Thursday, May 21, 2009

Modal analysis of X-Y table

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Problem:


The X-Y table shown in figure 1 is used in precision measurement and its structure and the dimensions are shown in figure 2 and 3. If the table is made of steel with the following material properties: modulus of elasticity E=200 Gpa, Poisson’s ratio ν=0.3, density ρ=7850 Kg/m3 and is fixed at the bottom of column. Please use elements SHELL63 and BEAM3 to do the ANSYS analysis and find the first 10 modes of free vibrations. Compare the differences of results by using different elements.


Solution:

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Solution 2 – Using Shell63

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1. Preparation:

For creating modeling, we create areas -> extrude to volumes -> Cut volume by work plane -> Glue all of volumes -> Delete volumes, keep areas.

First, we need create lines for extruding:


Where:
B1 x H1 x T1 are dimensions of cross session of vertical tubes.
B2 x H2 x T2 are dimensions of cross session of horizontal tube.
W x H are dimensions of X-Y table.

We use unit of meter for length.

2. APDL code:
! CREATE PARAMETERS
! - VERTICAL TUBES
B1=50E-3 ! DIMENSION OF CROSS SECTION
H1=50E-3 ! DIMENSION OF CROSS SECTION
T1=2E-3 ! THICKNESS OF STEEL
! - HORIZONTAL TUBE
B2=50E-3 ! DIMENSION OF CROSS SECTION
H2=35E-3 ! DIMENSION OF CROSS SECTION
T2=2E-3 ! THICKNESS OF STEEL
W=0.35 ! WIDTH OF X-Y TABLE
H=0.25 ! HEIGHT OF X-Y TABLE
YM=200E9 ! YOUNG MODULUS OF STEEL
PRT=0.3 ! POISSON RATIO
RHO=7850 ! MASS DENSITY OF STEEL
ESZ=1E-2 ! ELEMENT SIZE FOR MESHING
NMODE=10 ! MODES OF FREE VIBRATIONS
/PREP7
ET,1,SHELL63 ! DEFINE ELEMENT TYPE
R,1,T1 ! THICKNESS FOR VERTICAL TUBES
R,2,T2 ! THICKNESS FOR HORIZONTAL TUBE
MP,EX,1,YM ! DEFINE YOUNG MODULUS
MP,PRXY,1,PRT ! DEFINE POINSSON RATIO
MP,DENS,1,RHO ! DEFINE MASS DENSITY
! - CREATE KEYPOINTS
K,1,0,0 ! CREATE KEYPOINT 1
K,2,0,H-(H2-T2)/2 ! CREATE KEYPOINT 2
K,3,W,H-(H2-T2)/2 ! CREATE KEYPOINT 3
K,4,W,0 ! CREATE KEYPOINT 4
K,5,-(B1-T1)/2,H ! CREATE KEYPOINT 5
K,6,W+(B1-T1)/2,H ! CREATE KEYPOINT 6
! - CREATE LINES THAT IS THE PATH FOR EXTRUDING AREAS
L,1,2 ! CREATE LINES 1 (PATH OF LEFT VERTICAL TUBE)
L,5,6 ! CREATE LINES 2 (PATH OF HORIZONTAL TUBE)
L,3,4 ! CREATE LINES 3 (PATH OF RIGHT VERTICAL TUBE)
! - CREATE CROSS SECTION OF VERTICAL TUBES
WPROTA,0,-90 ! ROTATE WORK PLANE AROUND X-AXIS
RECT,-(B1-T1)/2,(B1-T1)/2,-(H1-T1)/2,(H1-T1)/2 ! CREATE AREA
WPOFFS,W,0,0 ! MOVE WORK PLANE TO RIGHT SIDE
RECT,-(B1-T1)/2,(B1-T1)/2,-(H1-T1)/2,(H1-T1)/2 ! CREATE AREA
WPCSYS,-1,0 ! ALIGN WP TO ORIGIN OF CARTESIAN CS
! - CREATE CROSS SECTION OF HORIZONTAL TUBE
WPOFFS,-(B1-T1)/2,H,0 ! MOVE WORK PLANE UP
WPROTA,0,0,90 ! ROTATE WORK PLANE AROUND Y-AXIS
RECT,-(B2-T2)/2,(B2-T2)/2,-(H2-T2)/2,(H2-T2)/2 ! CREATE AREA
WPCSYS,-1,0 ! ALIGN WP TO ORIGIN OF CARTESIAN CS
/VIEW,1,1,2,3 ! CHANGE THE VIEW POINT
APLOT ! SHOW AREAS


