Modelling and Analysis Laboratory Manual

Computer Aided Modeling and Analysis Laboratory

Department Of Mechanical Engineering, Don Bosco Institute of Technology 1



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In thermo mechanical members and structures, finite-element analysis (FEA) is typically

invoked to compute displacement and temperature fields from known applied loads and heat

fluxes. FEA has emerged in recent years as an essential resource for mechanical and

structural designers. Its use is often mandated by standards such as the ASME Pressure Vessel Code, by insurance requirements, and even by law. Its acceptance has benefited from

rapid progress in related computer hardware and software, especially computer-aided design (CAD) systems. Today, a number of highly developed, user-friendly finite-element codes are

available commercially. The purpose of this chapter is to introduce finite-element theory and practice. The next three chapters

focus on linear elasticity and thermal response, both static and dynamic, of basic structural members. After that, nonlinear thermo mechanical response is considered.

In FEA practice, a design file developed using CAD is often “imported” into finite element

codes, from which point little or no additional effort is required to develop the finite-element

model and perform sophisticated thermo mechanical analysis and simulation. CAD integrated

with an analysis tool, such as FEA, is an example of computer-aided engineering (CAE).

CAE is a powerful resource with the potential of identifying design problems much more

efficiently and rapidly than by “trial and error.” A major FEM application is the

determination of stresses and temperatures in a component or member in locations where

failure is thought most likely. If the stresses or temperatures exceed allowable or safe values,

the product can be redesigned and then reanalyzed. Analysis can be diagnostic, supporting

interpretation of product-failure data. Analysis also can be used to assess performance, for example, by determining whether the design-stiffness coefficient for a rubber spring is

attained.

OVERVIEW OF THE FINITE-ELEMENT METHOD

Consider a thermo elastic body with force and heat applied to its exterior boundary. The finite-element method serves to determine the displacement vector u(X,t) and the

temperature T( X,t ) as functions of the un deformed position X and time t . The process of creating a finite-element model to support the design of a mechanical system can be viewed

as having (at least) eight steps:

1. The body is first discretized, i.e., it is modeled as a mesh of finite elements connected

at nodes.

2. Within each element, interpolation models are introduced to provide approximate

expressions for the unknowns, typically u(X,t) and T(X,t), in terms of their nodal

values, which now become the unknowns in the finite-element model.

3. The strain-displacement relation and its thermal analog are applied to the

approximations for u and T to furnish approximations for the (Lagrangian) strain and

the thermal gradient.

4. The stress-strain relation and its thermal analog (Fourier’s Law) are applied to obtain approximations to stress S and heat flux q in terms of the nodal values of u and T.

5. Equilibrium principles in variational form are applied using the various approximations within each element, leading to element equilibrium equations.

6. The element equilibrium equations are assembled to provide a global equilibrium

equation



7. Prescribed kinematic and temperature conditions on the boundaries ( constraints ) are

applied to the global equilibrium equations, thereby reducing the number of degrees

of freedom and eliminating “rigid-body” modes.

8. The resulting global equilibrium equations are then solved using computer algorithms.

The output is post processed. Initially, the output should be compared to data or benchmarks,

or otherwise validated, to establish that the model correctly represents the underlying mechanical system. If not satisfied, the analyst can revise the finite-element model and repeat

the computations. When the model is validated, post processing, with heavy reliance on graphics, then serves to interpret the results, for example, determining whether the underlying

design is satisfactory. If problems with the design are identified, the analyst can then choose to revise the design. The revised design is modeled, and the process of validation and

interpretation is repeated.

MESH DEVELOPMENT

Finite-element simulation has classically been viewed as having three stages: preprocessing,

analysis, and post processing. The input file developed at the preprocessing stage consists of

several elements:

1. control information (type of analysis, etc.)

