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ME 2404 COMPUTER AIDED SIMULATIONAND ANALYSIS LABORATORY

LAB MANUAL

CONTENTS

SL. NO

NAME OF THE EXPERIMENTPAGE NO.

INTRODUCTION

ANALYSIS (SIMPLE TREATMENT ONLY)

1Stress analysis of a plate with a circular hole.

2Stress analysis of rectangular L bracket

3Stress analysis of an axi-symmetric component

4

a) Stress Analysis On Cantilever Beam Subjected To Point Load

b) Stress analysis of simply supported beam.

c) Stress analysis of fixed beam.

5 Model frequency analysis of 2D component

6

a) Mode Frequency Analysis Of Cantilever Beam

b) Mode frequency analysis of simply supported beam

c) Model frequency analysis of fixed beam

7Harmonic analysis of a 2D component

8 Thermal stress analysis of a 2D component

9Conductive heat transfer analysis of a 2D component

10Convective heat transfer analysis of a 2D component

11 Simulation of Air conditioning system with condenser temperature and evaporator temperatures as input to get COP using MAT Lab

12Simulation of hydraulic system using MAT Lab.

13Simulation of cam and follower mechanism using MAT Lab.

INTRODUCTION to FEA and ANSYS

What is FEA?

Finite Element analysis is a way to simulate loading conditions on a design and determine the designs response to those conditions.

The design is modeled using discrete building blocks called elements. Each element has exact equations that describe how it responds to a

certain load. The “Sum” of the response of all elements in the model gives the

total response of the design. The elements have a finite number of unknowns, hence the name

finite elements. The finite element model, which has a finite number of unknowns,

can only approximate the response of the physical system which has infinite unknowns.

How good is the approximation?

Unfortunately, there is no easy answer to this question, it depends entirely on what you are simulating and the tools you use for the simulation.

Why is FEA needed?

To reduce the amount of prototype testing. Computer Simulation allows multiple “what if“scenarios to be tested

quickly and effectively. To simulate designs those are not suitable for prototype testing. E.g.

Surgical Implants such as an artificial knee.

About ANSYS:

ANSYS14.5 is a complete FEA software package used by engineers worldwide in virtually all fields of engineering. ANSYS is a virtual Prototyping technique used to iterate various scenarios to optimize the product.

General Procedure of Finite Element Analysis:

1. Creation of geometry or continuum using preprocessor.2. Discretization of geometry or continuum using preprocessor.3. Checking for convergence of elements and nodes using preprocessor.4. Applying loads and boundary conditions using preprocessor.5. Solving or analyzing using solver6. Viewing of Results using postprocessor.

Build Geometry:

Construct a two (or) three dimensional representation of the object to be modeled and tested using the work plane co-ordinate system in Ansys.

Define Material Properties:

Define the necessary material from the library that composes the object model which includes thermal and mechanical properties.

Generate Mesh:

Now define how the model system should be broken down into finite pieces.

Apply Loads:

The last task in preprocessing is to restrict the system by constraining the displacement and physical loading.

Obtain Solution:

The solution is obtained using solver available in ANSYS. The computer can understand easily if the problem is solved in matrices.

Present the Result:

After the solution has been obtained there are many ways to present Ansys result either in graph or in plot.

Specific Capabilities of ANSYS Structural Analysis:

Structural analysis is probably the most the common application of the finite element method such as piston, machine parts and tools.

Static Analysis:

It is the used to determine displacement, stress etc. under static loading conditions. Ansys can compute linear and non-linear types (e.g. the large strain hyper elasticity and creep problems).

Transient Dynamic Analysis:

It is used to determine the response of a structure to time varying loads.

Buckling Analysis:

It is used to calculate buckling load and to determine the shape of the component after applying the buckling load. Both linear buckling and non – linear buckling analysis are possible.

Thermal Analysis:

The steady state analysis of any solid under thermal boundary conditions calculates the effect of steady thermal load on a system (or) component that includes the following.a) Convection.b) Radiation.c) Heat flow rates.d) Heat fluxes.e) Heat generation rates.f) Constant temperature boundaries.

Fluid Flow: The ANSYS CFD offers comprehensive tools for analysis of two-

dimensional and three dimensional fluid flow fields.