! - EXTRUSS AREA TO VOLUME
NUMCMP,ALL ! COMPRESS ALL OF NUMBERS
VDRAG,1, , , , , ,1 ! EXTRUDE AREA TO VOLUME (LEFT TUBE)
VDRAG,2, , , , , ,3 ! EXTRUDE AREA TO VOLUME (HORIZONTAL TUBE)
VDRAG,3, , , , , ,2 ! EXTRUDE AREA TO VOLUME RIGHT TUBE)
! - CUT HORIZONTAL TUBE
VSEL,S,LOC,X,W/2 ! SELECT VOLUME OF HORIZONTAL TUBE
WPOFFS,(B1-T1)/2,H,0 ! MOVE WORK PLANE UP
WPROTA,0,0,90 ! ROTATE WORK PLANE AROUND Y-AXIS
VSBW,ALL ! CUT VOLUME BY WP
WPOFFS,0,0,W-(B1-T1) ! MOVE WORK PLANE UP
VSBW,ALL ! CUT VOLUME BY WP
WPCSYS,-1,0 ! ALIGN WP TO ORIGIN OF CARTESIAN CS
VSEL,ALL ! RE-SELECT ALL VOLUMES
VGLUE,ALL ! GLUING ALL VOLUMES
VDELE,ALL ! DELETE VOLUMES, KEEP AREAS
APLOT ! SHOW ALL OF AREAS
! - MESHING
TYPE,1 ! SELECT ELEMENT TYPE 1
MAT,1 ! SELECT MATERIAL PROPERTY 1
ESIZE,ESZ ! DEFINE ELEMENT SIZE
REAL,1 ! SELECT REAL CONSTANT 1
MESHAPE,0,2D
MESHKEY,1 ! SIDED MESH
ASEL,S,LOC,X,-B1/2,B1/2 ! SELECT LEFT TUBE
AMESH,ALL ! MESH AREAS OF LEFT TUBE
ASEL,ALL
ASEL,S,LOC,X,W-B1/2,W+B1/2 ! SELECT RIGHT TUBE
AMESH,ALL ! MESH AREAS OF RIGHT TUBE
ASEL,ALL
ASEL,S,LOC,X,W/2 ! SELECT HORIZONTAL TUBE
REAL,2 ! SELECT REAL CONSTANT 2
AMESH,ALL ! MESH AREAS OF HORIZONTAL TUBE
ASEL,ALL
/SOLUTION
LSEL,S,LOC,Y,0 ! SELECT LINES ON THE BASE
DL,ALL, ,ALL ! BOUNDARY CONDITION
LSEL,ALL
ANTYPE,MODAL ! ANALYSIS TYPE IS MODEL ANALYSIS
MODOPT,LANB,NMODE ! SOLVER IS BLOCK LAMCZOR AND 10 MODES
SOLVE
/POST1
*DO,I,1,NMODE,1 ! BEGIN OF DO LOOP
SET,1,I ! READ IN THE RESULT OF 1ST LOAD STEP AND ITH SUBSTEPS
PLDISP,2 ! DISPLAY DEFORMED AND UN-DEFORMED SHAPE
/WAIT,1 ! FREEZE SCREEN 1 SECOND
*ENDDO ! END OF DO LOOP
3. Result:

First set:
Second set:
Third set:
List of result:

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Solution 2 – Using Beam3


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Tuesday, May 5, 2009

Theory of Finite Element Analysis

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Finite Element Analysis (FEA) is a numerical method which provides solutions to problems that would otherwise be difficult to obtain. In terms of fracture, FEA most often involves the determination of stress intensity factors. FEA, however, has applications in a much broader range of areas; for example, fluid flow and heat transfer. While this range is growing, one thing will remain the same: the theory of how the method works.

