMPORTANT: Convert all dimensions and forces into SI units.

STARTING ANSYS

Click on ANSYS 6.1 in the programs menu. Select Interactive. The following menu that comes up. Enter the working directory. All your files will be stored in this directory. Also enter 64 for Total Workspace and 32 for Database. Give your file a jobname. Click on Run.

MODELING THE STRUCTURE

Go to the ANSYS Utility Menu Click Workplane>WP Settings The following window comes up



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Check the Cartesian and Grid Only buttons Enter the values shown in the figure above. Go to the ANSYS Main Menu



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Click Preprocessor>-Modeling->Create>Keypoints>On Working Plane The following window comes up

Now we will pick the end points of the trusses. Select the keypoints on the workplane grid. Your points should look like this. If you cannot see the complete workplane then go to Utility Menu>PlotCntrls>Pan Zoom Rotate and zoom out to see the entire workplane



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Now create lines connecting the keypoints Click on Preprocessor>-Modeling->Create>-Lines->Lines>Straight Line Create lines by picking keypoints to make the figure shown below



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MATERIAL PROPERTIES

Go to the ANSYS Main Menu Click Preprocessor>Material Props>Material Models. In the window that comes up which is shown below, for Material Model 1, choose Structural>Linear>Elastic>Isotropic.



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Enter 1 for the Material Property Number and click OK. The following window comes up.

Fill in 2e11 for the Young's modulus and 0.3 for minor Poisson's Ratio. Click OK. Now the material 1 has the properties defined in the above table. We will use this material for the transmission tower.

ELEMENT PROPERTIES

SELECTING ELEMENT TYPE: Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Add... The following window opens.



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Type 1 in the Element type reference number Click on Structural Link and select 2D spar. Click OK. Close the 'Element types' window. So now we have selected Element type 1 to be a structural Link- 2D spar element. The trusses will be modeled as elements of type 1, i.e. structural link element. This finishes the selection of element type. Now we need to define the cross sectional area for this element. Go to Preprocessor>Real Constants In the "Real Constants" dialog box that comes up click on Add In the "Element Type for Real Constants" that comes up click OK. The following window comes up.

Type 6.25e-3 for cross sectional area and click on OK. We have now defined the cross sectional area of the link element.

MESHING:

DIVIDING THE TOWER INTO ELEMENTS: Go to Preprocessor>Meshing>Size Cntrls>ManualSize>Lines>All Lines. In the menu that comes up type 1 in the field for 'Number of element divisions'. This divides each of the lines in your figure into 1 element.



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Click on OK. Now when you mesh the figure ANSYS will automatically divide each line into 1 element. Now go to Preprocessor>-Meshing->Mesh>lines Select all the lines and click on OK in the "Mesh Lines" dialog box. Now each line is a truss element (Element 1).

BOUNDARY CONDITIONS AND CONSTRAINTS

APPLYING BOUNDARY CONDITIONS The tower is constrained in the X and Y directions at the bottom left corner and in the Y direction at the bottom right corner. Go to Main Menu. Click on Preprocessor>Loads>Define Loads>Apply>Structural>Displacement>On Keypoints Select the keypoint on which you want to apply displacement constraints. The following window comes up.



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Select UX and UY for the bottom left corner and UY for the bottom right corner and click OK. The default displacement value is taken to be zero. APPLYING FORCES Go to Main Menu. Click on Preprocessor>Loads>Define Loads>Apply>Structural>Forces/Moment>On Nodes. Select the top node. Click on OK in the 'Apply F/M on Nodes' window. The following window will appear.

Select FX and enter 5000 as the Force/Moment value. Click on OK. The figure on the ANSYS Graphics window will look like the following.



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Now the Modeling of the problem is done.

SOLUTION

Go to ANSYS Main Menu>Solution>-Analysis Type->New Analysis. Select static and click on OK. Go to Solution>-Solve->Current LS. Wait for ANSYS to solve the problem. Click on OK and close the 'Information' window.

POST-PROCESSING

Listing the results. Go to ANSYS Main Menu. Click on General Postproc>List Results>Nodal Solution. The following window will come up.



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Select DOF solution and All U's. Click on OK. The nodal displacements will be listed as follows.

Similarly you can list the stresses for each element by clicking Gen Postprcessing>List Results>Element Solution. Now select LineElem Results. The following table will be listed.



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MODIFICATION

You can also plot the displacements and stress. Go to General Postproc>Plot Results>-Contour Plot->Element Solution. The following window will come up.

Select a stress to be plotted and click OK. The output will be like this.



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Problem-4: Analysis 2-D Beam structure Introduction: In this example you will learn to use the 2-D Beam element in ANSYS. Physical Problem: Structural analysis of the frame shown below. Problem Description:

The structure is made up of beams. You may recall that a beam is a structural element whose length is very large compared to the other two dimensions. Units: Use S.I. units ONLY Geometry: The members have a annular cross-section. The cross sections (A) of each of the truss members is 5.5e-3 sq meter. The polar radius of gyration (R) is 5.5e-2 meter. (hint: Use the values of A and R to find Izz then find the value of the outer diameter (The beam height)) Material: Assume the structure is made of steel with modulus of elasticity E=210 GPa. Boundary conditions: All the DOFs are constrained at the bottom end, i.e. the bottom end is a built-in end. Loading: The structure is loaded at the ends of the two arms. The load is in the negative Y direction. The load value is 5000 N each. Objective:

To determine deflections at the points of application of load. To determine the maximum stress in the structure. Also determine the maximum possible load the frame can take. Look up for the value of yield stress for steel. Assume a factor of safety of 1.25.


IMPORTANT: Convert all dimensions and forces into SI units.



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STARTING ANSYS

Click on ANSYS 6.1 in the programs menu. Select Interactive. The following menu that comes up. Enter the working directory. All your files will be stored in this

Click on Run. MODELING THE STRUCTURE



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The following window comes up

Now we will pick the end points of the trusses. 5 meters is now 1 X 5 units, since each cell in the grid is 1 unit across, 5 meters is 5 cells wide. Using this conversion select the keypoints on the workplane grid. Your points should look like this.



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If you cannot see the complete workplane then go to Utility Menu>PlotCntrls>Pan Zoom Rotate and zoom out to see the entire workplane.

Now create lines connecting the keypoints

Click on Preprocessor>Modeling>Create>Lines>Lines>Straight Line Create lines by picking keypoints to make the figure shown below.



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MATERIAL PROPERTIES

Go to the ANSYS Main Menu Click Preprocessor>Material Props>Material Models. In the window that comes up choose Structural>Linear>Elastic>Isotropic. The following window will appear.

Double Click Isotropic. The following window comes up.



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Fill in 2.1e11 for the Young's modulus and 0.3 for Poisson's Ratio. Click OK Now the material 1 has the properties defined in the above table. We will use this material for the structure.

ELEMENT PROPERTIES: SELECTING ELEMENT TYPE:

Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Add... The following window opens.

Type 1 in the Element type reference number. Click on Structural Beam and select 2D elastic. Click OK. Close the 'Element types' window. So now we have selected Element type 1 to be a structural Beam- 2D elastic element. The trusses will be modeled as elements of type 1, i.e. structural beam element. This finishes the selection of element type. Now we need to define the cross sectional area, the second moment of inertia etc. for this element. Go to Preprocessor>Real Constants. In the "Real Constants" dialog box that comes up click on Add In the "Element Type for Real Constants" that comes up click OK. The following window comes up



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Type in 5.5e-3 for cross sectional area, calculate Izz from the value of the cross-sectional area and polar

We have now defined the geometric properties of the beam element. MESHING: DIVIDING THE STRUCTURE INTO ELEMENTS:



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Now each line is a truss element (Element 1).

BOUNDARY CONDITIONS AND CONSTRAINTS: APPLYING BOUNDARY CONDITIONS

The tower is constrained in the DOFs at the bottom node. Go to Main Menu Click on Preprocessor>Loads>Define Loads>Apply>Structural>Displacement>On Keypoints.

Select the keypoint on which you want to apply displacement constraints. The following window comes up.

Select All DOF and click OK. APPLYING FORCES

Go to Main Menu Click on Preprocessor>Loads>Define Loads>Apply>Forces/Moment>On Nodes. Select the top right node and the top left node. Click on OK in the 'Apply F/M on Nodes' window. The following window will appear. Enter the value of the force.



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The figure looks like this now.

Now the Modeling of the problem is done SOLUTION: Go to ANSYS Main Menu>Solution>Analysis Type>New Analysis. Select static and click on OK. Go to Solution>Solve>Current LS Wait for ANSYS to solve the problem. Click on OK and close the 'Information' window POST-PROCESSING: Listing the results Go to ANSYS Main Menu Click on General Postprocessing>List Results>Nodal Solution. The following window will come up.



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Similarly you can list the stresses for each element by clicking Gen Postprocessing>List Results>Element Solution. Now select LineElem Results. The following table will be listed.

MODIFICATIONS: You can also plot the displacements and stress.



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Go to General Postprocessing>Plot Results>Contour Plot>Element Solution. The following window will come up.

Select a stress to be plotted and click OK. The output will be like this.



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Problem-5: Analysis of a Steel Bracket Introduction: In this example you will learn to use the Solid 8 Node element in ANSYS. Physical Problem: Structural analysis of the Steel Bracket shown in the figure. This is a typical bracket used to support bookshelves. Problem Description:

We will model the bracket as a solid 8 node plane stress element. By a plane stress element we are assuming that there are no stresses in the thickness direction of the bracket. Geometry: The thickness of the bracket is 3.125 mm Material: Assume the structure is made of steel with modulus of elasticity E=200 GPa. Boundary conditions: The bracket is fixed at its left edge. Loading: The bracket is loaded uniformly along its top surface. The load is 2625 N/meter. Objective:

Plot deformed shape Determine the principal stress and the von Mises stress. (Use the stress plots to determine these) Remodel the bracket without the fillet at the corner, and see how principal stress and von Mises stress change.


IMPORTANT: Convert all dimensions and forces into SI units



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STARTING ANSYS:

Click on ANSYS 6.1 in the programs menu. Select Interactive. The following menu that comes up. Enter the working directory. All your files will be stored in this directory. Also enter 64 for Total Workspace and 32 for Database. Click on Run.

MODELING THE STRUCTURE:

Go to the ANSYS Utility Menu Click Workplane>WP Settings The following window comes up



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Check the Cartesian and Grid Only buttons Enter the values shown in the figure above.

Go to the ANSYS Main Menu Preprocessor>Modeling>Create>Keypoints>On Working Plane Outline a part of the bracket as shown in the figure.



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Note: To turn on the numbering: ANSYS Utility Menu>Plot Controls>Numbering...

Now create lines between keypoints, then create area inside. Go to Preprocessor>Modeling>Create>Areas>Arbitrary>By Lines.



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Now go to Preprocessor>Modeling>Create>Lines>Line Fillet. The following window comes up. Select the two lines between which you want the fillet and click OK.

In the box that comes up enter 0.025 for fillet radius and click OK.



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Now go to Preprocessor>Modeling>Create>Areas>Arbitrary>By Lines to fill the fillet area.



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Go to Preprocessor>Modeling>Create>Areas>Circles>Solid Circle and create the two circles with centre at the midpoint of the right edge and the bottom edge of the bracket and the diameter equal to the length of that edge.



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Now go to Preprocessor>Modeling>Operate>Booleans>Add>Areas and select all areas you have created to make a single area.



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Now go to Workplane>WP Settings and change the Snap Incr and grid settings to 0.00625. We do this so that we can make the small inner circle whose radius is 0.00625 meter. Go to Preprocessor>Modeling>Create>Areas>Circles>Solid Circle and create the a circle with center at the midpoint of the right edge of the horizontal rectangle and the radius equals to 0.00625. Do the same thing for the vertical rectangle.



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Now go to Preprocessor>Modeling>Operate>Booleans>Subtract>Areas. First select the base area from which the smaller area will be subtracted. Say OK. Now select the smaller circles and say OK. the

smaller circles will now be subtracted and the figure will look like this:



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MATERIAL PROPERTIES:

Go to the ANSYS Main Menu>Preprocessor>Material Props>Material Models. From this window, select Structural>Linear>Elastic>Isotropic.