Magnetic: Magnetic analysis is done using Ansys / Electromagnetic program. It

can calculate the magnetic field in device such as power generators,

electric motor etc. Interest in magnetic analysis is finding magnetic flux, magnetic density, power loss and magnetic forces.

Acoustic / Vibrations:

Ansys is the capable of modeling and analyzing vibration system. Acoustic is the study of the generation, absorption and reflection of pressure waves in a fluid application.

Few examples of acoustic applications are a) Design of concert house, where an even distribution of sound

pressure is possible.b) Noise cancellation in automobile. c) Underground water acoustics.d) Noise minimization in machine shop.e) Geophysical exploration.

Coupled Fields:

A coupled field analysis is an analysis that takes into account the interation between two (or) more fields of engineering analysis. Pressure vessels, Induction heating and Micro electro mechanical systems are few examples.

Result: Thus the basics of FEA and ANSYS are studied.

Ex.No:1 STRESS ANALYSIS OF A PLATE WITH A CIRCULAR HOLEAim

To perform stress analysis of a plate with a circular hole.

Procedure1. Utility Menu > File>Change Job Name > Enter Job Name.

Utility Menu > File > Change Title > Enter New Title.

2. Preference > Structural > OK.

3. Preprocessor > Element Type > Add/Edit/Delete > Solid > Quad 4Node 182 > ok > Select > Options > Plane Stress/Thickness.> OK> Close

4. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 2E5, PRXY = 0.3.> Materials > Exit

5. Preprocessor > Real Constant > Add/Edit/Delete > plane 182> OK > thickness = 20> OK > Close

6. Preprocessor > Modeling > Create > Areas > Rectangular by Dimension > X1, X2 = 0, 200 Y1, Y2 = 0, 100 > OK

7. Preprocessor > Modeling > Create > Areas > Circle > Solid circle > WP X = 100, WP Y = 50 & Radius = 20 > OK.

8. Preprocessor > Modeling > Operate > Boolean > Subtract > Areas > Select rectangle >Apply > Select Circle > Ok.

9. Preprocessor > Meshing > Size control> manual size>Element Edge Length = 2 > OK 10.Meshing >Mesh > Areas > Free.> Pick all

11. Solution > Analysis Type > New Analysis > Static > OK.

12. Solution > Define Load > Apply > Structural > Displacement > On lines > Select Bottom Line > UY > Displacement value = 0 > OK.

13. Solution > Define Load > Apply >Structural>Displacement >On Lines > Select left line > Ok > UX > Displacement value = 0 > OK. 14. Solution > Define load> Apply> Pressure > On Line > Select Right line > Ok > Value [100] > Ok.

15. Solution > Solve > Current LS > Ok.

16. Utility menu > Plot control > Style > Symmetry Expansion > Periodic > Reflect about XY.> Ok

17. Plot control > Animate > deformed shape > Ok.18. General Postprocessor > Plot Result > Deformed shape >Def+undeformed> Ok.

19. General post processor>Plot Result > Contour Plot > Nodal Solution > Stress > Von Mises > Ok.20. General post processor> Plot result > Nodal solution> DOF solution> Displacement vector sum> OK

Result:Thus the stress analysis of a plate with a circular hole is performed.

Ex: 1

EX.NO:2 STRESS ANALYSIS OF RECTANGULAR L - BRACKET

Aim To perform stress analysis of rectangular L- bracket and to determine the

maximum stress and maximum deflection.

Procedure1. Utility Menu > File> Change Job Name > Enter Job Name.



3. Preprocessor > Element Type > Add/Edit/Delete > Solid> Quad 8Node 182> OK> Select > Options > Plane Stress/Thickness.>OK>close

4. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 2E5, PRXY = 0.3.

5. Preprocessor > Real Constant > Add/Edit/Delete > plane 182> thickness = 0.5 >OK > close

6. Preprocessor > modeling > create > Area > rectangle > By dimension>X1=0 ,X2=6, Y1= -1, Y2=1ApplyX1=4, X2=6, Y1= -1, Y2= -3 > OK > close

7. Preprocessor > modeling > create > Areas > circle > solid circle > WP X = 0WP Y = 0Radius = 1ApplyWP X = 5WP Y = -3 Radius = 1 Ok.

8. Preprocessor > modeling > operate > Booleans > Add > Areas > pick all9. Preprocessor > modeling > create > Area > circle > solid circle > WP X = 0WP Y = 0

Radius = 0.4ApplyWP X = 5WP Y = -3 Radius = 0.4 Ok.10. Preprocessor > modeling > operate > Booleans > Subtract > Areas > Select the rectangle > Apply > select two circles > Ok.