The most efficient method of learning is by example. Therefore, I would like to present to you a simple FEA problem: the case of a three-member truss. The method of solution to this problem should demonstrate the basic concepts of FEA which are present in any analysis.

Before introducing specific quantities for our example, let's first take a look at our structure:

The overall objective of our analysis will be to determine the displacements of the truss members given the load P.

The first thing we must do is choose our elements. For our situation this is easy: each truss member should be one element. Further division would accomplish nothing, since each truss member can only support axial loads.

Let us now examine a single truss member:

Nodes are located at each end of the bar, each of which can have displacements in the x and y directions. The displacements are denoted u1, u2, u3, and u4. Corresponding forces due to these displacements are F1, F2, F3, and F4. The bar has a uniform cross-sectional area A and Young's Modulus E. The general relationship between force and displacement is Fi = kij*uj, where Fi is the force in direction i, uj is the displacement in direction j, and kij is the "stiffness" coefficient relating Fi to uj. In our particular example of a horizontal truss element, we have the following system of equations:

F1 = k11u1 + k12u2 + k13u3 + k14u4

F2 = k21u2 + k22u2 + k23u3 + k24u4

F3 = k31u1 + k32u2 + k33u3 + k34u4

F4 = k41u1 + k42u2 + k43u3 + k44u4

Alternatively, in matrix form:

The matrix kij is called the "stiffness matrix". It is the matrix which defines the geometric and material properties of the bar. Stiffness matrices are a fundamental part of FEA. These matrices always define inherent properties of the system being studied. For the system at hand, we need to determine the stiffness matrix. The way we will go about doing this may seem a little strange at first, but try to follow the reasoning as it does make sense. Let's begin by assuming u1 = 1 and u2 = u3 = u4 = 0. Then our matrix takes the form:

Each force Fi is equal to kj1. Now, recall from mechanics of materials that the displacement of a rod is given by u = FL/AE. With displacement u1 = 1, force 1 is F1 = AE/L. To maintain equilibrium, we must also have a force F3 = -AE/L:


Since our Fi's equal our ki1's, we have:

It important to remember that our element can support only axial loads. Therefore, displacements u2 and u4 can not give rise to stresses in the bar since these displacements are perpendicular to the axis of the bar. Thus, the stiffness coefficients of these displacements must be zero: ki2 = ki4 = 0. Finally, a displacement u3 = 1 will result in forces just opposite to those from u1 = 1, so ki3 = -ki1. Our stiffness matrix is:

It must be emphasized that the stiffness matrix just derived is only valid for bars parallel to the x-axis. Through a similar derivation it can be shown that the stiffness matrix for any bar oriented at an angle "theta" to the x-axis is:

where c = cos"theta" and s = sin"theta". Note that when "theta" = 0, this stiffness matrix reduces to the one we derived for a horizontal bar.

Now knowing the stiffness matrix for any axially loaded bar, we can apply it to a real situation with specific quantities. Consider the following truss:

The displacements and external forces are:

Note the symbols we are using: R is an external force on the truss; F is an internal force resulting from the stresses imposed on the structure during a displacement. Knowing the orientations of each element, we can set up matrices for them. Using "theta" = 90 degrees for element 1, "theta" = 135 degrees for element 2, and "theta" = 0 degrees for element 3 we obtain the following matrices:

We can now generate a set of equilibrium equations for each node. Consider the following figure:

The nodal forces (resulting from element displacements) must be equal and opposite to the externally applied forces. Note that we have all forces drawn in positive x and y directions. Thus, for equilibrium at node 1:

x - direction: R2 - F2(element3) - F2(element2) = 0

y - direction: R1 - F1(element3) - F1(element2) = 0

We want to solve for R1 and R2. Obtaining the nodal forces F2(element3), F2(element2), F1(element3), and F1(element2) from our previously determined matrices we get:

R1 = AE/L ( 3u1/2 - u2/2 - u3 - u5/2 + u6/2 )

R2 = AE/L ( -u1/2 + u2/2 + u5/2 - u6/2 ).