Enter 1 for the Material Property Number and click OK. The following window comes up.



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Fill in 2e11 for the Young's modulus and 0.3 for minor Poisson's Ratio. Click OK Now the material 1 has the properties defined in the above table. We will use this material for the structure.

ELEMENT PROPERTIES:

SELECTING ELEMENT TYPE:

Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Add... The following window opens.

Type 1 in the Element type reference number. Click on Structural Solid and select Quad 8 node 82. Click OK. Close the 'Element types' window. Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Options... The following window opens.



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Select Plane strs w/thk for K3 and click OK. So now we have selected Element type 1 to be a Structural Solid 8 node element. The bracket will now be modeled as elements of this type. Now we need to define the thickness for this element. Go to Preprocessor>Real Constants In the "Real Constants" dialog box that comes up click on Add In the "Element Type for Real Constants" that comes up click OK. The following window comes up.

Fill in the relevant values and click on OK. We have now defined the thickness of the element.

MESHING:

DIVIDING THE BRACKET INTO ELEMENTS: Go to Preprocessor>Meshing>Size Controls>Manual Size>Lines>Picked Lines. Pick all the lines on the outer boundary of the figure and click OK. In the menu that comes up type 0.0125 in the field for 'Element edge length'.



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Click on OK. Repeat the process to divide the lines forming the small inner circle. In this case enter 0.001 in the field for 'Element edge length'. Now go to Preprocessor>Meshing>Mesh>Areas>Free. Select the area and click on OK in the "Mesh Areas" dialog box. Now the bracket is divided into Solid 8 node elements.

BOUNDARY CONDITIONS AND CONSTRAINTS:

APPLYING BOUNDARY CONDITIONS The bracket is fixed at the left edge. Go to Main Menu Preprocessor>Loads>Define Loads>Apply>Structural>Displacement>On Lines. Select the line on the left edge and click OK. The following window comes up:



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Select All DOF and click OK.

APPLYING FORCES Go to Main Menu Preprocessor>Loads>Define Loads>Apply>Structural>Pressure>On Line. Select the top line. Click on OK in the 'Apply PRES on lines' window. The following window will appear:



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Enter the value of the pressure as shown above. Click OK.

The model should look like the one below.



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Now the Modeling of the problem is done. SOLUTION:

Go to ANSYS Main Menu>Solution>Analysis Type>New Analysis. Select static and click on OK. Go to Solution>Solve>Current LS. Wait for ANSYS to solve the problem. Click on OK and close the 'Information' window.

POST-PROCESSING:

Listing the results. Go to ANSYS Main Menu Click on General Postprocessing>List Results>Nodal Solution. The following window will come up.



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Similarly you can list the stresses for each element by clicking General Postprcessing>List Results>Element Solution. Now select LineElem Results. The following table will be listed.



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MODIFICATIONS:

You can also plot the displacements and stress. Go to General Postprocessing>Plot Results>Deformed shape. The following window comes up.

Select Def+undeformed and click OK. The output will be like the figure below.

Select a stress (SEQV) to be plotted and click OK. The output will be like this.



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Problem-6- Analysis of a steel bracket The problem to be modeled in this example is a simple bracket shown in the following figure. This bracket is to be built from a 20 mm thick steel plate. A figure of the plate is shown below.

This plate will be fixed at the two small holes on the left and have a load applied to the larger hole on the right.

Preprocessing: Defining the Problem 1. Give the Bracket example a Title

Utility Menu > File > Change Title 2. Form Geometry

Again, Boolean operations will be used to create the basic geometry of the Bracket. a. Create the main rectangular shape

The main rectangular shape has a width of 80 mm, a height of 100mm and the bottom left corner is located at coordinates (0,0)

Ensure that the Preprocessor menu is open. (Alternatively type /PREP7 into the command line window)

Now instead of using the GUI window we are going to enter code into the 'command line'. Now I will explain the line required to create a rectangle:

BLC4, XCORNER, YCORNER, WIDTH, HEIGHT BLC4, X coord (bottom left), Y coord (bottom

left), width, height Therefore, the command line for this rectangle is BLC4,0,0,80,100

b. Create the circular end on the right hand side The center of the circle is located at (80,50) and has a radius of 50 mm

The following code is used to create a circular area: CYL4, XCENTER, YCENTER, RAD1 CYL4, X coord for the center, Y coord for the center, radius

Therefore, the command line for this circle is CYL4,80,50,50 c. Now create a second and third circle for the left hand side using the following

dimensions:



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parameter circle 2 circle 3

XCENTER 0 0

YCENTER 20 80

RADIUS 20 20

d. Create a rectangle on the left hand end to fill the gap between the two small circles.

XCORNER -20

YCORNER 20

WIDTH 20

HEIGHT 60

e. Your screen should now look like the following...

f. g. Boolean Operations - Addition

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

Modeling > Operate > Booleans > Add > Areas In the 'Add Areas' window, click on 'Pick All'

(Alternatively, the command line code for the above step is AADD,ALL) You should now have the following model:



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h. Create the Bolt Holes

We now want to remove the bolt holes from this plate. Create the three circles with the parameters given below:

parameter circle 1 circle 2 circle 3

WP X 80 0 0

WP Y 50 20 80

radius 30 10 10

Now select Preprocessor > Modeling > Operate > Booleans > Subtract > Areas

Select the base areas from which to subract (the large plate that was created)

Next select the three circles that we just created. Click on the three circles that you just created and click 'OK'. (Alternatively, the command line code for the above step is ASBA,6,ALL) Now you should have the following:



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3. Define the Type of Element

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

(Alternatively, the command line code for the above step is ET,1,PLANE82 followed by KEYOPT,1,3,3)

o Under the Extra Element Output K5 select nodal stress. 4. Define Geometric Contants

o Preprocessor > Real Constants > Add/Edit/Delete o Enter a thickness of 20mm.

(Alternatively, the command line code for the above step is R,1,20) 5. Element Material Properties

o Preprocessor > Material Props > Material Library > Structural > Linear > Elastic > Isotropic We are going to give the properties of Steel. Enter the following when prompted: EX 200000 PRXY 0.3

(The command line code for the above step is MP,EX,1,200000 followed by MP,PRXY,1,0.3)

6. Mesh Size o Preprocessor > Meshing > Size Cntrls > Manual Size > Areas > All Areas o Select an element edge length of 5. Again, we will need to make sure the model

has converged. (Alternatively, the command line code for the above step is AESIZE,ALL,5,)

7. Mesh o Preprocessor > Meshing > Mesh > Areas > Free and select the area when

prompted (Alternatively, the command line code for the above step is AMESH,ALL)



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8. 9. Saving Your Job

Utility Menu > File > Save as...

Solution Phase: Assigning Loads and Solving You have now defined your model. It is now time to apply the load(s) and constraint(s) and solve the the resulting system of equations.

1. Define Analysis Type o 'Solution' > 'New Analysis' and select 'Static'.

(Alternatively, the command line code for the above step is ANTYPE,0) 2. Apply Constraints

As illustrated, the plate is fixed at both of the smaller holes on the left hand side. o Solution > Define Loads > Apply > Structural > Displacement > On Nodes o Instead of selecting one node at a time, you have the option of creating a box,

polygon, or circle of which all the nodes in that area will be selected. For this case, select 'circle' as shown in the window below. (You may want to zoom in to select the points Utilty Menu / PlotCtrls / Pan, Zoom, Rotate...) Click at the center of the bolt hole and drag the circle out so that it touches all of the nodes on the border of the hole.



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o Click on 'Apply' in the 'Apply U,ROT on Lines' window and constrain all

DOF's in the 'Apply U,ROT on Nodes' window. o Repeat for the second bolt hole.

3. Apply Loads As shown in the diagram, there is a single vertical load of 1000N, at the bottom of the large bolt hole. Apply this force to the respective keypoint ( Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints Select a force in the y direction of -1000) The applied loads and constraints should now appear as shown below.



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4. Solving the System

Solution > Solve > Current LS

Post-Processing: Viewing the Results We are now ready to view the results. We will take a look at the deflected shape and the stress contours once we determine convergence has occured.

1. Convergence using ANSYS As shown previously, it is necessary to prove that the solution has converged. Reduce the mesh size until there is no longer a sizeable change in your convergence criteria.

2. Deformation o General Postproc > Plot Results > Def + undeformed to view both the

deformed and the undeformed object. The graphic should be similar to the following



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o Observe the locations of deflection. Ensure that the deflection at the bolt hole is

indeed 0. 3. Deflection

o To plot the nodal deflections use General Postproc > Plot Results > Contour Plot > Nodal Solution then select DOF Solution - USUM in the window.

o Alternatively, obtain these results as a list. (General Postproc > List Results >

Nodal Solution...) o Are these results what you expected? Note that all translational degrees of

freedom were constrained to zero at the bolt holes.



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4. Stresses

o General Postproc > Plot Results > Nodal Solution... Then select von Mises Stress in the window.

o You can list the von Mises stresses to verify the results at certain nodes

General Postproc > List Results. Select Stress, Principals SPRIN



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Problem-7 (Cylindrical Shell Under Pressure)- to be taught in the class. Analysis Type(s): Static Analysis (ANTYPE = 0) Element Type(s): Plastic Axisymmetric Conical Shell Element (SHELL51) A long cylindrical pressure vessel of mean diameter d and wall thickness t has closed ends and is subjected to an internal pressure P. Determine the axial stress σy and the hoop stress σz in the vessel at the mid thickness of the wall.

Material Properties E = 30 x 106 psi υ = 0.3

Geometric Propertiest = 1 in d = 120 in

Loading P = 500 psi

Analysis Assumptions and Modeling Notes

An arbitrary axial length of 10 inches is selected. Nodal coupling is used in the radial direction. An axial force of 5654866.8 lb ((Pπd2)/4) is applied to simulate the closed-end effect.

Results Comparison

Target[1] ANSYS Ratio Stressy , psi 15,000. 15,000.[2] 1.000 Stressz , psi 29,749. 29,750. 1.000



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Additional Static Problems from worked out examples (Students may do these problems, if they have completed the above problems before schedule) Problem-1 (Statically Indeterminate Reaction Force Analysis)

Analysis Type(s): Static Analysis (ANTYPE = 0)

Element Type(s): Spar Elements (LINK1) A prismatic bar with built-in ends is loaded axially at two intermediate cross-sections by forces F1 and F2. Determine the reaction forces R1 and R2.

Material Properties

E = 30 x 106psi

Geometric Properties = 10 in.

a = b = 0.3

Loading F1 = 2F2 = 1000 lb


Nodes are defined where loads are to be applied. Since stress results are not to be determined, a unit cross-sectional area is arbitrarily chosen.

Results Comparison

Target ANSYS Ratio R1 , lb 900.0 900.0 1.000 R2 , lb 600.0 600.0 1.000



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Problem-2 (Deflection of a Hinged Support) Analysis Type(s): Static Analysis (ANTYPE = 0) Element Type(s): 2-D Spar Elements (LINK1)

A structure consisting of two equal steel bars, each of length and cross-sectional area A, with hinged ends is subjected to the action of a load F. Determine the stress, σ, in the bars and the deflection, δ, of point 2. Neglect the weight of the bars as a small quantity in comparison with the load F.

Material Properties E = 30 x 106 psi

Geometric Properties

= 15 ft A = .05 in Θ = 30°

Loading F = 5000 lb

Results Comparison

Target ANSYS Ratio Stress, psi 10,000. 10,000. 1.000 Deflection, in -0.120 -0.120 1.000



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3) Determine the force in each member of the following truss. Indicate if the member is in tension or compression. The cross-sectional area of each member is 0.01 m, the Young's modulus is 200x109 N/m2 and Poisson ratio is 0.3.

4) Determine the force in each member of the following truss using ANSYS. Indicate if the member is in tension or compression. Use the same LINK1 element as in the example. The cross-sectional area of each member is 0.02 m2, Young's modulus is 200x109 N/m2 and Poisson's ratio is 0.3. Verify your results by calculating the forces manually.



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5) Determine the nodal deflections, reaction forces, and stress for the truss system shown below (E = 200GPa, A = 3250mm2).