11. Preprocessor > modeling > create > line > line fillet > select 2 lines(horizontal, vert)>apply > fillet radius = 0.4 > Ok.

12. Preprocessor > modeling > create > Areas > Arbitrary > By lines > select the three lines(Horizontal,vertical,fillet radius)> OK

13.Preprocessor>modelling>Operate>Boolean>add > area>pick all14. Preprocessor > meshing > size control > manual size>Areas > all area>Element edge length = 0.1 > Ok

15. Preprocessor > meshing> mesh> Area > Free> pick all

16.Utility menu>Plot control>style>Size and shape>Display of element >On > OK

17. Solution > define loads > apply > Structural > displacement > online > select the left hole (Inside) > apply > All DOF > Displacement value = 0 > ok.

18. Solution > define loads> apply >structural > pressure > on line > pick all (Select bottom left of the circle> apply.Load value = 50 > OK

19. Solution > define loads> apply >structural > pressure > on line > pick all (Select bottom right of the circle> apply.Load value = 50 > apply

20. Solution > solve current LS > ok

21. General post processor > plot result > Deformed shaped > Deformed + Undeformed > Ok.

22. General post processor > plot result > contour plot > nodal solution > stress > von mises > Ok.

19. General post processor >List result > reaction solution > all items > Ok> close

Result:Thus the stress analysis of rectangular L-bracket with circular hole is

obtained.(i) Maximum stress = ………. N/mm2 (ii) Maximum deflection =

…….mm.

Ex 2 L BRACKET

EX.NO:3 STRESS ANALYSIS OF AN AXI – SYMMETRIC COMPONENT

AimTo obtain the stress distribution of an axisymmetric component. The model

will be that of a closed tube made from steel. Point loads will be applied at the centre of the top and bottom plate.

Procedure1. Utility Menu > File>Change Job Name > Enter Job Name.



3. Preprocessor > Element type > Add/Edit/ delete > solid > 8node 183 > OK>options>element shape > Triangular>,Element behavior >axisymmetric>OK>close.

4. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 2E5, PRXY = 0.3.>Material > Exit

5. Preprocessor>Modeling>create>Areas>Rectangle> By dimensions

Rectangle X1 X2 Y1 Y2 1 0 20 0 5 2 15 20 0 100 3 0 20 95 100 > OK

6. Preprocessor > Modeling > operate > Booleans > Add > Areas > pick all 7. Preprocessor > meshing > size control > manual size >Areas > All Area>Element edge length = 2 mm > Ok

8.Meshing > mesh > Areas > free> pick all.

9. Solution > Analysis Type>New Analysis>Static 10. Solution > Define loads > Apply .Structural > displacement > symmetry B.C > on lines. (Pick the two edger on the left at X = 0)11. Utility menu > select > Entities > by location > Y = 50 >ok.

(Select nodes and by location in the scroll down menus. Click Y coordinates and type 50 in to the input box.)12.Solution>define load>apply> structural> Displacement>on nodes>pick all>UY>50>OK13. Utility menu > select > Entities > select all14. Solution > Define loads > Apply > Structural > Force/Moment > on key points > Pick the top left corner of the area > apply> UY>100 >Ok.

15. Solution > Define Loads > apply > Structural > Force/moment > on key points > Pick the bottom left corner of the area > Apply >UY> -100> OK

16. Solution > Solve > Current LS

17. Utility Menu > select > Entities > Select nodes > by location > Y coordinates and type 45, 55 in the min., max. box, and click ok.

18. General postprocessor > List results > Nodal solution > stress > Y components .

19. Utility menu > plot controls > style > Symmetry expansion > 2D Axisymmetric > ¾ expansion.

Result:Thus the stress distribution of the axisymmetric component is studied.

EX 3 AXISYMMETRIC

Ex. No: 4 (a) STRESS ANALYSIS ON CANTILEVER BEAM SUBJECTED TO POINT LOAD

Aim:

To obtain stress analysis of cantilever beam subjected to point load and to determine max. stress and max. deflection.

