Laboratory Manual

LABORATORY RECORD

ED 7211

ANALYSIS AND SIMULATION LAB

M.E. ENGINEERING DESIGN

DEPARTMENT OF MECHANICAL ENGINEERING

RAJALAKSHMI ENGINEERING COLLEGE

THANDALAM-602105.

BONAFIDE CERTIFICATE

Name :

Class :

Roll No :

REGISTER NO.

Certified that this is the bonafide record of work done by

the above student in the

Laboratory during the year 20 - 20

Signature of Lab-in-charge Signature of Head of the

Department

Submitted for the Practical Examination held on:

Examiner

ED 7211

RAJALAKSHMI ENGINEERING COLLEGE

ANALYSIS AND SIMULATION LAB

List of Experiments

S.No. Date Name of the ExperimentPage No.

Signature

1.Stress analysis of beams(Candilever,Simply supported& fixed end)

2. Stress Analysis of a plate with circular hole

3. Stress Analysis of Rectangular L Bracket

4.Stress Analysis of an Axis- symmetric component

5.Mode frequency analysis of beams(Candilever,Simply supported,Fixed beam)

6. Mode frequency analysis of 2 –D element

7.Convective heart transfer analysis of a 2D component

8.Conductive heart transfer analysis of a 2D component

9. Thermal analysis of a 2D component

10. Harmonic analysis of a 2D component

INTRODUCTION

Finite element analysis is the core of computer aided engineering dictates the

modern mechanical industry.

Finite element analysis is a technique to simulate material properties, loading

conditions and environment of any system and to approximately determine response of the

system to those conditions.

NEED FOR FEA:

1. To reduce the amount of prototype testing

2. To simulate design that is not suitable for prototype testing

3. Cost Saving and time saving

4. Create a more reliable, better quality and competitive designs

ABOUT ANSYS:

ANSYS is a complete FEA simulation software package developed by ANSYS Inc.

a corporate of USA.

It is used by engineers worldwide in virtually all fields of engineering:-

Structural

Thermal

Fluid (CFD, Acoustics)

Low & High Frequency Electromagnetic

APPLICATIONS OF ANSYS:

ANSYS software can be applied in

Engineering structures

Automobiles

Aerospace

Biomedical

Buildings and other civil structures

Electronics

Heavy equipments & machinery

OTHER PRODUCTS:

ANSYS LS DYNA - Non Linear structural problems

ANSYS Professional - Linear Structural Thermal Analysis

ANSYS Design Space Analysis - Linear structural & steady state thermal analysis

ANSYS CFX - Fluid flow problems

CAPABILITIES OF ANALYSIS:

1. Structural Analysis: It is used to calculate the deformation, strain, stress and

reaction. The Eigen value problems where we find natural frequency of systems

also come under this category.

2. Thermal Analysis: It is used to determine the temperature distribution, amount of

heat lost or gained, thermal gradient and thermal flow.

3. Electromagnetic Analysis: It is used to calculate magnetic field effects in a

system. Often used with contact analysis.

4. Computational Field Analysis: It is used to determine the fluid flow

characteristics in a system like flow velocity, pressure, head loss, and even

temperature distribution.

5. Coupled Field Analysis: It is used to calculate the mutual interaction of two or

more of the above said disciplines. This is done by either sequential method or

simultaneous method.

STATIC STRESS ANALYSIS OF BEAMS (CANTILEVER, SIMPLY SUPPORTED, FIXED BEAM)

STATIC STRESS ANALYSIS OF CANTILEVER BEAM

Ex No:

Date:

Aim:

To perform static stress analysis for the given beams (Cantilever beam) using ansys.