6) Consider the square plate of uniform thickness with a circular hole with dimensions shown in the figure below. The thickness of the plate is 1 mm. The Young's modulus E =107 MPa and the Poisson ratio is 0.3. A uniform pressure p=1 MPa acts on the boundary of the hole. Assume that plane stress conditions prevail. The stress and displacement fields are to be determined using ANSYS.

7) Consider the square plate of uniform thickness with a circular hole with dimensions shown in the figure below. The plate is uniaxially loaded with a uniform pressure p=1 MPa. In addition, the plate is made of a Glass/Epoxy composite material with the fibers oriented in same direction as the applied load. The material properties are as follows: Young's modulus in the fiber direction Ex = 59.3 GPa Young's modulus in the transverse direction Ey = 22 GPa In-plane shear modulus Gxy = 8.96 GPa Major Poisson's ratio nxy = 0.26 Minor Poisson's ratio nyx= 0.047 The circumferential stress concentration on the boundary of the hole is to be determined using ANSYS.



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Modal analysis- Some Theory

Definition of Modal Analysis

You use modal analysis to determine the vibration characteristics (natural frequencies and mode shapes) of a structure or a machine component while it is being designed. It also can be a starting point for another, more detailed, dynamic analysis, such as a transient dynamic analysis, a harmonic response analysis, or a spectrum analysis.

Uses for Modal Analysis

You use modal analysis to determine the natural frequencies and mode shapes of a structure. The natural frequencies and mode shapes are important parameters in the design of a structure for dynamic loading conditions. They are also required if you want to do a spectrum analysis or a mode superposition harmonic or transient analysis. You can do modal analysis on a prestressed structure, such as a spinning turbine blade. Another useful feature is modal cyclic symmetry, which allows you to review the mode shapes of a cyclically symmetric structure by modeling just a sector of it. Modal analysis in the ANSYS family of products is a linear analysis. Any nonlinearities, such as plasticity and contact (gap) elements, are ignored even if they are defined. You can choose from several mode extraction methods: Block Lanczos (default), subspace, PowerDynamics, reduced, unsymmetric, damped, and QR damped. The damped and QR damped methods allow you to include damping in the structure. Details about mode extraction methods are covered later in this section.

Overview of Steps in a Modal Analysis

The procedure for a modal analysis consists of four main steps: 1. Build the model. 2. Apply loads and obtain the solution. 3. Expand the modes. 4. Review the results.

When building your model, remember these points:

• Only linear behavior is valid in a modal analysis. If you specify nonlinear elements, they are treated as linear. For example, if you include contact elements, their stiffnesses are calculated based on their initial status and never change.

• Material properties can be linear, isotropic or orthotropic, and constant or temperature-dependent. You must define both Young's modulus (EX) (or stiffness in some form) and density (DENS) (or mass in some form) for a modal analysis. Nonlinear properties are ignored. If applying element damping, you must define the required real constants for the specific element type (COMBIN7, COMBIN14, COMBIN37, and so on).

Analysis Types and Options

Option Command GUI Path New Analysis ANTYPE Main Menu> Solution> Analysis Type> New Analysis Analysis Type: Modal (see Note below)

ANTYPE Main Menu> Solution> Analysis Type> New Analysis> Modal

Mode Extraction Method MODOPT Main Menu> Solution> Analysis Type> Analysis Options

Number of Modes to Extract MODOPT Main Menu> Solution> Analysis Type> Analysis Options



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Option Command GUI Path

No. of Modes to Expand (see Note below)

MXPAND Main Menu> Solution> Analysis Type> Analysis Options

Mass Matrix Formulation LUMPM Main Menu> Solution> Analysis Type> Analysis Options

Prestress Effects Calculation PSTRES Main Menu> Solution> Analysis Type> Analysis Options

Loads Applicable in a Modal Analysis

Load Type Category Cmd Family GUI Path Displacement (UX, UY, UZ, ROTX, ROTY, ROTZ)

Constraints D Main Menu> Solution> Define Loads> Apply> Structural> Displacement

In an analysis, loads can be applied, removed, operated on, or listed.

Applying Loads Using Commands

Load Commands for a Modal Analysis lists all the commands you can use to apply loads in a modal analysis. Load Commands for a Modal Analysis

Load Type

Solid Model or FE

Entity Apply Delete List Operate Apply Settings

Solid Model Keypoints DK DKDELE DKLIST DTRAN - Solid Model Lines DL DLDELE DLLIST DTRAN - Solid Model Areas DA DADELE DALIST DTRAN -

Displacement

Finite Elem Nodes D DDELE DLIST DSCALE DSYM, DCUM Applying Loads Using the GUI

All loading operations (except List; see Listing Loads as below) are accessed through a series of cascading menus. From the Solution menu, you select the operation (apply, delete, and so on), then the load type (displacement, force, and so on), and then the object to which you are applying the load (key point, line, node, and so on). For example, to apply a displacement load to a line, follow this GUI path: GUI: Main Menu> Solution> Define Loads> Apply> Structural> Displacement> On lines

Listing Loads

To list existing loads, follow this GUI path: GUI: Utility Menu> List>Loads> load type

Modal Analysis Methods Used in our curriculum • Subspace method

The subspace method is used for large symmetric eigenvalue problems. Several solution controls are available to control the subspace iteration process. When doing a modal analysis with a large number of constraint equations, use the subspace method with the frontal solver instead of the JCG solver, or use the Block Lanczos mode extraction method.

• Reduced (Householder) method The reduced method is faster than the subspace method because it uses reduced (condensed) system matrices to calculate the solution. However, it is less accurate because the reduced mass matrix is approximate.



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Problem-8 - Modal Analysis of a Cantilever Beam – illustrated stepwise

Introduction This example was created using ANSYS 7.0 The purpose of this example is to outline the steps required to do a simple modal analysis of the cantilever beam shown below.

Preprocessing: Defining the Problem The simple cantilever beam is used in all of the Dynamic Analysis Examples. If you haven't created the model in ANSYS, please use the links below. Both the command line codes and the GUI commands are shown in the respective links.

Solution: Assigning Loads and Solving 1. Define Analysis Type

Solution > Analysis Type > New Analysis > Modal ANTYPE,2

2. Set options for analysis type: o Select: Solution > Analysis Type > Analysis Options..

The following window will appear



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o As shown, select the Subspace method and enter 5 in the 'No. of modes to

extract' o Check the box beside 'Expand mode shapes' and enter 5 in the 'No. of modes to

expand' o Click 'OK'

Note that the default mode extraction method chosen is the Reduced Method. This is the fastest method as it reduces the system matrices to only consider the Master Degrees of Freedom (see below). The Subspace Method extracts modes for all DOF's. It is therefore more exact but, it also takes longer to compute (especially when the complex geometries).

o The following window will then appear



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For a better understanding of these options see the Commands manual.

o For this problem, we will use the default options so click on OK. 3. Apply Constraints

Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Fix Keypoint 1 (ie all DOFs constrained).


Postprocessing: Viewing the Results 1. Verify extracted modes against theoretical predictions

o Select: General Postproc > Results Summary... The following window will appear



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The following table compares the mode frequencies in Hz predicted by theory and ANSYS.

Mode Theory ANSYS Percent Error

1 8.311 8.300 0.1

2 51.94 52.01 0.2

3 145.68 145.64 0.0

4 285.69 285.51 0.0

5 472.22 472.54 0.1

Note: To obtain accurate higher mode frequencies, this mesh would have to be refined even more (i.e. instead of 10 elements, we would have to model the cantilever using 15 or more elements depending upon the highest mode frequency of interest).

2. View Mode Shapes o Select: General Postproc > Read Results > First Set

This selects the results for the first mode shape o Select General Postproc > Plot Results > Deformed shape . Select 'Def +

undef edge' The first mode shape will now appear in the graphics window.

o To view the next mode shape, select General Postproc > Read Results > Next Set . As above choose General Postproc > Plot Results > Deformed shape . Select 'Def + undef edge'.

o The first four mode shapes should look like the following:



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3. Animate Mode Shapes

o Select Utility Menu (Menu at the top) > Plot Ctrls > Animate > Mode Shape The following window will appear

o Keep the default setting and click 'OK' o The animated mode shapes can be seen.



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Using the Reduced Method for Modal Analysis This method employs the use of Master Degrees of Freedom. These are degrees of freedom that govern the dynamic characteristics of a structure. For example, the Master Degrees of Freedom for the bending modes of cantilever beam are

For this option, a detailed understanding of the dynamic behavior of a structure is required. However, going this route means a smaller (reduced) stiffness matrix, and thus faster calculations. The steps for using this option are quite simple.

• Instead of specifying the Subspace method, select the Reduced method and specify 5 modes for extraction.

• Complete the window as shown below

Note:For this example both the number of modes and frequency range was specified. ANSYS then extracts the minimum number of modes between the two.

• Select Solution > Master DOF > User Selected > Define • When prompted, select all nodes except the left most node (fixed).

The following window will appear:



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• Select UY as the 1st degree of freedom (shown above).

The same constraints are used as above. The following table compares the mode frequencies in Hz predicted by theory and ANSYS (Reduced).

Mode Theory ANSYS Percent Error

1 8.311 8.300 0.1

2 51.94 52.01 0.1

3 145.68 145.66 0.0

4 285.69 285.71 0.0

5 472.22 473.66 0.3

As you can see, the error does not change significantly. However, for more complex structures, larger errors would be expected using the reduced method.



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Problem-9 (Natural Frequency of a Spring-Mass System) Analysis Type(s): Mode-frequency Analysis (ANTYPE = 2) Element Type(s): Spring Element (COMBIN14)

Mass Element (MASS21) An instrument of weight W is set on a rubber mount system having a stiffness k. Determine its natural frequency of vibration f.

Material Properties k = 48 lb/in W = 2.5 lb

Loading g = 386 in/sec2


The spring length is arbitrarily selected. One master degree of freedom is chosen at the mass in the spring length direction. The weight of the lumped mass element is divided by gravity in order to obtain the mass. Mass = W/g = 2.5/386 = .006477 lb-sec2/in.

Results Comparison

Target ANSYS Ratio f, Hz 13.701 13.701 1.000



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Problem-10 (Natural Frequency of a Motor-Generator) Analysis Type(s): Mode-frequency Analysis (ANTYPE = 2) Element Type(s): Pipe Elements (PIPE16)

Mass Elements (MASS21) A small generator of mass m is driven off a main engine through a solid steel shaft of diameter d. If the polar moment of inertia of the generator rotor is J, determine the natural frequency f in torsion. Assume that the engine is large compared to the rotor so that the engine end of the shaft may be assumed to be fixed. Neglect the mass of the shaft also.

Material PropertiesE = 31.2 x 106 psi m = 1 lb-sec2/in

Geometric Properties d = .375 in

= 8.00 in J = .031 lb-in-sec2

Analysis Assumptions and Modeling Notes One rotational master degree of freedom is selected at the mass. The wall thickness of the pipe is defined as half the diameter to obtain a solid cross-section. Results Comparison Target ANSYS Ratio f, Hz 48.781 48.781 1.00



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Problem-11 (Fundamental Frequency of a Simply Supported Beam) Analysis Type(s): Mode-frequency Analysis (ANTYPE = 2) Element Type(s): Beam Elements (BEAM3) Determine the fundamental frequency f of a simply-supported beam of length and uniform cross-section as shown below.

Material Properties E = 30 x 106 psi w = 1.124 lb/in ρ = w/Ag = .000728 lb-sec2/in4


= 80 in A = 4 in2 h = 2 in I = 1.3333 in4

Loading

g = 386 in/sec2

Analysis Assumptions and Modeling Notes Three lateral master degrees of freedom are selected. A partial solution is done to demonstrate the method. Results Comparison

Target ANSYS Ratio f, Hz 28.766 28.767 1.00



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Problem-12 (Automobile Suspension System Vibration) Analysis Type(s): Mode-frequency Analysis (ANTYPE = 2) Element Type(s): Beam Elements (BEAM3)

Spring Elements (COMBIN14) Mass Element (MASS21)

An automobile suspension system is simplified to consider only two major motions of the system:

• Up and down linear motion of the body • Pitching angular motion of the body

If the body is idealized as a lumped mass with weight W and radius of gyration r, determine the corresponding coupled frequencies f1 and f2.