Procedure:

1. Utility Menu > Change Job Name > Enter Job Name. Utility Menu > File > Change Title > Enter New Title.


3. Preprocessor > Element type > Add/Edit/ delete > beam > 2node 188 > close.

4. Preprocessor > Section> Beam >common section> B=10 > H=10> Ok


6. Preprocessor > Modeling > create > nodes > Inactive CSNode 1X=0Y=0

Node 2X= 20Y=0

Node 3X= 40Y=0

Node 4X= 60Y=0

Node 5X= 80Y=0

Node 6X= 100Y=07.List > nodes > coordinate only > ok

8.Preprocessor > modeling > create > elements > Auto numbered thru’ nodes > select Node 1 & 2Node 2 & 3Node 3 & 4Node 4 & 5Node 5 & 6 > ok. 9.Solution > define loads > apply > structural > displacement > on nodes > select node 1 > apply > all DOF > displacement = 0 > ok.

10.Solution > Force/moment > on nodes > node 6 > apply > FY > -100 > ok.

11.Solution > solve > current L.S > ok.

12.General post processor > plot result > deform shape > Deformed + Undeformed > ok.

13.General post processor > element table > define table > add > user table for item

by sequence num > SMIS 1 > apply



by sequence num > SMIS 4 > Ok.

14.General post processor >Plot result > line element result > SMIS1 > SMIS3 > first result >Evaluate table data > Ok.

15. General postprocessor > list result > nodal solution > DOF solution > UY > displacement result

16.General postprocessor >Plot results> contour plot > nodal solution> displacement vector sum > Ok.

Result:Thus the stress analysis on cantilever beam subjected point load is performed.

EX.4 (a) CANTI LEVER BEAM

Ex. No: 4(b) STRESS ANALYSIS OF SIMPLY SUPPORTED BEAM.

Aim:To perform Stress analysis of simply supported beam.

Procedure:

1. Utility Menu > File >Change Job Name > Enter Job Name. Utility Menu > File > Change Title > Enter New Title.


3. Preprocessor > Element type > Add/Edit/ delete > beam > 2node 188> close.

4. Preprocessor > Section > Beam > common section>B=10> H=10 > OK



Node 2X= 25Y=0

Node 3X= 50Y=0

Node 4X= 75Y=0

Node 5X= 100Y=07.List > nodes > coordinate only > ok

8.Preprocessor > modeling > create > elements > Auto numbered thru’ nodes > select Node 1 & 2Node 2 & 3Node 3 & 4Node 4 & 5Node 5 & 6 > ok. 9.Solution > define loads > apply > structural > displacement > on nodes > select node 1 > apply> All DOF> Select node 1& 5 > apply > UY > displacement = 0 > ok.

10. Solution > Force/moment > on nodes > node 3 > apply > FY > -100 > ok.




by sequence num > SMIS 1 > Apply

by sequence num > SMIS 3 > Apply



14.Plot result > line element result > SMIS1 > SMIS 2 > first result >Evaluate table data > Ok.

15.General postprocessor > Plot results>contour plot > nodal solution > DOF solution >Ok.

Result:Thus the stress analysis of simply supported beam is obtained.

EX 4 ( b ) SIMPLY SUPPORTED BEAM

Ex. No: 4(c) STRESS ANALYSIS OF FIXED BEAM.

Aim:To perform stress analysis of fixed beam subjected to point load.

Procedure:1. Utility Menu >File> Change Job Name > Enter Job Name.




4. Preprocessor > Section > Beam >common section > B=10> H=10 >Ok



Node 2X= 25Y=0

Node 3X= 50Y=0

Node 4X= 75Y=0

Node 5X= 100Y=0

7.List > nodes > coordinate only > ok

8. Preprocessor > modeling > create > elements > Auto numbered thru’ nodes > select

Node 1 & 2Node 2 & 3Node 3 & 4Node 4 & 5Node 5 & 6 > ok. 9.Solution > define loads > apply > structural > displacement > on nodes > select node 1 & node 5 > apply > all DOF > displacement = 0 > ok.10.Solution > Force/moment > on nodes > node 3 > apply > FY > -100 > ok.








14.Plot result > line element result > SMIS 1 > SMIS3 > first result > Ok.

15.General postprocessor > Plot results>contour plot > DOF Solution. > Ok.

Result:Thus the stress analysis of fixed beam is obtained.

EX 4 ( C) FIXED BEAM

Ex. No: 5 MODE FREQUENCY ANALYSIS OF 2D PLATE

Aim:To perform the model frequency analysis on 2D plate.