2. material properties (e.g., elastic modulus)

3. mesh (element types, nodal coordinates, connectivities)

4. applied force and heat flux data

5. supports and constraints (e.g., prescribed displacements)

6. initial conditions (dynamic problems)

In problems without severe stress concentrations, much of the mesh data can be developed conveniently using automatic-mesh generation. With the input file developed, the analysis

processor is activated and “raw” output files are generated. The postprocessor module typically contains (interfaces to) graphical utilities, thus facilitating display of output in the

form chosen by the analyst, for example, contours of the Von Mises stress. Two problems arise at this stage: Validation and interpretation.

The analyst can use benchmark solutions, special cases, or experimental data to validate the

analysis. With validation, the analyst gains confidence in, for example, the mesh. He or she

still may face problems of interpretation, particularly if the output is voluminous. Fortunately,

current graphical-display systems make interpretation easier and more reliable, such as by

displaying high stress regions in vivid colors. Postprocessors often allow the analyst to “zoom

in” on regions of high interest, for example, where rubber is highly confined. More recent

methods based on virtual-reality technology enable the analyst to fly through and otherwise

become immersed in the model.

The goal of mesh design is to select the number and location of finite-element nodes and

element types so that the associated analyses are sufficiently accurate.

Several methods include automatic-mesh generation with adaptive capabilities, which serve

to produce and iteratively refine the mesh based on a user-selected error tolerance. Even so,

satisfactory meshes are not necessarily obtained, so that model editing by the analyst may be necessary. Several practical rules are as follows:

1. Nodes should be located where concentrated loads and heat fluxes are applied.



2. Nodes should be located where displacements and temperatures are constrained or

prescribed in a concentrated manner, for example, where “pins” prevent movement.

3. Nodes should be located where concentrated springs and masses and their thermal

analogs are present.

4. Nodes should be located along lines and surface patches, over which pressures, shear

stresses, compliant foundations, distributed heat fluxes, and surface convection are

applied. 5. Nodes should be located at boundary points where the applied tractions and heat

fluxes experience discontinuities. 6. Nodes should be located along lines of symmetry.

7. Nodes should be located along interfaces between different materials or components. 8. Element-aspect ratios (ratio of largest to smallest element dimensions) should be no

greater than, for example, five. 9. Symmetric configurations should have symmetric meshes.

10. The density of elements should be greater in domains with higher gradients.

11. Interior angles in elements should not be excessively acute or obtuse, for example,

less than 45°or greater than 135°. 12. Element-density variations should be gradual rather than abrupt.

13. Meshes should be uniform in subdomains with low gradients.

14. Element orientations should be staggered to prevent “bias.”

In modeling a configuration, a good practice is initially to develop the mesh locally in

domains expected to have high gradients, and thereafter to develop the mesh in the

intervening low-gradient domains, thereby “reconciling” the high-gradient domains.

There are two classes of errors in finite-element analysis:

Modeling error ensues from inaccuracies in such input data as the material properties,

boundary conditions, and initial values. In addition, there often are compromises in the mesh,

for example, modeling sharp corners as rounded.

Numerical error is primarily due to truncation and round-off. As a practical matter, error in a

finite-element simulation is often assessed by comparing solutions from two meshes, the second of which is a refinement of the first.

The sensitivity of finite-element computations to error is to some extent controllable. If the

condition number of the stiffness matrix (the ratio of the maximum to the minimum eigen value) is modest, sensitivity is reduced. Typically, the condition number increases rapidly as

the number of nodes in a system grows. In addition, highly irregular meshes tend to produce

high-condition numbers. Models mixing soft components, for example, rubber, with stiff

components, such as steel plates, are also likely to have high-condition numbers. Where

possible, the model should be designed to reduce the condition number.



Types of supports

Degrees of Freedom to be

Restricted Sl.

No Type of support

Tx Ty Tz Rx Ry Rz

1 Fixed

� � � � � �

2 Roller

x-y plane � � � �

3 Hinged

Or pinned

x-y plane � � � � �

4 Simply support

x-y plane � � � �





TRUSS

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� and the Young’s modulus is 200x109

N/m2.

Step1: Start Algor – Start- Program files-Algor22- Fempro

Step2: Select the "File: New" command. The "New" dialog will appear. Select the

"FEA Model" icon and press the "New" button.