Material Properties E = 4 x 109 psf w = 3220 lb m = W/g = 100 lb-sec2/ft k1 = 2400 lb/ft k2 = 2600 lb/ft

Geometric PropertiesL1 = 4.5 ft L2 = 5.5ft r = 4 ft

Loading g = 32.2 ft/sec2


The beam geometric properties are input (all as unity) but not used for this solution. The torsional moment of inertia IT is calculated as IT = Wr2/g = 1600 lb-sec2-ft. A lateral master degree of freedom (MDOF) and a rotational MDOF are selected at the mass. The spring length is used only to define the spring direction.

Results Comparison

Target ANSYS Ratio f1 , Hz 1.0981 1.0981 1.000 f2 , Hz 1.4406 1.4406 1.000



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Harmonic Response Analysis- Some Theory Any sustained cyclic load will produce a sustained cyclic response (a harmonic response) in a structural system. Harmonic response analysis gives you the ability to predict the sustained dynamic behavior of your structures, thus enabling you to verify whether or not your designs will successfully overcome resonance, fatigue, and other harmful effects of forced vibrations. Uses for Harmonic Response Analysis Harmonic response analysis is a technique used to determine the steady-state response of a linear structure to loads that vary sinusoidally (harmonically) with time. The idea is to calculate the structure's response at several frequencies and obtain a graph of some response quantity (usually displacements) versus frequency. "Peak" responses are then identified on the graph and stresses reviewed at those peak frequencies. This analysis technique calculates only the steady-state, forced vibrations of a structure. The transient vibrations, which occur at the beginning of the excitation, are not accounted for in a harmonic response analysis.

Typical harmonic response system. Fo and Ω are known. uo and Φ are unknown (a). Transient and steady-state dynamic response of a structural system (b). Harmonic response analysis is a linear analysis. Any nonlinearities, such as plasticity and contact (gap) elements, will be ignored, even if they are defined. You can, however, have unsymmetric system matrices such as those encountered in a fluid-structure interaction problem. Harmonic analysis can also be performed on a prestressed structure, such as a violin string (assuming the harmonic stresses are much smaller than the pretension stress).

The Three Solution Methods Three harmonic response analysis methods are available: full, reduced, and mode superposition. The ANSYS Professional program allows only the mode superposition method. Before we study the details of how to implement each of these methods, let's explore the advantages and disadvantages of each method. The Full Method The full method is the easiest of the three methods. It uses the full system matrices to calculate the harmonic response (no matrix reduction). The matrices may be symmetric or unsymmetric. The advantages of the full method are: • It is easy to use, because you don't have to worry about choosing master degrees of freedom or mode shapes.



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• It uses full matrices, so no mass matrix approximation is involved. • It allows unsymmetric matrices, which are typical of such applications as acoustics and bearing problems. • It calculates all displacements and stresses in a single pass. • It accepts all types of loads: nodal forces, imposed (nonzero) displacements, and element loads

(Pressures and temperatures). • It allows effective use of solid-model loads. A disadvantage is that this method usually is more expensive than either of the other methods when you use the frontal solver. However, when you use the JCG solver or the ICCG solver, the full method can be very efficient. The Reduced Method The reduced method enables you to condense the problem size by using master degrees of freedom and reduced matrices. After the displacements at the master DOF have been calculated, the solution can be expanded to the original full DOF set. The advantages of this method are: • It is faster and less expensive compared to the full method when you are using the frontal solver. • Prestressing effects can be included. The disadvantages of the reduced method are: • The initial solution calculates only the displacements at the master DOF. A second step, known as the expansion pass, is required for a complete displacement, stress, and force solution. (However, the expansion pass might be optional for some applications.) • Element loads (pressures, temperatures, etc.) cannot be applied. • All loads must be applied at user-defined master degrees of freedom. (This limits the use of solid-model loads.) The Mode Superposition Method The mode superposition method sums factored mode shapes (eigenvectors) from a modal analysis to calculate the structure's response. Its advantages are: • It is faster and less expensive than either the reduced or the full method for many problems. • Element loads applied in the preceding modal analysis can be applied in the harmonic response analysis via the LVSCALE command, unless the modal analysis was done using PowerDynamics. • It allows solutions to be clustered about the structure's natural frequencies. This results in a smoother, more accurate tracing of the response curve. • Prestressing effects can be included. • It accepts modal damping (damping ratio as a function of frequency). Disadvantages of the mode superposition method are: • Imposed (nonzero) displacements cannot be applied. • When you are using PowerDynamics for the modal analysis, initial conditions cannot have previously-applied loads. Restrictions Common to All Three Methods All three methods are subject to certain common restrictions: • All loads must be sinusoidally time-varying. • All loads must have the same frequency. • No nonlinearities are permitted. • Transient effects are not calculated. You can overcome any of these restrictions by performing a transient dynamic analysis, with harmonic loads expressed as time-history loading functions. Transient Dynamic Analysis describes the procedure for a transient dynamic analysis.



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Analysis Types and Options

Option Command GUI Path New Analysis ANTYPE Main Menu> Solution> Analysis Type> New Analysis Analysis Type: Harmonic Response

ANTYPE Main Menu> Solution> Analysis Type> New Analysis> Harmonic

Solution Method HROPT Main Menu> Solution> Analysis Type> Analysis Options Solution Listing Format HROUT Main Menu> Solution> Analysis Type> Analysis Options Mass Matrix Formulation LUMPM Main Menu> Solution> Analysis Type> Analysis Options Equation Solver EQSLV Main Menu> Solution> Analysis Type> Analysis Options Applicable Loads in a Harmonic Response Analysis

Load Type Category Cmd Family

GUI Path

Displacement (UX, UY, UZ, ROTX, ROTY, ROTZ)

Constraints D Main Menu> Solution> Define Loads> Apply> Structural> Displacement

Force, Moment (FX, FY, FZ, MX, MY, MZ)

Forces F Main Menu> Solution> Define Loads> Apply> Structural> Force/Moment

Pressure (PRES) Surface Loads

SF Main Menu> Solution> Define Loads> Apply> Structural> Pressure

Temperature (TEMP), Fluence (FLUE)

Body Loads BF Main Menu> Solution> Define Loads> Apply> Structural> Temperature

Gravity, Spinning, etc. Inertia Loads

- Main Menu> Solution> Define Loads> Apply> Structural> Other

Load Commands for a Harmonic Response Analysis Load Type Solid Model

or FE Entity Apply Delete List Operate Apply Settings

Solid Model Keypoints DK DKDELE DKLIST DTRAN - Solid Model Lines DL DLDELE DLLIST DTRAN - Solid Model Areas DA DADELE DALIST DTRAN -

Displace- ment

Finite Elem Nodes D DDELE DLIST DSCALE DSYM, DCUM Solid Model Keypoints FK FKDELE FKLIST FTRAN - Force Finite Elem Nodes F FDELE FLIST FSCALE FCUM Solid Model Lines SFL SFLDEL

E SFLLIST SFTRAN SFGRAD

Solid Model Areas SFA SFADELE

SFALIST SFTRAN SFGRAD

Finite Elem Nodes SF SFDELE SFLIST SFSCALE SFGRAD, SFCUM

Pressure

Finite Elem Elements SFE SFEDELE

SFELIST SFSCALE SFGRAD, SFBEAM, SFFUN, SFCUM

Solid Model Keypoints BFK BFKDELE

BFKLIST BFTRAN - Tempera- ture, Fluence Solid Model Lines BFL BFLDEL

E BFLLIST BFTRAN -



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Load Type Solid Model

or FE Entity Apply Delete List Operate Apply Settings

Solid Model Areas BFA BFADELE

BFALIST BFTRAN -

Solid Model Volumes BFV BFVDELE

BFVLIST BFTRAN -

Finite Elem Nodes BF BFDELE BFLIST BFSCALE BFCUM

Finite Elem Elements BFE BFEDELE

BFELIST BFSCALE BFCUM

Inertia - - ACEL, OMEGA, DOMEGA, CGLOC, CGOMGA, DCGOMG

- -



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Problem-13: Harmonic Analysis of a Cantilever Beam - Illustrated

Introduction This example was created using ANSYS 7.0 The purpose of this example is to explain the steps required to perform Harmonic analysis the cantilever beam shown below.

We will now conduct a harmonic forced response test by applying a cyclic load (harmonic) at the end of the beam. The frequency of the load will be varied from 1 - 100 Hz. The figure below depicts the beam with the application of the load.

ANSYS provides 3 methods for conducting a harmonic analysis. These 3 methods are the Full , Reduced and Modal Superposition methods. This example demonstrates the Full method because it is simple and easy to use as compared to the other two methods. However, this method makes use of the full stiffness and mass matrices and thus is the slower and costlier option.


Solution: Assigning Loads and Solving



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1. Define Analysis Type (Harmonic)

Solution > Analysis Type > New Analysis > Harmonic ANTYPE,3

2. Set options for analysis type: o Select: Solution > Analysis Type > Analysis Options..

The following window will appear

o As shown, select the Full Solution method, the Real + imaginary DOF

printout format and do not use lumped mass approx. o Click 'OK'

The following window will appear. Use the default settings (shown below).


o Select Solution > Define Loads > Apply > Structural > Displacement > On Nodes The following window will appear once you select the node at x=0 (Note small changes in the window compared to the static examples):



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o Constrain all DOF as shown in the above window

4. Apply Loads: o Select Solution > Define Loads > Apply > Structural > Force/Moment > On

Nodes o Select the node at x=1 (far right) o The following window will appear. Fill it in as shown to apply a load with a

real value of 100 and an imaginary value of 0 in the positive 'y' direction

Note: By specifying a real and imaginary value of the load we are providing information on magnitude and phase of the load. In this case the magnitude of the load is 100 N and its phase is 0. Phase information is important when you have two or more cyclic loads being applied to the structure as these loads could be in or out of phase. For harmonic analysis, all loads applied to a structure must have the SAME FREQUENCY.

5. Set the frequency range o Select Solution > Load Step Opts > Time/Frequency > Freq and Substps... o As shown in the window below, specify a frequency range of 0 - 100Hz, 100

substeps and stepped b.c..



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By doing this we will be subjecting the beam to loads at 1 Hz, 2 Hz, 3 Hz, ..... 100 Hz. We will specify a stepped boundary condition (KBC) as this will ensure that the same amplitude (100 N) will be applyed for each of the frequencies. The ramped option, on the other hand, would ramp up the amplitude where at 1 Hz the amplitude would be 1 N and at 100 Hz the amplitude would be 100 N. You should now have the following in the ANSYS Graphics window

6. Solve the System

Solution > Solve > Current LS SOLVE

Postprocessing: Viewing the Results We want to observe the response at x=1 (where the load was applyed) as a function of frequency. We cannot do this with General PostProcessing (POST1), rather we must use TimeHist PostProcessing (POST26). POST26 is used to observe certain variables as a function of either time or frequency.

1. Open the TimeHist Processing (POST26) Menu Select TimeHist Postpro from the ANSYS Main Menu.

2. Define Variables In here we have to define variables that we want to see plotted. By default, Variable 1 is assigned either Time or Frequency. In our case it is assigned Frequency. We want to see the displacement UY at the node at x=1, which is node #2. (To get a list of nodes and their attributes, select Utility Menu > List > nodes).



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o Select TimeHist Postpro > Variable Viewer... and the following window

should pop up.

o Select Add (the green '+' sign in the upper left corner) from this window and the

following window should appear

o We are interested in the Nodal Solution > DOF Solution > Y-Component of

displacement. Click OK. o Graphically select node 2 when prompted and click OK. The 'Time History

Variables' window should now look as follows



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3. List Stored Variables

o In the 'Time History Variables' window click the 'List' button, 3 buttons to the left of 'Add' The following window will appear listing the data:

4. Plot UY vs. frequency

o In the 'Time History Variables' window click the 'Plot' button, 2 buttons to the left of 'Add' The following graph should be plotted in the main ANSYS window.