Similarly, from equilibrium of nodes 2 and 3 we obtain:

R3 = AE/L ( -u1 + u3 )

R4 = AE/L ( u4 - u6 )

R5 = AE/L ( -u1/2 + u2/2 + u5/2 - u6/2 )

R6 = AE/L ( u1/2 - u2/2 - u4 - u5/2 +3u6/2 )

We can now combine all of our external forces into one matrix:

Now recall what we are trying to do here: given a load P, we want to solve for the displacements at each node. Observing that node 2 is pinned and that node 3 is on a roller, the displacements u3, u4, and u5 must equal 0. These values are quite important because without them we wouldn't be able to solve the problem. As a matter of fact, values such as these are always needed in finite element analyses; they are known as "boundary conditions". Next, we must state the reactions which are known from our particular loading. We can see from the truss that R1 = 0, R2 = -P, and R6 = 0. Entering the known displacements and reactions into our matrix we get:

This matrix reduces to:

We can now finish our problem by solving this matrix for u1, u2, and u6:

u1 = - PL/AE

u2 = -4PL/AE

u6 = -PL/AE

This application of FEA to a simple three-member truss shows in general how the method works. Most applications to engineering problems, however, are much more complex. Such analyses require large numbers of elements and nodes in order to accurately represent the physical system being studied. These analyses inevitably require the application of a computer.

For Virginia Tech engineering students interested in FEA, one undergraduate course is available: ESM 4734 - An Introduction to the Finite Element Method. In this course, students study the theory and application of FEA to problems in various fields of engineering and applied sciences. The pre-requisite for the course is: ESM 2074 - Computational Methods.

References

Finite Element Analysis on Microcomputers, Nicholas M. Baran, McGraw-Hill Book Company, 1988.

Finite Element Primer, Bruce Irons and Nigel Shrive, John Wiley & Sons, Inc., 1983.

[Jason Midkiff]
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Finite Element Analysis: An introduction

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Finite Element Analysis (FEA) was first developed in 1943 by R. Courant, who utilized the Ritz method of numerical analysis and minimization of variational calculus to obtain approximate solutions to vibration systems. Shortly thereafter, a paper published in 1956 by M. J. Turner, R. W. Clough, H. C. Martin, and L. J. Topp established a broader definition of numerical analysis. The paper centered on the "stiffness and deflection of complex structures".

By the early 70's, FEA was limited to expensive mainframe computers generally owned by the aeronautics, automotive, defense, and nuclear industries. Since the rapid decline in the cost of computers and the phenomenal increase in computing power, FEA has been developed to an incredible precision. Present day supercomputers are now able to produce accurate results for all kinds of parameters.

What is Finite Element Analysis?

FEA consists of a computer model of a material or design that is stressed and analyzed for specific results. It is used in new product design, and existing product refinement. A company is able to verify a proposed design will be able to perform to the client's specifications prior to manufacturing or construction. Modifying an existing product or structure is utilized to qualify the product or structure for a new service condition. In case of structural failure, FEA may be used to help determine the design modifications to meet the new condition.

There are generally two types of analysis that are used in industry: 2-D modeling, and 3-D modeling. While 2-D modeling conserves simplicity and allows the analysis to be run on a relatively normal computer, it tends to yield less accurate results. 3-D modeling, however, produces more accurate results while sacrificing the ability to run on all but the fastest computers effectively. Within each of these modeling schemes, the programmer can insert numerous algorithms (functions) which may make the system behave linearly or non-linearly. Linear systems are far less complex and generally do not take into account plastic deformation. Non-linear systems do account for plastic deformation, and many also are capable of testing a material all the way to fracture.

How Does Finite Element Analysis Work?