Step3: Select Linear – Statics Stress With Linear Material model

Step4: Select the New Button on the lower right corner



Step5: Create a new Analysis file truss1 and save-this will open us the Working environment

Step6: Go for Plane 1 <XY-Top> , right click it and select Sketch. Now the Drawing

Environment opens.

Step7: Go for Geometry in the menu bar for the creation of line element.



Step 8: uncheck the USE AS CONSTRUCTION box..

Step9: Start the line by pressing enter i.e start from 0,0. Then

Y=2.8– Enter

Then X=1.5 and Y=2.0- Enter

Then X=0 , y=0

Step 10: Go for Plane 1 <XY-Top> , right click it and select Sketch again to come out of the

Drawing Environment.



Step 11: A New Part 1 Will be generated. In Part1 kindly select Element type and right click

to define it . Select Truss from the list.

Step 12: Select the Element Definition, right click for Modify Element Definition, A new

dialog box opens give cross-sectional area as 0.01m2.

Step13: Select Material, right click to modify material and select AISI !005 Steel from the

List to assign it to the truss



Step14: Defining boundary condition:

Select the Vertices from Selection-Select from menu bar

Move the cursor to near the A node and select it. The Node will highlight, then right click in

the screen and select Nodal Boundary condition and select fixed.



Select Node C and right click for applying Force

.

Select Y Direction and give -2800 as nodal force – ve sign for force acting downwards



Move the cursor to near the C node and select it. The Node will highlight, then right click in

the screen and select Nodal Boundary condition and select all expect TY.

Step15: Go for Analysis in the menu bar and select Parameters. Select all in the output tab.

Step 15: Go for Analysis in the menu bar and select Perform Analysis.



Step16: After analysis completes, Go for Results Element Force And moments –Axial

Forces. The Model will show the results.

Select Nodes from selection-select menu bar

Select all nodes and right click to add probe to selection:

FORCE ACTING ON

AB=1700 AC=2000

BC=-2500(Compression)

Step17: After analysis completes, Go for Reaction Vector –Reaction Forces-Y. The Model

will show the results.



Reaction Force

FY: 2800

Repeat for X Direction

FX: 1500

-1500

Step 18: Select Automatic result generation to complete the tutorial



Exercises on Truss 1) Determine the reaction force, displacement and elemental stress for the truss shown

below.

Given: Material : Mild steel

E = 209X103N/mm

2

A= 100mm2

2) Determine the reaction force displacement and elemental stress for the truss shown below.

Given:

Material : Mild steel

E = 209X103N/mm

2

Section A –A : 10X10 Sq

3) Determine the reaction force displacement and elemental stress for the truss shown below

Given: E = 209X103N/mm2

A = 0.01m2

100



BARS Determine the Displacement in the direction of force applied for a bar of constant cross

Section area.








Step5: Create a new Analysis file Bar 1 and save-this will open us the Working environment.

Step6: To create the bar we need to draw line in YZ plane, to go to drawing environment

right Click the YZ plane and select sketch.

Step 7: To create Line go for geometry – add line

Step 8: Remove the construction only and press enter to start line and enter the value 1m in Y direction to complete the Line.

Step 9: To come out of the drawing double Click the YZ plane.

Step 10: Assign Beam Element for the Line by right Clicking the Element type.

Step 11: To assign the Cross section, right click the Element definition, select the colum

value –the Cross-Section Libraries iron will appear. Press this icon to get the library. Select round from the List and assign value 0.1m. Click ok to accept.



Step12: Select Material, right click to modify material and select AISI 1005 Steel from the

List to assign it to the truss





Click on the left side end vertices and right Click to add the constraints. Select Fixed.

Click on the Right side end vertices and Right click to add force

Assign the value 1000N.



With this step we conclude the boundary condition assigning.





Step 16: Now the result window opens. Select the Results- displacement -magnitude to know

the displacement

Select Automatic result generation to complete the tutorial



Exercises on Bars

Case 1: Bars with uniform cross section 1) A circular rod of dia 20 mm and 500 mm long is subjected to a tensile force of 45 KN.