Procedure:1. Utility Menu > Change Job Name > Enter Job Name.

Utility Menu > File > Change Title > Enter New Title.2. Preference > Structural > OK. 3. Preprocessor > Element type > Add/Edit/ delete > Solid 8node 183 > options > plane stress with thickness > close.

4. Preprocessor > Real Constant > Add/Edit/Delete > thickness = 1 > Ok

5. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 2.068 E5, PRXY = 0.3 & Density = 7.83E-6.

6. Preprocessor>Modeling>create>Areas>Rectangle> By dimensions X1 , X2 = 0, 250 Y 1 , Y2, = 0,757. Preprocessor > meshing > size control >manual contl> Areas > all area>Element edge length = 15 mm > Ok8.meshing > mesh > Areas > free> pick all.9.Solution > Analysis Type > New Analysis > modal > OK.

10.Solution > Analysis option >No. of modes to extract=5>Expand mode shape = yes > No. of mode to expand =5> OK

11. Solution > define load > apply > structural > displacement > on lines > select left side line > all DOF > Ok.

12. Solve > current L.S > close

13. General postprocessor > result summary.

14. General postprocessor > first set > plot result > deform shape > deformed + undeformed > next set > plot result > deformed + undeformed > Ok.

Result:Thus the modal frequency analysis on 2D element is performed.

EX 5 MODE FREQUENCY ANALYSIS OF 2D PLATE

Ex. No: 6(a)MODE FREQUENCY ANALYSIS OF CANTILEVER BEAM Aim:

To obtain the mode frequency analysis on Cantilever beam and to determine its natural frequency.

Procedure:1. Utility Menu > File > Change Job Name > Enter Job Name.



3. Preprocessor > Element type > Add/Edit/ delete > beam > 2 node 188> close.

4. Preprocessor > Section > Beam >common section> B=10,> H=10> Ok

5. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 2.068 E5, PRXY = 0.3 & Density = 7.83E-6. 6. Preprocessor > Modeling > create > key points > inactive CS Key point no.1 = (0, 0) Key point no.2 = (1000, 0)

7. Preprocessor > Modeling > create > lines > straight lines > select 1&2.

8. Meshing > size control>manual size>lines > all lines > Element edge length > = 100 mm >ok 9. Meshing > mesh> lines > pick all10. Solution > analysis type > new analysis > modal > ok 11.Solution> Analysis type>analysis options >mode extraction method>No.of mode to extract=5> expand mode shape=yes> No.of mode=5> OK 5

12. Solution > define loads > apply > structural > displacement > on key points > select first point > apply > all DOF > displacement = 0 > Ok.


14.General postprocessor > result summary.

15.General postprocessor > read result > first set > Ok.

16.General postprocessor > plot result > deform shape > deformed + undeformed > Ok.

17. General postprocessor > plot control > animate > modal shape.

Result:Thus the mode frequency analysis of Cantilever beam is obtained.

EX 6 (A) MODE FREQUENCY ANALYSIS OF A CANTILEVER BEAM

Ex. No: 6(b) MODE FREQUENCY ANALYSIS OF SIMPLY SUPPORTED BEAM

Aim:To perform the model frequency analysis on simply supported beam.

Procedure:1. Utility Menu >File> Change Job Name > Enter Job Name.



3. Preprocessor > Element type > Add/Edit/ delete > beam > 2Node 188 > close.

4. Preprocessor > Section>Beam> common section >B=10,>H=10 > OK


6. Preprocessor > Modeling > create > key points > inactive CS Key point no.1 = (0, 0) Key point no.2 = (1000, 0)


8. Meshing > lines > Element edge length = 100 mm > ok9.meshing > mesh >lines > pick all10.Solution > analysis type > new analysis > modal > ok >11.Solution>Analysis type> analysis options > mode extraction method> No.of mode to extract=5> expand mode shape=yes> No.mode to expand=5 OK

12. Solution > define loads > apply > structural > displacement > on key points > select first point & second point > apply > UY > displacement = 0 > Ok.





17.General postprocessor > plot control > animate > modal shape.

Result:Thus the mode frequency analysis of simply supported beam is

obtained.

EX. 6 (B) MODE FREQUENCY ANALYSIS OF SIMPLY SUPPORTED BEAM

Ex. No: 6(c) MODEL FREQUENCY ANALYSIS OF FIXED BEAM.