SOFTWARE USED:

ANSYS 13.0

PROCEDURE:

1. PREFERENCE:

Preference → structural → ok

2. PREPROCESSOR

Element type

Preprocessor → element type →add → beam → 2node188→ok

Material property

Preprocessor → material property → material model → structural → linear → elastic → isotropic→ Ex (2e3) →PRxy (0.3)→ ok → close

Modeling

Preprocessor → section → beam → common section → section ID-1 → select sub type → rectangle → enter L*B (150* 300) → ok

Preprocessor→ modeling → create → keypoints→ in active CS

Preprocessor → modeling → create → line → straight line → draw the lines b/w the keypoints

Meshing

Preprocessor→ meshing → mesh tool → set line option → pick line 1→ apply → ok

Select mesh → pick all →ok

3. SOLUTION:

Solution → define loads → apply →structural → displacement → on keypoints → select first keypoint → ok → select all DOF → ok

Solution → define loads → apply → structural → force / moment → select third keypoint 3 → select Fy (30e3) → ok

Solution → define loads → apply → structural → pressure → select element1 → enter value 40

Solution → solve → current LS → ok

4. GENERAL POSTPROCESSOR:

General postproc → plot result → deformed shape &undeformed shape

KEYPOINTS X Y Z1 0 0 02 2000 0 03 1000 0 0

General postprocessor → list result → nodal solution

Utility menu → plot ctrl → animation deformed shape

RESULT:

Thus the stress analysis of cantilever beam is performed using ANSYS and results are plotted.

Calculation:

To find deflection,

Apply Maculay’s equation:

M x=−30 x−20

2¿

EId2 yd x2 =−30 x

−202

¿

EIdydx

=−30x2

2+c1−

103

¿

EIy=−30x3

6+c1 x+c2−

1012

¿

At x=3 , y=0

EI (0 )=−3033

6+c1(3)+c2−

2024

¿

At x=3 ,dydx

=0

EI (0 )=−3033

2+c1−

206

¿

c1=161.66

¿c1∈(1 )−−−→ c2=−336.4

EIy=−30x3

6+161.66 x−336.64−20

4¿

At x=1 , y= yb

EI yb=−3013

6+161.66(1)−336.64

yb=−2.53 mm

At x=0 , y= yc

EI yc=−3003

6+161.66(0)−336.64

yc=−4.74 mm

M c=30∗1=30 kNm

M A=30∗1+20∗2∗1=130 kNm

σ c=30∗106∗300∗12

150∗2∗30 03 =13.33N

mm2

σ a=130∗1 06∗300∗12

150∗2∗30 03 =57.77N

mm2

RA=(20∗2 )+30=70 KN

STATIC STRESS ANALYSIS OF SIMPLY SUPPORTED BEAM

Aim:

To perform static stress analysis for the given beams (Simply supported) using ansys.

SOFTWARE USED:

ANSYS 13.0

PROCEDURE:

PREFERENCE:


PREPROCESSOR

Element type


Material property


Modeling



Key points X Y1. 0 02. 3000 03. 4000 04. 6000 0


Meshing



5. SOLUTION:









RESULT:

Thus the stress analysis of simply supported beam is performed using ANSYS and results are plotted.

Calculation:

1) Simply supported Beam

To find reaction

RA - Reaction at ARB - Reaction at B

Taking moment about A,

Rb x6=20 x3 x32+(20 x 4 )

Rb=28.33 kN

Ra+Rb=(20∗3 )+20

¿51.67 KN

To find deflection:

Apply Maculey’s method,

Bending moment at section ‘x’ from end B at distance ‘x’

Mx = RB x -20(x-2)-20/2 (x-3)2

Mx =EI d2y/dy2

EId2y/dy2 = RB x -20(x-2)-20/2 (x-3)2

EI dy/dx = RB x2/2 +C1-20/2(x-2)2-(10/3)(x-3)4

EI y = (RBx3)/6+C1x+C2(-20/6)(x-2)3(-10/2)(x-3)4

At x=0 , y=0

EI(0)=C2 --- C2=0

At x=6,y=0

EI(0) = RB/6(6)3 -20(4)3 +C1(6)-(10/12)3

C1=-143.73

EI(y)=(RBx3)/6-20/6(x-2)3-143.73x-(10/12)(x-3)4

At x=2,y=yo

EIyD = 28.33/6 * 23-143.73(2)

YD = -5.96mm

At x=3,y=yc

Yc=-5.11mm

Mc=28.11*3-20*1=65kNm

Md=RB*2=56.66kNm

σ D=M D∗ y

I=

56.6∗106∗(264 /2)132∗2643

12

σ D=36.95 N /mm2

σ C=M D∗y

I=

65∗106∗(264 /2)132∗2643

12

σ C=42.38 N /mm2

STATIC STRESS ANALYSIS OF FIXED BEAM

Aim:

To perform static stress analysis for the given beams (Fixed supported) using ansys.