The modulus of elasticity for steel may be taken as 200 KN/mm2. Find stress, strain and

elongation of the bar due to applied load.

( Ans: A=314.159 mm2 , stress = 143.24 mm

2, strain = 0.0007162 , Elongation =

0.358 mm)

2) A bar of 800 mm length is attached rigidly at A and B as shown in figure. Force of 30KN

and 60 KN act as shown on the bar. If E= 200GPa, determine the reactions at the two

ends. If the bar dia is 25 mm, Find the stresses and change in length of each portion.

A C D B

275 150 375

(Ans: RA = 8.4375KN, RB = 21.5625KN)

Case 2: Bars with cross section varying in steps 3) For the bar shown below, determine total extension, Max and Min stress developed in the

bar, Max and Min strain in the bar and reaction force. E= 198714.72 N/mm2

P P

160 240 160

(Ans: Deformation = 0.285 mm)

4) The composite bar shown in fig is subjected to a tensile force of 30kN. Young’s modulus

of brass and steel are 99777.6N/mm2 and 2x 10

5 N/mm

2 respectively. Find the extension

of the bar. (Ans: 0.186mm)

P = 30KN

400 300

D1=25mm D2 = 20mm D3 = 25mm

60KN

30KN

D1=30mm, Steel D2= 20mm, Brass



Case 3: Compound bars

5) A compound bar of length 500 mm consists of a strip of Aluminum 50 mm wide X 20

mm thick and a strip of steel 50 mm wide X 15 mm thick rigidly joined at ends. If the bar

is subjected to a load of 50KN, find the stress developed in each materials and extension

of the bar. Take EAl = 1x105N/mm2 and ESteel = 2x105N/mm2

( Ans: Deformation = 0.1 mm)

50KN

500mm

6) A compound bar consist of a circular rod of steel of dia 20 mm rigidly fitted into a copper

tube of internal dia 20 mm and thickness 5 mm. If the bar is subjected to a load of 100

KN, find the stresses developed in two materials. Take Esteel 2X105 N/mm

2 and Ecopper =

1.2 X105 N/mm2. Length of both the bars is 100 mm.

Case 4: Bars with taper cross section

7) A 1.5 meter long steel bar is having uniform dia of 40 mm for a length of 1 m and in the next 0.5 m its dia gradually reduces from 40 mm to 20 mm. Determine the elongation of

this bar when subjected to an axial tensile load of 160 KN. Given: E = 200GPa.

8) Find the extension of the bar shown in figure under an axial load of 20KN. Take E= 2GPa.

(Ans: extension=0.444mm)

Aluminum

Steel



Beams 1) Determine the maximum bending stress and strain developed in the beam and Maximum

deflection of the beam due to applied load. Also plot SFD and BMD for the cantilever

beam shown below. Material used is Steel AISI 4130.






Step5: Create a new Analysis file : Cantilever BEAM and save-this will open us the Working environment.



Step6: To create the bar we need to draw line in XY plane, to go to drawing environment

right Click the XY plane and select sketch.

Step 7: To create Line go for geometry – add line

Step 8: Remove the construction only and press enter to start line and enter the value 1m in X

direction to complete the Line, then enter 2 m in X direction , again enter 3.5 m in X

direction, press ESC twice to exit from sketch. Or

Step 9: To come out of the drawing double Click the XY plane.


Step 11: To assign the Cross section, right click the Element definition, and select the

column value –the Cross-Section Libraries icon will appear.

Press this icon to get the library. Select Rectangular section and enter the values as below.

b=.01m h=.008m



Press Ok. Again OK.

Step12: Select Material, right click to modify material and select AISI 4130 Steel from the

List to assign it to the beam.






Click on the 2nd and 3rd vertices and Right click to add force. Assign the value -20KN in Y

direction. Similarly select 4th

vertices assign -10KN load in Y direction.

With this step we conclude the boundary condition assigning.