FEA uses a complex system of points called nodes which make a grid called a mesh (Figure 2). This mesh is programmed to contain the material and structural properties which define how the structure will react to certain loading conditions. Nodes are assigned at a certain density throughout the material depending on the anticipated stress levels of a particular area. Regions which will receive large amounts of stress usually have a higher node density than those which experience little or no stress. Points of interest may consist of: fracture point of previously tested material, fillets, corners, complex detail, and high stress areas. The mesh acts like a spider web in that from each node, there extends a mesh element to each of the adjacent nodes. This web of vectors is what carries the material properties to the object, creating many elements. (Theory)

A wide range of objective functions (variables within the system) are available for minimization or maximization:
  • Mass, volume, temperature
  • Strain energy, stress strain
  • Force, displacement, velocity, acceleration
  • Synthetic (User defined)
There are multiple loading conditions which may be applied to a system. Next to Figure 3, some examples are shown:
  • Point, pressure (Figure 3), thermal, gravity, and centrifugal static loads
  • Thermal loads from solution of heat transfer analysis
  • Enforced displacements
  • Heat flux and convection
  • Point, pressure and gravity dynamic loads
Each FEA program may come with an element library, or one is constructed over time. Some sample elements are:
  • Rod elements
  • Beam elements
  • Plate/Shell/Composite elements
  • Shear panel
  • Solid elements
  • Spring elements
  • Mass elements
  • Rigid elements
  • Viscous damping elements
Many FEA programs also are equipped with the capability to use multiple materials within the structure such as:
  • Isotropic, identical throughout
  • Orthotropic, identical at 90 degrees
  • General anisotropic, different throughout
Types of Engineering Analysis

Structural analysis consists of linear and non-linear models. Linear models use simple parameters and assume that the material is not plastically deformed. Non-linear models consist of stressing the material past its elastic capabilities. The stresses in the material then vary with the amount of deformation as in Figure 4.

Vibrational analysis is used to test a material against random vibrations, shock, and impact. Each of these incidences may act on the natural vibrational frequency of the material which, in turn, may cause resonance and subsequent failure.

Fatigue analysis helps designers to predict the life of a material or structure by showing the effects of cyclic loading on the specimen. Such analysis can show the areas where crack propagation is most likely to occur. Failure due to fatigue may also show the damage tolerance of the material (Figure 5).

Heat Transfer analysis models the conductivity or thermal fluid dynamics of the material or structure (Figure 1). This may consist of a steady-state or transient transfer. Steady-state transfer refers to constant thermoproperties in the material that yield linear heat diffusion.

Results of Finite Element Analysis

FEA has become a solution to the task of predicting failure due to unknown stresses by showing problem areas in a material and allowing designers to see all of the theoretical stresses within. This method of product design and testing is far superior to the manufacturing costs which would accrue if each sample was actually built and tested.
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Tuesday, April 7, 2009

Two Dimensional Truss

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Introduction

This tutorial was created using ANSYS 7.0 to solve a simple 2D Truss problem. This is the first of four introductory ANSYS tutorials.

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 (such as 'Bridge Truss Tutorial')

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.

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

keypointcoordinate
x y
100
218003118
336000
454003118
572000
690003118
7108000

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

* From the 'ANSYS Main Menu' select:
Preprocessor > Modeling > Create > Keypoints > In Active CS
The following window will then appear:
To define the first keypoint which has the coordinates x = 0 and y = 0:

Enter keypoint number 1 in the appropriate box, and enter the x,y coordinates: 0, 0 in their appropriate boxes (as shown above).

Click 'Apply' to accept what you have typed.

* Enter the remaining keypoints using the same method.

Note: When entering the final data point, click on 'OK' to indicate that you are finished entering keypoints. If you first press 'Apply' and then 'OK' for the final keypoint, you will have defined it twice!

If you did press 'Apply' for the final point, simply press 'Cancel' to close this dialog box.

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

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

* In the main menu select: Preprocessor > Modeling > Create > Lines > Lines > In Active Coord. The following window will then appear:
* Use the mouse to pick keypoint #1 (i.e. click on it). It will now be marked by a small yellow box.