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Note that we get peaks at frequencies of approximately 8.3 and 51 Hz. This corresponds with the predicted frequencies of 8.311 and 51.94Hz. To get a better view of the response, view the log scale of UY.

o Select Utility Menu > PlotCtrls > Style > Graphs > Modify Axis The following window will appear



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o As marked by an 'A' in the above window, change the Y-axis scale to

'Logarithmic' o Select Utility Menu > Plot > Replot o You should now see the following



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This is the response at node 2 for the cyclic load applied at this node from 0 - 100 Hz.

o For ANSYS version lower than 7.0, the 'Variable Viewer' window is not available. Use the 'Define Variables' and 'Store Data' functions under TimeHist Postpro. See the help file for instructions.



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Problem-14 (Harmonic Response of a Dynamic System) Analysis Type(s): Reduced Harmonic Response Analysis (ANTYPE = 3) Element Type(s): Combination Element (COMBIN40) A machine of weight W is supported on springs of total stiffness k. If a harmonic disturbing force of magnitude F1 and frequency f (equal to the natural frequency of the machine, fn) acts on the machine, determine the displacement response in terms of the peak amplitude Ao and phase angle Φ. Assume a viscous damping coefficient c.

Material Properties W = 193 lb k = 200 lb/in c = 6 lb-sec/in

Loading g = 386 in/sec2

F1 = 10 lb

Analysis Assumptions and Modeling Notes The mass of the machine is m = W/g = 0.5 lb-sec2/in. Hence the frequency of the disturbing force (f) becomes f = fn = sqrt(km)/sqrt(km)/2π = 3.1831 Hz. The node locations are arbitrarily selected. One master degree of freedom is selected at the mass in the force direction. Results Comparison Target ANSYS Ratio Ao, in 0.0833 0.0833 1.000 angle, deg -90.0 -90.0 1.000



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Problem-15 (Harmonic Response of a Two-Mass-Spring System) Analysis Type(s): Reduced Harmonic Response Analysis (ANTYPE = 3) Element Type(s): Spring Elements (COMBIN14)

Mass Elements (MASS21) Determine the response amplitude (Xi) and phase angle (Φi) for each mass (mi) of the system in Two-mass-spring System Problem Sketch when excited by a harmonic force (F1sin ωt) acting on mass m1.

Material Properties m1 = m2 = 0.5 lb-sec2/in k1 = k2 = kc = 200 lb/in

Loading F1 = 200 lb

Analysis Assumptions and Modeling Notes The spring lengths are arbitrarily selected and are used only to define the spring direction. Two master degrees of freedom (MDOF) are selected at the masses in the spring direction. A frequency range from zero to 7.5 Hz with a solution at 7.5/30 = 0.25 Hz intervals is chosen to give an adequate response curve. POST26 is used to get an amplitude versus frequency display. Results Comparison Target ANSYS Ratio X1 , in (f = 1.5 Hz)[1] 0.82272 0.82272 1.000



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Target ANSYS Ratio Angle1, deg (f = 1.5 Hz) 0.0000 0.0000 - X2 , in (f = 1.5 Hz)[1] 0.46274 0.46274 1.000 Angle2, deg (f = 1.5 Hz) 0.000 0.0000 - X1 , in (f = 4.0 Hz) 0.51145 0.51146 1.000 Angle1, deg (f = 4.0 Hz) 180.00 180.00 1.000 X2 , in (f = 4.0 Hz) 1.2153 1.2153 1.000 Angle2, deg (f = 4.0 Hz) 180.00 180.00 1.000 X1 , in (f = 6.5 Hz) 0.58513 0.58512 1.000 Angle1, deg (f = 6.5 Hz) 180.00 180.00 1.000 X2 , in (f = 6.5 Hz) 0.26966 0.26965 1.000 Angle2, deg (f = 6.5 Hz) 0.0000 0.0000 -

1. X1 = UX @ m1 (node 2) X2 = UX @ m2 (node 3)



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Problem-16 (Harmonic Response of a Guitar String) Analysis Type(s):

Static Analysis (ANTYPE = 0); Mode-frequency Analysis (ANTYPE = 2) Mode Superposition Harmonic Response Analysis (ANTYPE = 3)

Element Type(s):

Spar Elements (LINK1)

A uniform stainless steel guitar string of length L and diameter d is stretched between two rigid supports by a tensioning force F1, which is required to tune the string to the E note of a C scale. The string is then struck near the quarter point with a force F2. Determine the fundamental frequency, f1. Also, show that only the odd-numbered frequencies produce a response at the midpoint of the string for this excitation.

Material Properties E = 190 x 109 Pa ρ = 7920 kg/m3


= 710 mm c = 165 mm d = 0.254 mm

Loading F1 = 84 NF2 = 1 N

Analysis Assumptions and Modeling Notes Enough elements are selected so that the model can be used to adequately characterize the string dynamics. The stress stiffening capability of the element is used. Harmonic response analysis is used to determine the displacement response to the lateral force F2. The harmonic response is displayed with the time-history postprocessor, POST26, to show the excitation of the odd-numbered frequencies at peak displacement amplitudes. Results Comparison Target ANSYS Ratio Modal f, Hz 322.2 322.3 1.000

f1, (322.2 Hz) Response Response, 320 < f < 328 - f2, (644.4 Hz) No Response No Response - f3, (966.6 Hz) Response Response, 968 < f < 976 - f4, (1288.8 Hz) No Response No Response - f5, (1611.0 Hz) Response Response,1624 < f < 1632 -

POST26

f6, (1933.2 Hz) No Response No Response -



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Problem-17 (Harmonic Response of a Spring-mass System) Analysis Type(s):

Mode-frequency Analysis (ANTYPE = 2) Harmonic Response Mode Superposition Analysis (ANTYPE = 3)

Element Type(s):

Combination Element (COMBIN40)

Determine the natural frequencies of the spring-mass system shown and the displacement response when excited by a harmonic load of variable frequency from 0.1 to 1.0 Hz, with an amplitude of Fo.

Material Properties k1 = 6 N/m k2 = 16 N/m m1 = m2 = 2 kg

Loading Fo = 50 N

Analysis Assumptions and Modeling Notes COMBIN40 combination elements are used to represent the springs and masses. Node locations are arbitrary. Results Comparison Target ANSYS Ratio Y1 , m (@ .226 Hz) -1371.7 -1371.7 1.000 Y2 , m (@ .226 Hz) -458.08 -458.08 1.000 Y1 , m (@ .910 Hz) -0.8539 -0.8539 1.000 Y2 , m (@ .910 Hz) 0.1181 0.1181 1.000



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Nonlinear Structural Analysis- some Theory

Structural Nonlinearity? You encounter structural nonlinearities on a routine basis. For instance, whenever you staple two pieces of paper together, the metal staples are permanently bent into a different shape. (See Common Examples of Nonlinear Structural Behavior (a).) If you heavily load a wooden shelf, it will sag more and more as time passes. (See Figure (b).) As weight is added to a car or truck, the contact surfaces between its pneumatic tires and the underlying pavement change in response to the added load. (See Figure (c).) If you were to plot the load-deflection curve for each of these examples, you would discover that they all exhibit the fundamental characteristic of nonlinear structural behavior - a changing structural stiffness.

Causes of Nonlinear Behavior Nonlinear structural behavior arises from a number of causes, which can be grouped into these principal categories:

• Changing status



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• Geometric nonlinearities • Material nonlinearities

Changing Status (Including Contact)

Many common structural features exhibit nonlinear behavior that is status-dependent. For example, a tension-only cable is either slack or taut; a roller support is either in contact or not in contact. Status changes might be directly related to load (as in the case of the cable), or they might be determined by some external cause.

Geometric Nonlinearities If a structure experiences large deformations, its changing geometric configuration can cause the structure to respond nonlinearly. An example would be the fishing rod shown in A Fishing Rod Demonstrates Geometric Nonlinearity. Geometric nonlinearity is characterized by "large" displacements and/or rotations.

Material Nonlinearities

Nonlinear stress-strain relationships are a common cause of nonlinear structural behavior. Many factors can influence a material's stress-strain properties, including load history (as in elastoplastic response), environmental conditions (such as temperature), and the amount of time that a load is applied (as in creep response). ANSYS employs the "Newton-Raphson" approach to solve nonlinear problems. In this approach, the load is subdivided into a series of load increments. The load increments can be applied over several load steps

Load and Displacement Directions Consider what happens to loads when your structure experiences large deflections. In many instances, the loads applied to your system maintain constant direction no matter how the structure deflects. In other cases, forces will change direction, "following" the elements as they undergo large rotations.



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Problem-18- Non Linear Analysis of a Cantilever Beam

Introduction This example was created using ANSYS 7.0 The purpose of this example is to outline the steps required to do a simple nonlinear analysis of the beam shown below.

There are several causes for nonlinear behaviour such as Changing Status (ex. contact

elements), Material Nonlinearities and Geometric Nonlinearities (change in response due to large deformations). This example will deal specifically with Geometric Nonlinearities. To solve this problem, the load will be added incrementally. After each increment, the stiffness matrix will be adjusted before increasing the load. The solution will be compared to the equivalent solution using a linear response.

Preprocessing: Defining the Problem 1. Give example a Title

Utility Menu > File > Change Title ... 2. Create Keypoints

Preprocessor > Modeling > Create > Keypoints > In Active CS We are going to define 2 keypoints (the beam vertices) for this structure to create a beam with a length of 5 inches: Keypoint Coordinates (x,y)

1 (0,0) 2 (5,0)

3. Define Lines Preprocessor > Modeling > Create > Lines > Lines > Straight Line Create a line between Keypoint 1 and Keypoint 2.

4. Define Element Types Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the BEAM3 (Beam 2D elastic) element. This element has 3 degrees of freedom (translation along the X and Y axis's, and rotation about the Z axis). With only 3 degrees of freedom, the BEAM3 element can only be used in 2D analysis.



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5. Define Real Constants

Preprocessor > Real Constants... > Add... In the 'Real Constants for BEAM3' window, enter the following geometric properties:

i. Cross-sectional area AREA: 0.03125 ii. Area Moment of Inertia IZZ: 4.069e-5

iii. Total beam height HEIGHT: 0.125 This defines an element with a solid rectangular cross section 0.25 x 0.125 inches.


i. Young's modulus EX: 30e6 ii. Poisson's Ratio PRXY: 0.3

If you are wondering why a 'Linear' model was chosen when this is a non-linear example, it is because this example is for non-linear geometry, not non-linear material properties. If we were considering a block of wood, for example, we would have to consider non-linear material properties.

7. Define Mesh Size Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > All Lines... For this example we will specify an element edge length of 0.1 " (50 element divisions along the line).

8. Mesh the frame Preprocessor > Meshing > Mesh > Lines > click 'Pick All' LMESH,ALL


Solution > New Analysis > Static ANTYPE,0

2. Set Solution Controls o Select Solution > Analysis Type > Sol'n Control...

The following image will appear:



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Ensure the following selections are made (as shown above)

A. Ensure Large Static Displacements are permitted (this will include the effects of large deflection in the results)

B. Ensure Automatic time stepping is on. Automatic time stepping allows ANSYS to determine appropriate sizes to break the load steps into. Decreasing the step size usually ensures better accuracy, however, this takes time. The Automatic Time Step feature will determine an appropriate balance. This feature also activates the ANSYS bisection feature which will allow recovery if convergence fails.

C. Enter 5 as the number of substeps. This will set the initial substep to 1/5 th of the total load. The following example explains this: Assume that the applied load is 100 lb*in. If the Automatic Time Stepping was off, there would be 5 load steps (each increasing by 1/5 th of the total load):

20 lb*in 40 lb*in 60 lb*in 80 lb*in 100 lb*in

Now, with the Automatic Time Stepping is on, the first step size will still be 20 lb*in. However, the remaining substeps will be determined based on the response of the material due to the previous load increment.

D. Enter a maximum number of substeps of 1000. This stops the program if the solution does not converge after 1000 steps.