Step 16: Now the result window opens. Select the Results- displacement -magnitude to know

the displacement



ANSWERS Displacement: 0.326395m

Bending stress in local 3 direction=-2.3841E-6N/m2



Max strain about axis 3= -6.938E-18m/m

Min Value” -0.00430442m/m

Reaction forces: RA=50000N

Procedure To Draw SFD and BMD:

i) For SFD 1) Keep the displacement plot in the results area.

2) Results options> Deselect Show displaced model

3) Right click >select vector plot.

4) With Rectangular select on select elements- select complete line.( Selection>shape>

rectangle and selection> select> Elements)



5) With elements selected go to Inquire >Add shear diagram( axis 2) to get SFD.

6) To get the magnitude of Shear force at required points…..

Results>Element forces and Moments>Local 2 force

7) With rectangular selection ON and Vertices selection ON, Select all vertices >right

click> Add probes to selection.

8) View >display >features.

9) Inquire > clear beam diagram

For BMD 1) Keep the displacement plot in the results area.

2) Results options> Deselect Show displaced model

3) Right click >select vector plot. 4) With Rectangular select ON, select elements- select complete line-

(Selection>shape> rectangle and selection> select> Elements) 5) With elements selected go to Inquire >Add Moment diagram (axis 3) to get BMD.

6) To get the magnitude of bending moment at required points….. Results>Element forces and Moments>Local 3 moment

7) With rectangular selection ON and Vertices selection ON, Select all vertices >right

click> Add probes to selection. 8) View >display >features.

9) Inquire > clear beam diagram



2) Determine the maximum stress and strain developed in the beam and Maximum

deflection of the beam due to applied load. Also plot SFD and BMD for the beam shown

below. Material used is Steel AISI 4130.


deflection of the beam due to applied load. Also plot SFD and BMD for the beam shown

below. Material used is Steel AISI 4130.


deflection of the beam due to applied load. Also plot SFD and BMD for the beam

shown below. Material used is Steel AISI 4130.

5) Plot SFD and BMD for the beam shown below.

6) Plot SFD and BMD for the beam shown below.







1) Stress Analysis of a Rectangular Plate with a Circular Hole

Determine the maximum stress for a rectangular plate of 50mm x 80mm with hole of 10mm

diameter in the center is loaded in axial tension. Thickness of the plate is 10mm. Take E =

200GPa

Theoretical calculation:

Results Comparison

FEM Theoretical

Deformation 0.009284 mm -

Stress 59.99 N/ mm2 62.5 N / mm

2

Procedure:

Steps to create geometry:

1) Select XY plane > Sketch

2) Select>Create rectangle> Press enter to define one corner of the rectangle( 0, 0,0)

3) Type X and Y coordinate of opposite corner of the rectangle ( 80, 50,0)

4) To crate hole: Select>circle by center and radius> Type center point of the circle (40,

25, 0)> press Enter. To define radius: check use relative> Enter DX= 5 mm( Radius of circle)

5) Exit from the sketch.

Steps to solve the problem:

1) To generate 2-D mesh: Right click >1< XY top> > Create 2D mesh> Enter 500 in the mesh density tab > Apply,



2) Define Element type (Plate), Element definition >Check design variable> Enter thickness

(10 mm) and Material type (200GPa).

3) To Apply BC and Load: select all nodes in left side > Apply fixed BC > Select all nodes in

right side> Apply Nodal force as shown below. (1000N/ No of nodes)> OK.

4) Perform Analysis. 5) Find out displacement, Maximum stress etc…



Exercises: Determine the maximum stress for a rectangular plate of 50mm x 80mm with hole of 10mm

diameter in the center is loaded in axial tension. Thickness of the plate is 10mm. Take E = 210GPa. Take Axial Load P = 100KN. Validate your results with theoretical results.

P P



2) Thermal Analysis -2D problem with conduction and convection

boundary conditions

Determine Temperature Distribution in a Plate. Material Alumina, 99.9%, AL203. Assume

thickness of the plate 10mm. Plot temperature distribution curve.