Aim:To perform the model frequency analysis on Fixed beam.

Procedure:

1. Utility Menu > File>Change Job Name > Enter Job Name. Utility Menu > File > Change Title > Enter New Title.



4. Preprocessor > Section > Beam >common section > B=10> H=10 > OK


6. Preprocessor > Modeling > create > key points > inactive CS Key point no.1 = (0, 0) Key point no.2 = (1000, 0)


8.Meshing > mesh > lines > Element edge length > = 100 mm > ok9.mesh >lines> pick all

10.Solution > analysis type > new analysis > modal > ok 11.Solution > Analysis type> analysis options > mode extraction method> No.of mode to extract=5> expand mode shape=yes> No.of modes to expand=5> OK12. Solution > define loads > apply > structural > displacement > on key points > select first point & second point > apply > all DOF > displacement = 0 > Ok.





17.General postprocessor > plot control > animate > modal shape.

Result:Thus the mode frequency analysis on fixed beam is performed.

EX 6 ( C ) MODE FREQUENCY ANALYSIS OF FIXED BEAM

Ex. No: 7 HARMONIC ANALYSIS ON 2D PLATE

Aim:To perform the harmonic analysis on 2D plate. We conduct a harmonic

forced response test by applying a cyclic load at the end of the plate. Procedure:1. Utility Menu > File>Change Job Name > Enter Job Name.



3. Preprocessor > Element type > Add/Edit/ delete > Solid 8node 183 > options > plane stress with thickness > close.

4. Preprocessor > Real Constant > Add/Edit/Delete > thickness = 1 > Ok


6. Preprocessor>Modeling>create>Areas>Rectangle> By dimensions0, 250 0, 75

7. Preprocessor > meshing > mesh > size control > Areas > Element edge length = 15 mm > Ok 8. Meshing> mesh > Areas > free> pick all.

9.Solution > Analysis Type > New Analysis > harmonic > OK > analysis options > real + imaginary (full solution method).

10.Solution > define loads > apply > structural > force/moment > on nodes > click right corner > FY real value = 100 & Imaginary value = 0 > Ok.

11.Solve > current L.S > ok.

12.Load step option > time frequency > frequency & sub steps > 0,200 > 200 > stepped > Ok.

13.Time history postprocessor > variable viewer > add > nodal solution > DOF solution > Y-component of displacement > click right corner > ok > graph data > Ok.14.Utility Menu > plot controls > style > graphs > modify axis ( change the Y-axis scale to logarithmic)15.Utility menu > plot > replot.Result:

Thus the harmonic analysis on 2D plate is performed.

EX 7 HARMONIC ANALYSIS ON 2D PLATE

Ex. No: 8 THERMAL STRESS ANALYSIS OF A 2D COMPONENT

Aim:To perform the thermal stress analysis of a 2D component.

Procedure:1. Preference > thermal > Ok.

2. Preprocessor > Element type > Add/edit /delete > LINK (Thermal Mass Link 3D conduction33) > close.

3. Preprocessor > real constant > add > Area = 4e-4

4. Preprocessor > material properties > Material Models > Thermal conductivity > Isotropic > KXX: 60.5

5. Preprocessor > Modeling > Create > Keypoints > In Active CS...

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

6. Preprocessor > modeling > create > lines > lines > In active coordinate system > select 1 & 2.

7. Preprocessor > Meshing > Mesh tool > Size Controls > Manual Size > element edge length = 0.1 > mesh > Areas > Free > Pick All

8. Preprocessor > Physics > Environment > Write In the window that appears, enter the TITLE Thermal and click OK.

9. Preprocessor > Physics > Environment > Clear > OK

10. Preprocessor > Element Type > Switch Elem Type (Choose Thermal to Structural from the scroll down list.)

11. Preprocessor > Material Properties > Material Models > Structural > Linear > Elastic > Isotropic > EX: 200e9, PRXY: 0.3

12. Preprocessor > Material Props > Material Models > Structural > Thermal Expansion Coefficient > Isotropic > ALPX = 12e-6

13. Preprocessor > Physics > Environment > Write > In the window that appears, enter the TITLE Struct.

14. Solution > Analysis Type > New Analysis > Static

15. Solution > Physics > Environment > Read > Choose thermal and click OK.

(If the Physics option is not available under Solution, click Unabridged Menu at the bottom of the Solution menu. This should make it visible).