E. Enter a minimum number of substeps of 1. F. Ensure all solution items are writen to a results file.

3. Apply Constraints Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Fix Keypoint 1 (ie all DOFs constrained).

4. Apply Loads

Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints Place a -100 lb*in moment in the MZ direction at the right end of the beam (Keypoint 2)


The following will appear on your screan for NonLinear Analyses



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This shows the convergence of the solution.

General Postprocessing: Viewing the Results 1. View the deformed shape

General Postproc > Plot Results > Deformed Shape... > Def + undeformed PLDISP,1

2. View the deflection contour plot

General Postproc > Plot Results > Contour Plot > Nodal Solu... > DOF solution, UY PLNSOL,U,Y,0,1



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3. List Horizontal Displacement

If this example is performed as a linear model there will be no nodal deflection in the horizontal direction due to the small deflections assumptions. However, this is not realistic for large deflections. Modeling the system non-linearly, these horizontal deflections are calculated by ANSYS. General Postproc > List Results > Nodal Solution...> DOF solution, UX

Other results can be obtained as shown in previous linear static analyses.



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Problem-19 : NonLinear Materials

Introduction This example was completed using ANSYS 7.0 The purpose of the example is to describe how to include material nonlinearities in an ANSYS model. For instance, the case when a large force is applied resulting in a stresses greater than yield strength. In such a case, a multilinear stress-strain relationship can be included which follows the stress-strain curve of the material being used. This will allow ANSYS to more accurately model the plastic deformation of the material.

For this analysis, a simple tension speciment 100 mm X 5 mm X 5 mm is constrained at the bottom and has a load pulling on the top. This specimen is made out of a experimental substance called "WhoKilledKenium". The stress-strain curve for the substance is shown above. Note the linear section up to approximately 225 MPa where the Young's Modulus is constant (75 GPa). The material then begins to yield and the relationship becomes plastic and nonlinear.

Preprocessing: Defining the Problem 1. Give example a Title

Utility Menu > File > Change Title ... /title, NonLinear Materials

2. Create Keypoints Preprocessor > Modeling > Create > Keypoints > In Active CS /PREP7 K,#,X,Y We are going to define 2 keypoints (the beam vertices) for this structure to create a beam with a length of 100 millimeters: Keypoint Coordinates (x,y)

1 (0,0)



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2 (0,100)

3. Define Lines Preprocessor > Modeling > Create > Lines > Lines > Straight Line Create a line between Keypoint 1 and Keypoint 2. L,1,2

4. Define Element Types Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the LINK1 (2D spar) element. This element has 2 degrees of freedom (translation along the X and Y axis's) and can only be used in 2D analysis.

5. Define Real Constants Preprocessor > Real Constants... > Add... In the 'Real Constants for LINK1' window, enter the following geometric properties:

i. Cross-sectional area AREA: 25 ii. Initial Strain: 0

This defines an element with a solid rectangular cross section 5 x 5 millimeters. 6. Define Element Material Properties

Preprocessor > Material Props > Material Models > Structural > Linear > Elastic > Isotropic In the window that appears, enter the following geometric properties for steel:


Now that the initial properties of the material have been outlined, the stress-strain data must be included.

Preprocessor > Material Props > Material Models > Structural > Nonlinear > Elastic > Multilinear Elastic The following window will pop up.

Fill in the STRAIN and STRESS boxes with the following data. These are points from the stress-strain curve shown above, approximating the curve with linear interpolation between the points. When the data for the first point is input, click Add Point to add another. When all the points



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have been inputed, click Graph to see the curve. It should look like the one shown above. Then click OK.

Curve Points Strain Stress

1 0 0 2 0.001 75 3 0.002 150 4 0.003 225 5 0.004 240 6 0.005 250 7 0.025 300 8 0.060 355 9 0.100 390 10 0.150 420 11 0.200 435 12 0.250 449 13 0.275 450

To get the problem geometry back, select Utility Menu > Plot > Replot. /REPLOT 7. Define Mesh Size

Preprocessor > Meshing > Manual Size > Size Cntrls > Lines > All Lines... For this example we will specify an element edge length of 5 mm (20 element divisions along the line).








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Ensure the following selections are made under the 'Basic' tab (as shown above)



C. Enter 20 as the number of substeps. This will set the initial substep to 1/20 th of the total load.


E. Enter a minimum number of substeps of 1. F. Ensure all solution items are writen to a results file. This means rather

than just recording the data for the last load step, data for every load step is written to the database. Therefore, you can plot certain parameters over time.

Ensure the following selection is made under the 'Nonlinear' tab (as shown below)

G. Ensure Line Search is 'On'. This option is used to help the Newton-Raphson solver converge.

H. Ensure Maximum Number of Iterations is set to 1000



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NOTE There are several options which have not been changed from their default values. For more information about these commands, type help followed by the command into the command line.


4. Apply Loads Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints Place a 10,000 N load in the FY direction on the top of the beam (Keypoint 2).


The following will appear on your screen for NonLinear Analyses



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General Postprocessing: Viewing the Results 1. To view the element in 2D rather than a line: Utility Menu > PlotCtrls > Style > Size

and Shape and turn 'Display of element' ON (as shown below).

2. View the deflection contour plot



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Time History Postprocessing: Viewing the Results As shown, you can obtain the results (such as deflection, stress and bending moment diagrams) the same way you did in previous examples using the General Postprocessor. However, you may wish to view time history results such as the deflection of the object over time.

1. Define Variables o Select: Main Menu > TimeHist Postpro. The following window should open

automatically.



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If it does not open automatically, select Main Menu > TimeHist Postpro > Variable Viewer

o Click the add button in the upper left corner of the window to add a variable.

o Select Nodal Solution > DOF Solution > Y-Component of displacement (as shown below) and click OK. Pick the uppermost node on the beam and click OK in the 'Node for Data' window.

o To add another variable, click the add button again. This time select Reaction

Forces > Structural Forces > Y-Component of Force. Pick the lowermost node on the beam and click OK.

o On the Time History Variable window, click the circle in the 'X-Axis' column for FY_3. This will make the reaction force the x-variable. The Time History Variables window should now look like this:

2. Graph Results over Time

o Click on UY_2 in the Time History Variables window.

o Click the graphing button in the Time History Variables window.



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o The labels on the plot are not updated by ANSYS, so you must change them

manually. Select Utility Menu > Plot Ctrls > Style > Graphs > Modify Axes and re-label the X and Y-axis appropriately.

This plot shows how the beam deflected linearly when the force, and subsequently the stress, was low (in the linear range). However, as the force increased, the deflection (proportional to strain) began to increase at a greater rate. This is because the stress in the beam is in the plastic range and thus no longer relates to strain linearly. When you verify this example analytically, you will see the solutions are very similar. The difference can be attributed to the ANSYS solver including large deflection calculations.

End of Non linear Analysis



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Transient Dynamic Analysis- Some Theory Transient dynamic analysis (sometimes called time-history analysis) is a technique used to determine the dynamic response of a structure under the action of any general time-dependent loads. You can use this type of analysis to determine the time-varying displacements, strains, stresses, and forces in a structure as it responds to any combination of static, transient, and harmonic loads. The time scale of the loading is such that the inertia or damping effects are considered to be important. If the inertia and damping effects are not important, you might be able to use a static analysis instead.

Preparing for a Transient Dynamic Analysis

A transient dynamic analysis is more involved than a static analysis because it generally requires more computer resources and more of your resources, in terms of the “engineering” time involved. You can save a significant amount of these resources by doing some preliminary work to understand the physics of the problem. For example, you can:

1. Analyze a simpler model first. A model of beams, masses, and springs can provide good insight into the problem at minimal cost. This simpler model may be all you need to determine the dynamic response of the structure.

2. If you are including nonlinearities, try to understand how they affect the structure's response by doing a static analysis first. In some cases, nonlinearities need not be included in the dynamic analysis.

3. Understand the dynamics of the problem. By doing a modal analysis, which calculates the natural frequencies and mode shapes, you can learn how the structure responds when those modes are excited. The natural frequencies are also useful for calculating the correct integration time step.

4. For a nonlinear problem, consider substructuring the linear portions of the model to reduce analysis costs.

Three methods are available to do a transient dynamic analysis: full, mode superposition, and reduced. The ANSYS Professional program allows only the mode superposition method.



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Problem-20: Transient Analysis of a Cantilever Beam

Introduction This example was created using ANSYS 7.0 The purpose of this example is to show the steps involved to perform a simple transient analysis.

Transient dynamic analysis is a technique used to determine the dynamic response of a structure under a time-varying load. The time frame for this type of analysis is such that inertia or damping effects of the structure are considered to be important. Cases where such effects play a major role are under step or impulse loading conditions, for example, where there is a sharp load change in a fraction of time. If inertia effects are negligible for the loading conditions being considered, a static analysis may be used instead. For our case, we will impact the end of the beam with an impulse force and view the response at the location of impact.

Since an ideal impulse force excites all modes of a structure, the response of the beam should contain all mode frequencies. However, we cannot produce an ideal impulse force numerically. We have to apply a load over a discrete amount of time dt.



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After the application of the load, we track the response of the beam at discrete time points for as long as we like (depending on what it is that we are looking for in the response). The size of the time step is governed by the maximum mode frequency of the structure we wish to capture. The smaller the time step, the higher the mode frequency we will capture. The rule of thumb in ANSYS is time_step = 1 / 20f where f is the highest mode frequency we wish to capture. In other words, we must resolve our step size such that we will have 20 discrete points per period of the highest mode frequency. It should be noted that a transient analysis is more involved than a static or harmonic analysis. It requires a good understanding of the dynamic behavior of a structure. Therefore, a modal analysis of the structure should be initially performed to provide information about the structure's dynamic behavior. In ANSYS, transient dynamic analysis can be carried out using 3 methods.

• The Full Method: This is the easiest method to use. All types of non-linearities are allowed. It is however very CPU intensive to go this route as full system matrices are used.

• The Reduced Method: This method reduces the system matrices to only consider the Master Degrees of Freedom (MDOFs). Because of the reduced size of the matrices, the calculations are much quicker. However, this method handles only linear problems (such as our cantilever case).

• The Mode Superposition Method: This method requires a preliminary modal analysis, as factored mode shapes are summed to calculate the structure's response. It is the quickest of the three methods, but it requires a good deal of understanding of the problem at hand.

We will use the Reduced Method for conducting our transient analysis. Usually one need not go further than Reviewing the Reduced Results. However, if stresses and forces are of interest than, we would have to Expand the Reduced Solution.





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o Select Solution > Analysis Type > New Analysis > Transient o The following window will appear. Select 'Reduced' as shown.

2. Define Master DOFs

o Select Solution > Master DOFs > User Selected > Define o Select all nodes except the left most node (at x=0).

The following window will open, choose UY as the first dof in this window

For an explanation on Master DOFs, see the section on Using the Reduced

Method for modal analysis. 3. Constrain the Beam

Solution Menu > Define Loads > Apply > Structural > Displacement > On nodes Fix the left most node (constrain all DOFs).

4. Apply Loads We will define our impulse load using Load Steps. The following time history curve shows our load steps and time steps. Note that for the reduced method, a constant time step is required throughout the time range.



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We can define each load step (load and time at the end of load segment) and save them in a file for future solution purposes. This is highly recommended especially when we have many load steps and we wish to re-run our solution. We can also solve for each load step after we define it. We will go ahead and save each load step in a file for later use, at the same time solve for each load step after we are done defining it.

a. Load Step 1 - Initial Conditions i. Define Load Step

We need to establish initial conditions (the condition at Time = 0). Since the equations for a transient dynamic analysis are of second order, two sets of initial conditions are required; initial displacement and initial velocity. However, both default to zero. Therefore, for this example we can skip this step.

ii. Specify Time and Time Step Options Select Solution > Load Step Opts > Time/Frequenc > Time -

Time Step .. set a time of 0 for the end of the load step (as shown

below). set [DELTIM] to 0.001. This will specify a time step size

of 0.001 seconds to be used for this load step.



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iii. Write Load Step File

Select Solution > Load Step Opts > Write LS File The following window will appear

Enter LSNUM = 1 as shown above and click 'OK'

The load step will be saved in a file jobname.s01 b. Load Step 2

i. Define Load Step Select Solution > Define Loads > Apply > Structural >

Force/Moment > On Nodes and select the right most node (at x=1). Enter a force in the FY direction of value -100 N.