SOFTWARE USED:

ANSYS 13.0

PROCEDURE:

PREFERENCE:


PREPROCESSOR

Element type


Material property


Modeling



Key points X Y1. 0 02. 4000 03. 6000 04. 8000 0


Meshing



7. SOLUTION:









RESULT:

Thus the stress analysis of Fixed supported beam is performed using ANSYS and results are plotted.

Calculation:

To find deflection,

M x=RB x−M B−80 ( x−2 )−402

¿

EIdydx

=RB x2

2−M B x+c1−

802

¿

EIy=RB x3

6−

M B x2

2+c1 x+c2−

806

¿

At x=0 , y=0

EI (0 )=RB (0 )−M B (0 )+C1 (0 )+C2

At x=0 ,dydx

=0−−−→ C2=0

EI (0 )=RB

2(0 )−M B (0 )+C1 (0 )=¿=¿C1=0

EIy=RB x3

6−

M B x2

2−80

6¿

At x=8 , y=0

EI (0)=RB(8)3

6−

M B(8)2

2−80

6¿

EI (0 )=85.33 RB−32 M B−2880−426.65

At x=8 ,dydx

=0

EI (0 )=RB 82

2−M B(8)−80

2(8−2)−40

2¿¿

32 M B−8M B=1866.66

Solving the above eq:

RB=97.49 kN ; M B=156.63 kNm

RA +RB=80+40∗4=¿=¿=¿ RA=142.51kN

M A=80∗6+40∗4∗2+156.63−97.49∗8−−−−→ M A=176.71 kNm

At x=2 , y= yd

EI yd=97.49∗23

6−156.63∗23

2−−−→ yd=1.22∗1 0−2 mm

At x=4 , y= yc

EI yc=97.49∗43

6−156.63∗43

2−80¿¿

σ d=M d y

I=38.35∗106¿¿

y=d2

, I=bd3

12

σ c=M c y

I=73.33∗1 06 ¿¿

M d=−156.63+97.49∗2−−→ M d=38.35 kNm

M c=−80∗2−156.63+97.49∗4−−→ M c=73.33 kNm

STRESS ANALYSIS OF A RECTANGULAR PLATE WITH CILRCULAR HOLE

Ex No:

Date:

Aim:

To perform static stress analysis for the given rectangular plate with circular hole using ANSYS

Description:

A steel plate of dimension ( 100 × 15 × 25mm ) having 3 holes of diameter 3mm, 5mm & 10mm. It is fixed at one end and force of 20 KN acts on the other end of the plate. The stress induced at the holes due to the force is to be determined.

DIAGRAM:

Φ3mmΦ5mm Φ10mm

Formula used:

1. Normal stress ¿P

(w−d ) t

P – Load (N).

w – Width of the plate (mm).

d – Diameter of hole (mm).

t – Thickness of the plate (mm).

2. Maximum stress ¿σ N × k f

Kf – Stress concentration factor.

PROCEDURE:

Pre-processing: Defining the Problem

1. Open ANSYS APDL from start menu.

2. Define the type of problem,

Pre-processor → preference → structural → ok

3. Define the Type of Element

Pre-processor → Element Type → Add/Edit/Delete [Add the element: Solid - PLANE82]

4. Define Geometric Properties

Pre-processor → Real Constants → Add/Edit/Delete [Enter a thickness as 15mm]

5. Element Material Properties

Pre-processor → Material Properties → Material models → Structural → linear → Elastic → Isotropic [We are going to give the properties of Steel. Enter the following when prompted: EX – 2.1E5, PRXY – 0.3]