16. Solution > Define Loads > Apply > Thermal > Temperature > On Keypoints > Set the temperature of Keypoint 1, the left-most point, to 348 Kelvin.


18. Main Menu > Finish

The thermal solution has now been obtained. If you plot the steady-state temperature on the link, you will see it is a uniform 348 K, as expected. This information is saved in a file labelled Jobname.rth, were .rth is the thermal results file. Since the jobname wasn't changed at the beginning of the analysis, this data can be found as file.rth. We will use these results in determining the structural effects.

19. Solution > Physics > Environment > Read

Choose struct and click OK.

20. Solution > Define Loads > Apply > Structural > Displacement > On Keypoints > Fix Keypoint 1 for all DOF's and Keypoint 2 in the UX direction.

21. Solution > Define Loads > Apply > Structural > Temperature > From Thermal Analysis

As shown below, enter the file name File.rth. This couple the results from the solution of the thermal environment to the information prescribed in the structural environment and uses it during the analysis.

22. Preprocessor > Loads > Define Loads > Settings > Reference Temp

For this set the reference temperature to 273 degrees Kelvin


24. General Postprocessor > Element Table > Define Table > Add > CompStr > By Sequence Num > LS > LS, 1.

25. General Postprocessor > Element Table > List Elem Table > COMPSTR > Ok.

1. Hand Calculations

Hand calculations were performed to verify the solution found using ANSYS:

As shown, the stress in the link should be a uniform 180 MPa in compression.

Result:

Thus the thermal stress analysis of 2D component is performed and the stress in each element ranges from -0.180e9 Pa, or 180 MPa in compression.

Ex. No: 9 CONDUCTIVE HEAT TRANSFER ANALYSIS OF 2D COMPONENT

Aim:To solve a simple conduction problem, The thermal conductivity (k) of the

material is 10 w/m° C and the block is assumed to be infinitely long.Procedure:1. Preference > thermal >h method> Ok.2. Preprocessor > Element type > Add/edit /delete > Select thermal mass solid, Quad 4 node 55 (Plane 55) > close. Options> plane thickness > ok3. Real constant>add/edit/delete> add> thk=0.5>ok> close4. Preprocessor > material properties > Material Models > Thermal conductivity > Isotropic > KXX = 10 (thermal Conductivity ) 5. Preprocessor > modeling > create > Areas > Rectangle > By 2 corners> X= 0, Y = 0, width=1, height=1> Ok.6. Preprocessor > Meshing > Mesh > Size Controls > Manual Size > area> all area> element edge length=0.05>ok

7. Meshing > mesh > Areas > free> Pick All

8. Solution > Analysis type > New analysis > steady – state > Ok.

9. Solution > Define loads > Apply > Thermal > Temperature > On lines> Click two vertical lines right and left > apply>temp 100 C > ok> all DOF > value of temp 100 > ok 9.Solution> define loads> apply > thermal> temp> on lines> click top and bottom lines> apply >temp500 > ok10.Solution > solve > Current LS.

11. General Preprocessor > Plot results > Contour Plot > Nodal Solution > DOF solution > nodal temperature (TEMP) > Ok.Result:Thus conductive heat transfer analysis is performed.

EX 9 CONDUCTIVE HEAT TRANSFER ANALYSIS

Dimension 1x 1 m

Ex. No: 10 CONVECTIVE HEAT TRANSFER ANALYSIS OF 2D COMPONENT

Aim:To perform the thermal analysis on a given block with convective heat

transfer coefficient (h) of 10 W/m° C and the thermal conductivity (k) of the material is 10 W/m° C.Procedure:1. Preference > thermal >h method> Ok.

2. Preprocessor > Element type > Add/edit /delete > Select thermal mass solid, Quad 4 node 55 (Plane 55) > close.

3. Preprocessor > material properties > Material Models > Thermal conductivity > Isotropic > KXX = 10 (thermal Conductivity )

4. Preprocessor > modeling > create > Areas > Rectangle > By 2 Corners > X = 0, Y = 0, width = 1, Height = 1 > Ok.

5. Preprocessor > Meshing > Mesh > Size Controls > Manual Size >areas> all area> element edge length = 0.05 >ok

6. Meshing> mesh > Areas > Free > Pick All

7. Solution > Analysis type > New analysis > steady state > Ok.

8. Solution > Define loads > Apply > Thermal > Temperature > on lines > click the top of the rectangular box > temperature > 500 > apply > click the left side of the rectangular box > ok > temperature > 100 > Ok.