Time Step .. and set a time of 0.001 for the end of the load step iii. Write Load Step File

Solution > Load Step Opts > Write LS File Enter LSNUM = 2

c. Load Step 3



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i. Define Load Step

Select Solution > Define Loads > Delete > Structural > Force/Moment > On Nodes and delete the load at x=1.


Time Step .. and set a time of 1 for the end of the load step iii. Write Load Step File

Solution > Load Step Opts > Write LS File Enter LSNUM = 3

5. Solve the System o Select Solution > Solve > From LS Files

The following window will appear.

o Complete the window as shown above to solve using LS files 1 to 3.

Postprocessing: Viewing the Results To view the response of node 2 (UY) with time we must use the TimeHist PostProcessor (POST26).

1. Define Variables In here we have to define variables that we want to see plotted. By default, Variable 1 is assigned either Time or Frequency. In our case it is assigned Frequency. We want to see the displacement UY at the node at x=1, which is node #2. (To get a list of nodes and their attributes, select Utility Menu > List > nodes).

o Select TimeHist Postpro > Variable Viewer... and the following window should pop up.



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o Select Add (the green '+' sign in the upper left corner) from this window and the

following window should appear

o We are interested in the Nodal Solution > DOF Solution > Y-Component of

displacement. Click OK. o Graphically select node 2 when prompted and click OK. The 'Time History

Variables' window should now look as follows



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2. List Stored Variables

o In the 'Time History Variables' window click the 'List' button, 3 buttons to the left of 'Add' The following window will appear listing the data:

3. Plot UY vs. frequency

o In the 'Time History Variables' window click the 'Plot' button, 2 buttons to the left of 'Add' The following graph should be plotted in the main ANSYS window.



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A few things to note in the response curve

There are approximately 8 cycles in one second. This is the first mode of the cantilever beam and we have been able to capture it.

We also see another response at a higher frequency. We may have captured some response at the second mode at 52 Hz of the beam.

Note that the response does not decay as it should not. We did not specify damping for our system.

Expand the Solution

For most problems, one need not go further than Reviewing the Reduced Results as the response of the structure is of utmost interest in transient dynamic analysis. However, if stresses and forces are of interest, we would have to expand the reduced solution. Let's say we are interested in the beam's behaviour at peak responses. We should then expand a few or all solutions around one peak (or dip). We will expand 10 solutions within the range of 0.08 and 0.11 seconds.

1. Expand the solution o Select Finish in the ANSYS Main Menu o Select Solution > Analysis Type > ExpansionPass... and switch it to ON in

the window that pops open. o Select Solution > Load Step Opts > ExpansionPass > Single Expand >

Range of Solu's o Complete the window as shown below. This will expand 10 solutions withing

the range of 0.08 and 0.11 seconds



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2. Solve the System


3. Review the results in POST1 Review the results using either General Postprocessing (POST1) or TimeHist Postprocessing (POST26). For this case, we can view the deformed shape at each of the 10 solutions we expanded.

Damped Response of the Cantilever Beam We did not specify damping in our transient analysis of the beam. We specify damping at the same time we specify our time & time steps for each load step. We will now re-run our transient analysis, but now we will consider damping. Here is where the use of load step files comes in handy. We can easily change a few values in these files and re-run our whole solution from these load case files.

• Open up the first load step file (Dynamic.s01) for editing Utility Menu > File > List > Other > Dynamic.s01. The file should look like the following..

• /COM,ANSYS RELEASE 5.7.1 UP20010418 14:44:02 08/20/2001 • /NOPR • /TITLE, Dynamic Analysis • _LSNUM= 1 • ANTYPE, 4 • TRNOPT,REDU,,DAMP • BFUNIF,TEMP,_TINY • DELTIM, 1.000000000E-03 • TIME, 0.00000000 • TREF, 0.00000000 • ALPHAD, 0.00000000 • BETAD, 0.00000000 • DMPRAT, 0.00000000 • TINTP,R5.0, 5.000000000E-03,,, • TINTP,R5.0, -1.00000000 , 0.500000000 , -1.00000000 • NCNV, 1, 0.00000000 , 0, 0.00000000 , 0.00000000 • ERESX,DEFA • ACEL, 0.00000000 , 0.00000000 , 0.00000000 • OMEGA, 0.00000000 , 0.00000000 , 0.00000000 , 0 • DOMEGA, 0.00000000 , 0.00000000 , 0.00000000 • CGLOC, 0.00000000 , 0.00000000 , 0.00000000 • CGOMEGA, 0.00000000 , 0.00000000 , 0.00000000 • DCGOMG, 0.00000000 , 0.00000000 , 0.00000000



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• • D, 1,UX , 0.00000000 , 0.00000000 • D, 1,UY , 0.00000000 , 0.00000000 • D, 1,ROTZ, 0.00000000 , 0.00000000 • /GOPR • Change the damping value BETAD from 0 to 0.01 in all three load step files. • We will have to re-run the job for the new load step files. Select Utility Menu > file >

Clear and Start New. • Repeat the steps shown above up to the point where we select MDOFs. After selecting

MDOFs, simply go to Solution > (-Solve-) From LS files ... and in the window that opens up select files from 1 to 3 in steps of 1.

• After the results have been calculated, plot up the response at node 2 in POST26. The damped response should look like the following



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Problem-21: (Transient Response to a Constant Force) Analysis Type(s): Reduced Transient Dynamic Analysis (ANTYPE = 4) Element Type(s): Beam Elements (BEAM3)

Mass Element (MASS21) A steel beam of length L and geometric properties shown below, is supporting a concentrated mass, m. The beam is subjected to a dynamic load F(t) with a rise time tr and a maximum value F1. If the weight of the beam is considered negligible, determine the time of maximum displacement response tmax and the maximum displacement response ymax. Additionally, determine the maximum bending stress σbend in the beam.

Material Properties E = 30 x 103 ksi m = 0.0259067 kips-sec2/in

Geometric PropertiesI = 800.6 in4 h = 18 in

= 20 ft = 240 in

Loading F1 = 20 kips tr = 0.075 sec


The beam area is not used in this solution and is arbitrarily input as unity. The final time of 0.1 sec allows the mass to reach its largest deflection. One master degree of freedom (MDOF) is selected at the mass in the lateral direction. A static solution is done at the first load step. The

integration time step (0.004 sec) is based on 1/25 of the period to allow the abrupt change in acceleration to be followed reasonably well and to produce sufficient printout for the theoretical comparison. Symmetry could have been used in the model. The time of maximum response (0.092 sec) is selected for the expansion pass calculation.

Results Comparison

Target[1] ANSYS Ratio tmax , sec 0.092 0.092 1.00 Transient ymax , in 0.331 0.335 1.01

Expansion Pass Stressbend , ksi -18.6 -18.9 1.01



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Problem-22:(Transient Response of a Ball Impacting a Flexible Surface)

Analysis Type(s): Nonlinear Transient Dynamic Analysis (ANTYPE = 4) Element Type(s): 2-D Contact Surface Element (CONTAC26)

Mass Element (MASS21) 2-D Node-to-Surface Contact Element (CONTA175)

A rigid ball of mass m is dropped through a height h onto a flexible surface of stiffness k. Determine the velocity, kinetic energy, and displacement y of the ball at impact and the maximum displacement of the ball.

Material Propertiesm = 0.5 lb-sec2/in k = 1973.92 lb/in

Geometric Properties h = 1 in Loading g = 386 in/sec2


The node locations are arbitrarily selected. The final time of 0.11 seconds allows the mass to reach its largest deflection. The integration time step (0.11/110 0.001 sec) is based on 1/100 of the period (during impact), to allow the initial step acceleration change to be followed reasonably well and to produce sufficient printout for the theoretical comparison. At release h, the mass acceleration is 386 in/sec2. Therefore, a load step with a small time period is used to ramp to the appropriate acceleration while maintaining essentially zero velocity. Displacements and velocities are listed against time in POST26 and stored kinetic energy is obtained in POST1.



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The model is solved twice first using CONTAC26 and then using the Node-to-Surface CONTA175 element.

Results Comparison

Target ANSYS RatioTime,sec 0.07198 0.072 1.000y displacement, in -1.0000 -0.9991 0.999y velocity, in/sec -27.79 -27.76 0.999

CONTAC26 At Impact [1]

kinetic energy, lb-in 193.00 192.65 0.998Time,sec 0.10037 0.10100 1.006At "Zero" Velocity [2] max. y displacement, in -1.5506 -1.5503 1.000time,sec 0.07198 0.072 1.000y displacement, in -1.0000 -0.9991 0.999

CONTA175 At Impact [1]

y velocity, in/sec -27.79 -27.76 0.999time,sec 0.10037 0.10100 1.006At "Zero" Velocity [2] max. y displacement, in -1.5506 -1.5503 1.000

1. Target results are for t = 0.07198 sec. ANSYS results are reported for closest time point, t= 0.072 sec.

2. ANSYS results are from the time point closest (reported in POST26) to the change in velocity from negative to positive



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Problem-23 (Plastic Response to a Suddenly Applied Constant Force)

Analysis Type(s): Full Transient Dynamic, Plastic Analysis (ANTYPE = 4) Element Type(s): Spar Elements (LINK1)

Mass Elements (MASS21) A mass m supported on a thin rod of area A and length L is subjected to the action of a suddenly applied constant force F1. The stress-strain curve for the rod material is shown below. Determine the maximum deflection ymax and minimum deflection ymin of the mass, neglecting the mass of the rod.

Material Properties m = 0.0259 kips-sec2/in E = 30 x 103 ksi σyp = 162.9 ksi


= 100 in A = 0.278 in2

Loading F1 = 30 kips


The initial integration time step (0.004/10 = 0.0004 sec) is chosen small enough to allow the initial step change in acceleration to be followed reasonably well. The final integration time

step ((0.14-.004)/68 = 0.002 sec) is based on 1/60 of the period to produce sufficient printout for the theoretical comparison. The final time of 0.14 sec allows slightly more than 1 cycle of vibration to be followed. POST26 is used to extract results from the solution phase.



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Results Comparison

Target ANSYS[1] Ratio ymax, in 0.806 0.804 0.998 Time, sec 0.0669 0.0680 1.016 ymin, in 0.438 [2] 0.438 0.999 Time, sec 0.122 [2] 0.122 1.000



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Problem-24 (Transient Response of a Spring-mass System) Analysis Type(s): Mode-frequency Analysis (ANTYPE = 2)

Transient Dynamic Mode Superposition Analysis (ANTYPE = 4) Element Type(s): Combination Elements (COMBIN40) A system containing two masses, m1 and m2, and two springs of stiffness k1 and k2 is subjected to a pulse load F(t) on mass 1. Determine the displacement response of the system for the load history shown.

Material Properties k1 = 6 N/m k2 = 16 N/m m1 = 2 Kg m2 = 2 Kg

Loading F0 = 50 N td = 1.8 sec


COMBIN40 combination elements are used to represent the springs and masses. Node locations are arbitrary. The response of the system is examined for an additional 0.6 seconds after the load is removed.