Solution:

Step1: Start Algor – Start- Program files-Algor23.1- Fempro



Step3: Select –Thermal- Steady State Heat Transfer


Step5: Create a new Analysis file THERMAL 1 and save-this will open us the Working

environment

Step6: Go for Plane 1 <YZ-Right>, right click it and select Sketch. Now the Drawing

Environment opens.

Step7: Go for Geometry in the menu bar for the creation of rectangle element. Uncheck the

USE AS CONSTRUCTION BOX. Create a rectangle by pressing enter for the starting corner

and enter the values Z=20 and Y=10 and press enter to complete the rectangle.

Step8: Double click the YZ plane to come out of the drawing environment. Step9: Right Click the element type and Select Plate.

Step 10: Right Click Element definition and assign 0.1m for the thickness of the plate. Step11: Right Click Material and assign Alumina material from the material List.

Step 12: Right Click the YZ plane and select create 2D Mesh option and create Mesh; you can modify the mesh also.



Step 13: Select ---Selection-Shape-Rectangle.

Step 14: Select ---Selection-Select -Vertices

Step 15: Window the left side edge nodes and right Click to add Nodal Applied temperature.

Give value 100. Repeat the procedure for all other sides.

Steps 16: Select –Analysis- perform Analysis

Step 17 : Your window will change to Result window to show the results

Step 18: Select Automatic result generation to complete the tutorial



Exercises: 1) A furnace wall is made of inside silica brick (k = 1.5 W/mK) and outside magnesia brick

(k = 4.9 W/mK), each 10 cm thick. The inner and outer surfaces are exposed to fluids at temperatures of 820°C and 110°C respectively. The contact resistance is 0.001 m2K/W.

The heat transfer coefficient for inner and outside surfaces is equal to 35 W/m2K. Find the

heat flow through the wall per unit area per unit time and temperature distribution across

the wall.

Theoretical calculation:

2) A rod of 6 cm dia with k= 98 W/mK and 125 cm long is attached to an evaporation

chamber maintained at -15°C. The film coefficient of heat transfer is 40 W/m2K and the

ambient temp is 28°C. Compute and plot the temperature distribution along the length of

the fin and Find the length up to which there will be ice formation.



Dynamic analysis

1) Modal Analysis of Cantilever beam Determine the natural frequency and different modes of vibration (Simple Modal Analysis) of

a cantilever beam as shown below. Given:

E=2e11 N/m2,

I = 8.33e-06 m4

Area A = 0.01 m2 , Density = 7830 Kg/m3

Manual calculation:

Mode FEM (Frequency) Theory (Frequency)

1 81.1569 81.70

2 491.593 510.63

3 1263.48 1432.10

For hand calculation:




Step3: Select Linear –Natural frequency (Modal)

Step4: Select the New Button on the lower right corner Step5: Create a new Analysis file : Cantilever BEAM and save-this will open us the Working

environment. Step6: To create the bar we need to draw line in XY plane, to go to drawing environment

right Click the XY plane and select sketch. Step 7: To create Line go for geometry – add line

Step 8: Remove the construction only and press enter to start line and enter the value 1m in X direction to complete the Line, press ESC twice to exit from sketch.

Or

Step 9: To come out of the drawing double Click the XY plane.


Step 11: To assign the Cross section, right click the Element definition, and select the column

value –the Cross-Section Libraries icon will appear. Select Rectangular section and enter the

values as below.

b=0.1m h=0.1m

Press Ok. Again OK.

Step12: Select Material, right click to modify material, create custom material with

following properties.



E=2e11 N/m2

Density = 7830 Kg/m3





Step 16: Now the result window opens. To get the modes of vibration click the icon as below.

To animate the results: use the icons as shown below.

Results:

Mode FEM (Frequency)

1(2) 81.1569

3(4) 491.593

5 1263.48

Exercises:

Determine the natural frequency and different modes of vibration (Simple Modal Analysis) of

truss shown below.

Given:

E=2e11 N/m2,

Area A = 0.01 m2 , Density = 7830 Kg/m3

Results: Mode Frequency

1 57.1109

2 107.986

3 141.601

4 196.702

5 231.16