6. Create the main rectangular shape

Pre-processor → Modelling → Create → Areas → Rectangle → By 2 Corners [Enter the following values: WPX – 0, WPY – 0, Width – 100, Height – 25]

7. Create the circle

Pre-processor → Modelling → Create → Areas → Circle → Solid Circle [Enter the following values: Circle 1: WPX – 25, WPY – 12.5, Radius – 1.5; Circle 2: WPX – 50, WPY – 12.5, Radius – 2.5; Circle 3: WPX – 75, WPY – 12.5, Radius – 5]

20KN25mm

25mm 25mm 25mm 25mm

8. Subtraction of circles

Pre-processor → Modelling → Operate → Booleans → Subtract → Areas [First select the base area from which the area has to be deleted (rectangle) and then select the area that has to be deleted (circle).

9. Meshing

Pre-processor → mesh → mesh tool → Areas [set] → select the area → apply → enter element size as 10 → ok

Click mesh → select the area → ok

Solution Phase: Assigning Loads and Solving

10. Define constraints

Solution → define load → apply → structural → displacement → on lines [Select the line, select all DOF & click ok]

11. Define load

Solution → define load → apply → structural → pressure → on line [Select the line, enter the pressure value as 53.33 N/mm2]

12. Solving the System

Solution → Solve → Current LS

Post processing: Viewing the Results

13. Deflection

General Postproc → Plot Results → Nodal Solution → DOF solution → X-component

14. Stress

General Postproc → Plot Results → Nodal Solution → von mises stress

15. Stress values

General postproc → list result → nodal solution → von mises stress [Note down the stress values on respective nodes.

Result:

Thus static stress analysis for the given rectangular plate with circular hole has been successfully done using ANSYS.

Calculation:

1. For 3mm hole

σ N= P(w−d )t

P = 20KN ; d = 3mm

w = 25mm ; t = 15mm

σ N= 20 ×103

(25−3 ) ×15

= 60.61 N/mm2

From PSG Data book page 7.10, kf = 2.65 (d/w = 0.12)

σ max=σN × k f = 60.61 x 2.65

σ max=¿161.62 N/mm2

2. For 5mm hole

σ N= P(w−d )t

P = 20KN ; d = 5mm

w = 25mm ; t = 15mm

σ N= 20 ×103

(25−5 ) ×15

= 66.67 N/mm2

From PSG Data book, kf = 2.5 (d/w = 0.2)

σ max=σN × k f = 66.67 x 2.5

σ max=¿166.68 N/mm2

3) For 10mm hole

σ N= P(w−d )t

P = 20KN ; d = 10mm

w = 25mm ; t = 15mm

σ N= 20 ×103

(25−10 ) ×15

= 88.89 N/mm2

From PSG Data book, kf = 2.25 (d/w = 0.4)

σ max=σN × k f = 88.89 x 2.25

σ max=¿200 N/mm2

Observation:

SL.No Hole size Analytical Simulation

1 3 161.62 136.87

2 5 166.68 144.60

3 10 200 191.48

STRESS ANALYSIS OF A RECTANGULAR L-BRACKET

Ex No:

Date:

AIM:

To determine a deform shape and stress analysis for a given L- bracket using ANSYS software.

FORMULA USED:

M / I = σ b / Y

Where,

M - Bending moment

I - Moment of inertia of cross sectional area

σ b - Bending stress

Y - Distance from neutral axis to the outer most fiber

PROCEDURE:

The modules available in ANSYS are ,

i. Preferences ii. Pre- Processor iii. Solution iv. General Post Processor module

1. Preferences > Structural > OK

2. In Pre Processor module, element type for analysis in chosen by Pre- Processor > Element type > Add > Solid >Quad 4 node 182

2. The Material properties are given by Pre – Processor > Material modal > Linear > Elastic > Isotropic >Define young’s modulus and Poisson ratio

3. Modeling of the L section

Modeling > Create >Area > Rectangle > By 2 Corners > X=0,Y=0,Width= 335,Height= 70

Modeling > Create >Area > Rectangle > By 2 Corners > X=165,Y=70,Width= 70, Height = 150

Modeling > Create >Area > Circle > Solid circle > X=300, Y=220,Radius=45

Modeling >Operate > Boolean > Add > Areas> Pick all > OK

Modeling > Create >Area > Circle > Solid circle > X=300, Y=220,Radius=35

Modeling >Delete > Areas Only > Pick the areas to be deleted > OK

4. The area is divided into finite no. of elements by Meshing > mesh tool > Area >Pick all>Edge Length = 20 >mesh tool > mesh.