Results Comparison

Target ANSYS Ratio Y1 , m (@ t = 1.3s) 14.48 14.40 0.995 Y2 , m (@ t = 1.3s) 3.99 3.95 0.990 Y1 , m (@ t = 2.4s) 18.32 18.40 1.004 Y2 , m (@ t = 2.4s) 6.14 6.16 1.003



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Buckling Analysis- Some Theory

Buckling analysis is a technique used to determine buckling loads - critical loads at which a structure becomes unstable - and buckled mode shapes - the characteristic shape associated with a structure's buckled response. Two techniques are available in the ANSYS Multiphysics, ANSYS Mechanical, ANSYS Structural, and ANSYS Professional programs for predicting the buckling load and buckling mode shape of a structure: nonlinear buckling analysis, and eigenvalue (or linear) buckling analysis. Since these two methods frequently yield quite different results, let's examine the differences between them before discussing the details of their implementation. Nonlinear Buckling Analysis Nonlinear buckling analysis is usually the more accurate approach and is therefore recommended for design or evaluation of actual structures. This technique employs a nonlinear static analysis with gradually increasing loads to seek the load level at which your structure becomes unstable, as depicted in Buckling Curves (a). Using the nonlinear technique, your model can include features such as initial imperfections, plastic behavior, gaps, and large-deflection response. In addition, using deflection-controlled loading, you can even track the post-buckled performance of your structure (which can be useful in cases where the structure buckles into a stable configuration, such as "snap-through" buckling of a shallow dome). Eigenvalue Buckling Analysis Eigenvalue buckling analysis predicts the theoretical buckling strength (the bifurcation point) of an ideal linear elastic structure. (See Buckling Curves (b).) This method corresponds to the textbook approach to elastic buckling analysis: for instance, an eigenvalue buckling analysis of a column will match the classical Euler solution. However, imperfections and nonlinearities prevent most real-world structures from achieving their theoretical elastic buckling strength. Thus, eigenvalue-buckling analysis often yields unconservative results, and should generally not be used in actual day-to-day engineering analyses.



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Problem-25: Buckling - Illustrated

Introduction This example was created using ANSYS 7.0 to solve a simple buckling problem. Buckling loads are critical loads where certain types of structures become unstable. Each load has an associated buckled mode shape; this is the shape that the structure assumes in a buckled condition. There are two primary means to perform a buckling analysis:

1. Eigenvalue Eigenvalue buckling analysis predicts the theoretical buckling strength of an ideal elastic structure. It computes the structural eigenvalues for the given system loading and constraints. This is known as classical Euler buckling analysis. Buckling loads for several configurations are readily available from tabulated solutions. However, in real-life, structural imperfections and nonlinearities prevent most real-world structures from reaching their eigenvalue predicted buckling strength; ie. it over-predicts the expected buckling loads. This method is not recommended for accurate, real-world buckling prediction analysis.

2. Nonlinear Nonlinear buckling analysis is more accurate than eigenvalue analysis because it employs non-linear, large-deflection, static analysis to predict buckling loads. Its mode of operation is very simple: it gradually increases the applied load until a load level is found whereby the structure becomes unstable (ie. suddenly a very small increase in the load will cause very large deflections). The true non-linear nature of this analysis thus permits the modeling of geometric imperfections, load perterbations, material nonlinearities and gaps. For this type of analysis, note that small off-axis loads are necessary to initiate the desired buckling mode.



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This example will use a steel beam with a 10 mm X 10 mm cross section, rigidly constrained at the bottom. The required load to cause buckling, applied at the top-center of the beam, will be calculated.

Eigenvalue Buckling Analysis Preprocessing: Defining the Problem

1. Open preprocessor menu /PREP7

2. Give example a Title Utility Menu > File > Change Title ... /title,Eigen-Value Buckling Analysis

3. Define Keypoints Preprocessor > Modeling > Create > Keypoints > In Active CS ... K,#,X,Y We are going to define 2 Keypoints for this beam as given in the following table: Keypoints Coordinates (x,y)

1 (0,0) 2 (0,100)

4. Create Lines Preprocessor > Modeling > Create > Lines > Lines > In Active Coord L,1,2 Create a line joining Keypoints 1 and 2

5. Define the Type of Element Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the BEAM3 (Beam 2D elastic) element. This element has 3 degrees of freedom (translation along the X and Y axes, and rotation about the Z axis).


i. Cross-sectional area AREA: 100 ii. Area moment of inertia IZZ: 833.333

iii. Total Beam Height HEIGHT: 10 This defines a beam with a height of 10 mm and a width of 10 mm.


i. Young's modulus EX: 200000 ii. Poisson's Ratio PRXY: 0.3

8. Define Mesh Size Preprocessor > Meshing > Size Cntrls > ManualSize > Lines > All Lines... For this example we will specify an element edge length of 10 mm (10 element divisions along the line).




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Solution Phase: Assigning Loads and Solving

1. Define Analysis Type Solution > Analysis Type > New Analysis > Static ANTYPE,0

2. Activate prestress effects To perform an eigenvalue buckling analysis, prestress effects must be activated.

o You must first ensure that you are looking at the unabridged solution menu so that you can select Analysis Options in the Analysis Type submenu. The last option in the solution menu will either be 'Unabridged menu' (which means you are currently looking at the abridged version) or 'Abriged Menu' (which means you are looking at the unabridged menu). If you are looking at the abridged menu, select the unabridged version.

o Select Solution > Analysis Type > Analysis Options o In the following window, change the [SSTIF][PSTRES] item to 'Prestress ON',

which ensures the stress stiffness matrix is calculated. This is required in eigenvalue buckling analysis.


Solution > Define Loads > Apply > Structural > Displacement > On Keypoints Fix Keypoint 1 (ie all DOF constrained).

4. Apply Loads Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints The eignenvalue solver uses a unit force to determine the necessary buckling load. Applying a load other than 1 will scale the answer by a factor of the load. Apply a vertical (FY) point load of -1 N to the top of the beam (keypoint 2).

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



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5. Solve the System


6. Exit the Solution processor Close the solution menu and click FINISH at the bottom of the Main Menu. FINISH

Normally at this point you enter the postprocessing phase. However, with a buckling analysis you must re-enter the solution phase and specify the buckling analysis. Be sure to close the solution menu and re-enter it or the buckling analysis may not function properly.

7. Define Analysis Type Solution > Analysis Type > New Analysis > Eigen Buckling ANTYPE,1

8. Specify Buckling Analysis Options o Select Solution > Analysis Type > Analysis Options o Complete the window which appears, as shown below. Select 'Block Lanczos'

as an extraction method and extract 1 mode. The 'Block Lanczos' method is used for large symmetric eigenvalue problems and uses the sparse matrix solver. The 'Subspace' method could also be used, however it tends to converge slower as it is a more robust solver. In more complex analyses the Block Lanczos method may not be adequate and the Subspace method would have to be used.



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9. Solve the System


10. Exit the Solution processor Close the solution menu and click FINISH at the bottom of the Main Menu. FINISH

Again it is necessary to exit and re-enter the solution phase. This time, however, is for an expansion pass. An expansion pass is necessary if you want to review the buckled mode shape(s).

11. Expand the solution o Select Solution > Analysis Type > Expansion Pass... and ensure that it is on.

You may have to select the 'Unabridged Menu' again to make this option visible.

o Select Solution > Load Step Opts > ExpansionPass > Single Expand > Expand Modes ...

o Complete the following window as shown to expand the first mode

12. Solve the System


Postprocessing: Viewing the Results 1. View the Buckling Load

To display the minimum load required to buckle the beam select General Postproc > List Results > Detailed Summary. The value listed under 'TIME/FREQ' is the load (41,123), which is in Newtons for this example. If



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more than one mode was selected in the steps above, the corresponding loads would be listed here as well. /POST1 SET,LIST

2. Display the Mode Shape o Select General Postproc > Read Results > Last Set to bring up the data for

the last mode calculated. o Select General Postproc > Plot Results > Deformed Shape

Non-Linear Buckling Analysis Ensure that you have completed the NonLinear Example prior to beginning this portion of the example Preprocessing: Defining the Problem

1. Open preprocessor menu /PREP7

2. Give example a Title Utility Menu > File > Change Title ... /TITLE, Nonlinear Buckling Analysis

3. Create Keypoints Preprocessor > Modeling > Create > Keypoints > In Active CS K,#,X,Y We are going to define 2 keypoints (the beam vertices) for this structure to create a beam with a length of 100 millimeters: Keypoint Coordinates (x,y)

1 (0,0) 2 (0,100)

4. Define Lines Preprocessor > Modeling > Create > Lines > Lines > Straight Line Create a line between Keypoint 1 and Keypoint 2. L,1,2

5. Define Element Types



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Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the BEAM3 (Beam 2D elastic) element. This element has 3 degrees of freedom (translation along the X and Y axis's, and rotation about the Z axis). With only 3 degrees of freedom, the BEAM3 element can only be used in 2D analysis.


i. Cross-sectional area AREA: 100 ii. Area Moment of Inertia IZZ: 833.333

iii. Total beam height HEIGHT: 10 This defines an element with a solid rectangular cross section 10 x 10 millimeters.



8. Define Mesh Size Preprocessor > Meshing > Size Cntrls > Lines > All Lines... For this example we will specify an element edge length of 1 mm (100 element divisions along the line). ESIZE,1








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Ensure the following selections are made under the 'Basic' tab (as shown above)



C. Enter 20 as the number of substeps. This will set the initial substep to 1/20 th of the total load.


E. Enter a minimum number of substeps of 1. F. Ensure all solution items are writen to a results file.

Ensure the following selection is made under the 'Nonlinear' tab (as shown below)

G. Ensure Line Search is 'On'. This option is used to help the Newton-Raphson solver converge.

H. Ensure Maximum Number of Iterations is set to 1000



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NOTE There are several options which have not been changed from their default values. For more information about these commands, type help followed by the command into the command line.


4. Apply Loads Solution > Define Loads > Apply > Structural > Force/Moment > On Keypoints Place a -50,000 N load in the FY direction on the top of the beam (Keypoint 2). Also apply a -250 N load in the FX direction on Keypoint 2. This horizontal load will persuade the beam to buckle at the minimum buckling load. The model should now look like the window shown below.



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5. Solve the System


The following will appear on your screen for NonLinear Analyses


General Postprocessing: Viewing the Results 1. View the deformed shape

o To view the element in 2D rather than a line: Utility Menu > PlotCtrls > Style > Size and Shape and turn 'Display of element' ON (as shown below).



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o General Postproc > Plot Results > Deformed Shape... > Def + undeformed

PLDISP,1

o View the deflection contour plot



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Time History Postprocessing: Viewing the Results As shown, you can obtain the results (such as deflection, stress and bending moment diagrams) the same way you did in previous examples using the General Postprocessor. However, you may wish to view time history results such as the deflection of the object over time.

1. Define Variables o Select: Main Menu > TimeHist Postpro. The following window should open

automatically.



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If it does not open automatically, select Main Menu > TimeHist Postpro > Variable Viewer

o Click the add button in the upper left corner of the window to add a variable.

o Double-click Nodal Solution > DOF Solution > Y-Component of displacement (as shown below) and click OK. Pick the uppermost node on the beam and click OK in the 'Node for Data' window.

o To add another variable, click the add button again. This time select Reaction

Forces > Structural Forces > Y-Component of Force. Pick the lowermost node on the beam and click OK.

o On the Time History Variable window, click the circle in the 'X-Axis' column for FY_3. This will make the reaction force the x-variable. The Time History Variables window should now look like this:

2. Graph Results over Time

o Click on UY_2 in the Time History Variables window.

o Click the graphing button in the Time History Variables window.



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o The labels on the plot are not updated by ANSYS, so you must change them

manually. Select Utility Menu > Plot Ctrls > Style > Graphs > Modify Axes and re-label the X and Y-axis appropriately.

The plot shows how the beam became unstable and buckled with a load of approximately 40,000 N, the point where a large deflection occured due to a small increase in force. This is slightly less than the eigen-value solution of 41,123 N, which was expected due to non-linear geometry issues discussed above.



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Problem-26 (Buckling of a Bar with Hinged Ends (Line Elements)) Analysis Type(s): Buckling Analysis (ANTYPE = 1)

Static (prestress) Analysis (ANTYPE = 0) Element Type(s): Beam Elements (BEAM3) Determine the critical buckling load of an axially loaded long slender bar of length L with hinged ends. The bar has a cross-sectional height h, and area A.

Material Properties E = 30 x 106 psi


= 200 in A = 0.25 in2 h = 0.5 in

LoadingF = 1 lb


Only the upper half of the bar is modeled because of symmetry. The boundary conditions become free-fixed for the half symmetry model. A total of 10 master degrees of freedom in the X-direction are selected to characterize the buckling mode. The moment of inertia of the bar is calculated as I = Ah2/12 = 0.0052083 in4 . Results Comparison Target ANSYS Ratio Fcr, lb 38.553 38.553[1] 1.000

1. Fcr = Load Factor (1st mode). End of Structural Analysis workbook

ecc ansys workbook v2006 r1_structural analysis

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