* Now move the mouse toward keypoint #2. A line will now show on the screen joining these two points. Left click and a permanent line will appear.

* Connect the remaining keypoints using the same method.

* 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:
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

Define the Type of Element:

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

* From the Preprocessor Menu, select: Element Type > Add/Edit/Delete. The following window will then appear:
* Click on the 'Add...' button. The following window will appear:
* 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.

* Click on 'Close' in the 'Element Types' dialog box.

Define Geometric Properties:

We now need to specify geometric properties for our elements:

* In the Preprocessor menu, select Real Constants > Add/Edit/Delete
* Click Add... and select 'Type 1 LINK1' (actually it is already selected). Click on 'OK'. The following window will appear:
* As shown in the window above, enter the cross-sectional area (3250mm):

* Click on 'OK'.

* 'Set 1' now appears in the dialog box. Click on 'Close' in the 'Real Constants' window.

Element Material Properties:

You then need to specify material properties:

* In the 'Preprocessor' menu select Material Props > Material Models
* Double click on Structural > Linear > Elastic > Isotropic
We are going to give the properties of Steel. Enter the following field:
EX 200000

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

Mesh Size:

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

* In the Preprocessor menu select Meshing > Size Cntrls > ManualSize > Lines > All Lines
* In the size 'NDIV' field, enter the desired number of divisions per line. For this example we want only 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.

Mesh:

Now the frame can be meshed.

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

Your model should now appear as shown in the 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'
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:

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

* Click 'OK'.

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.

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

* 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:

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

* Select Define Loads > Apply > Structural > Force/Moment > on Keypoints.

* Select the first Keypoint (left end of the truss) and click 'OK' in the 'Apply F/M on KPs' window.
* Select FY in the 'Direction of force/mom'. This indicate that we will be applying the load in the 'y' direction

* Enter a value of -280000 in the 'Force/moment value' box and click 'OK'. Note that we are using units of N here, this is consistent with the previous values input.

* The force will appear in the graphics window as a red arrow.

* 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:

* In the 'Solution' menu select Solve > Current LS. This indicates that we desire the solution under the current Load Step (LS).
* The above windows will appear. Ensure that your solution options are the same as shown above and click 'OK'.

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

Postprocessing: Viewing the Results

1. Reaction Forces

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

* from the Main Menu select General Postproc > List Results > Reaction Solu.
* Select 'All struc forc F' as shown above and click 'OK'
2. Deformation:

* In the General Postproc menu, select Plot Results > Deformed Shape. The following window will appear.
* Select 'Def + undef edge' and click 'OK' to view both the deformed and the undeformed object.
* Observe the value of the maximum deflection in the upper 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!)

3. Deflection:

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

* From the 'General Postproc' menu select Plot results > Contour Plot > Nodal Solution. The following window will appear.
* Select 'DOF solution' and 'USUM' as shown in the above window. Leave the other selections as the default values. Click 'OK'.
* Looking at the scale, you may want to use more useful intervals. From the Utility Menu select Plot Controls > Style > Contours > Uniform Contours...

* Fill in the following window as shown and click 'OK'.
You should obtain the following.
* 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.
* 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.

* 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'.

4. 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'

* From the General Postprocessor menu select Element Table > Define Table

* Click on 'Add...'
* 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

* Click on 'OK' and close the 'Element Table Data' window.

* Plot the Stresses by selecting Element Table > Plot Elem Table

* The following window will appear. Ensure that 'SAXL' is selected and click 'OK'
* Because you changed the contour intervals for the Displacement plot to "User Specified" - you need to switch this back to "Auto calculated" to obtain new values for VMIN/VMAX.
Utility Menu > PlotCtrls > Style > Contours > Uniform Contours ...
* Again, you may wish to select more appropriate intervals for the contour plot

* List the Stresses
o From the 'Element Table' menu, select 'List Elem Table'
o From the 'List Element Table Data' window which appears ensure 'SAXL' is highlighted
o Click 'OK'
Note that the axial stress in Element 1 is 82.9MPa as predicted analytically.

[Source: University of Alberta]
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