9. Solution > Define loads > Apply > Thermal > convection > on lines > click the right side of the rectangular box > Ok.

10. Solve > current L.S > Ok.10. General Preprocessor > Plot results > Contour Plot > Nodal Solution > DOF solution > nodal temperature (TEMP) > Ok.Result:Thus convective heat transfer analysis is performed.

EX 10 CONVECTIVE HEAT TRANSFER ANALYSIS OF 2D COMPONENT

EX 11 SIMULATION OF AIR CONDITIONING SYSTEM

Problem Description :

The simple vapour compression refrigeration system works on freon 22 as a

refrigerant. The enthalpy of air refrigerant entering into the compressor is

1565kJ/kg. The enthalpy of the refrigrant extraction into the compressor is

1780kJ/kg and the enthalpy of the evaporator temp. is 260kJ/kg. Calculate the COP

of the refrigerator, compressor work and heat extracted by the refrigerant. Take

mass flow rate of the refrigerant 0.5kg/sec. Assume the condition of the refrigerant

is dry and saturated before entering the compressor.

Aim :

To perform the simulation of air conditioning system, to find the COP and the

power required for the system.

Formulae used :

Work done = h2 - h1 in kJ/kg

Refrigerating Effect = h1 - h4 in kJ/kg

COP = Refrigerating Effect / Work done

Power = m (h2 - h1) in kW.

Procedure:

1. Open the MATLAB application from programs

2. Choose a new simulation by selecting the ne icon or file new – option

3. A screen appears, go to the formulae section and the pick the necessary tool

boxes and drag and drop it in the working space.

4. The constants are given namely h1, h2, h4, P etc. and then they are selected,

right clicked and create subsystem is made.

5. Then the values of numbers are provided as gain.

6. They are dragged and dropped and placed in screen.

7. Then select the operation required and make the icons inside the work area.

8. Then to obtain the results, use the display box.

9. Now rename all the tool box icons and then create a subsystem and mask

wherever necessary.

10.Connect the input and output based on the equation (i.e) use the formulae

given.

11.Now click the run icon ad thereby the simulation runs and displays the result.

EX 12 SIMULATION OF HYDRAULIC SYSTEM


In a single acting reciprocating pump, the diameter of the cylinder is 152mm and

the stroke is 304mm. The water is raised through a height of 18m at a pump speed

of 40rpm. Calculate the theoretical power req. If the actual discharge is 0.212x10 -3

m3/s. Determine the coefficient of discharge.

Aim :

To perform the simulation of hydraulic system, to find the co efficient of discharge

and the power required for the system.

Formulae used :

Co efficient of Discharge = Q / Qt (Actual discharge / Theoretical discharge)

where, Qt = ALN / 60 in m3/sec (A is area of the cylinder, L is stroke, N is speed)

Power = w. Qt. H in kW. [Where, w=9810 kg/m3]

Procedure:





4. The constants are given namely A, L, N, H, w etc. and then they are selected,







wherever necessary.


given.


EX 13 SIMULATION OF CAM FOLLOWER MECHANISM


A cam is to be designed for a knife edge follower with the following data:

Cam lift = 40mm during 90° of cam rotation with SHM.

Dwell for the next 30°

During the next 60° of cam rotation, the follower returns to its original position

with SHM

Dwell during the remaining 180°

The radius of the base circle of the cam is 40mm.Determine the max. Velocity &

acceleration of the follower during its outstroke & return stroke, if the cam rotates

at 240rpm.

Aim :

To perform the simulation of cam follower mechanism, to find the max velocity and

acceleration of the follower during outstroke & return stroke for the system.

Formulae used :

Angular speed , ω = =2πN

60

Max. velocity of follower during outstroke , vomax = = πωs

2θo

Max. velocity of follower during return stroke , vrmax = = πωs

2θr

Max. acceleration during outstroke, aomax

= π2ω2s2θ

o2

Max. acceleration during return stroke, armax = π

2ω2s

2θ2r

Procedure:





4. The constants are given namely N, s, θo, θr, etc. and then they are selected,







wherever necessary.


given.


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