5. The boundary condition is defined at the both end after beam by arresting displacements in Y- direction by Solution > loads > structural > displacement > on areas >arrest All DOF.

6. Load is applied on the nodes at the top edge by solution> loads >Structural > Force > on nodes

7. Now the system is ready to solve and is done by solution > solve > current L.S.

8. By general post processor > Plot result > counter plot > nodal solution.> DOF solution > displacement

9. The stress is also obtained by General post processor > plot result > counter plot > Nodal solution> Von-Mises Stress

RESULT:

Thus the deformed shape and stress analysis for the L bracket is done

CALCULATION:

Resolving the forces 5KN

Horizontal component P H = P Sin 60°= 4.33 KN

Vertical component PV = P Cos 60° = 5 cos 60 ° = 2.5 KN

Bending moment due to PH :

MH = 4.33x150x100

MH = 649.5x103 N mm

Bending stress σ bh = My /Z

649.5x103x6/3.5x102 = 22.72 N/mm2

Bending moment due to Pv :

MV = 2.5x106x300 = 750 x 103 N mm

Bending moment due to Pv :

σ bv = MY /Z

σ bv = 750x103x6/35x102

= 26.33 N\ mm2

Direct stress due to Vertical force PV

σ dv = PV / A = 2.5x 103/ 70x35 = 1.02 N\ mm2

Total Stress = σ bh+σbv +σdv

= 50.16 N\ mm

STRESS ANALYSIS OF AN AXIS -SYMMETRIC COMPONENT

Ex No:

Date :

AIM:

To analyze a cylindrical pressure vessel (axis symmetric) for hoop stress and longitudinal stress

DIAGRAM:

PROCEDURE:

Preprocessing: Defining the Problem:

1. Give example a Title Utility Menu > File > Change Title...2. Open preprocessor menu ANSYS Main Menu > Preprocessor 3. Create Areas Preprocessor > Modeling > Create > Areas > Rectangle > By Dimensions4. Add Areas Together Preprocessor > Modeling > Operate > Booleans > Add > Areas 5. Define the Type of Element Preprocessor > Element Type > Add/Edit/Delete... For this problem we will use the PLANE2 [Axisymmetric]

6. Define Element Material Properties Preprocessor > Material Props > Material Models > Structural > Linear > Elastic > Isotropic

7. Define Mesh Size Preprocessor > Meshing > Size Cntrls>ManualSize> Areas > All Areas [An element edge length of 2mm]8. Mesh the frame Preprocessor > Meshing > Mesh > Areas > Free > click 'Pick All'

Solution Phase: Assigning Loads and Solving:

1. Define Analysis Type Solution > Analysis Type > New Analysis > Static2. Apply Constraints Solution > Define Loads > Apply > Structural > Displacement > Symmetry B.C. > On Lines Pick the two edges on the left, at x=0. Utility Menu > Select > Entities [Select Nodes and By Location from the scroll down menus. Click Y coordinates and type in 50] Solution > Define Loads > Apply > Structural > Displacement > On Nodes > Pick All Constrain the nodes in the y-direction (UY).3. Utility Menu > Select > Entities

[In the select entities window, click Select All to reselect all nodes.]4. Apply Loads

Solution > Define Loads > Apply > Structural > Force/Moment > On Key points [Pick the top left corner of the area and click OK. Apply a load of 100 in the FY direction. Pick the bottom left corner of the area and click OK. Apply a load of -100 in the FY direction. ]5. Solve the System

Solution > Solve > Current LS


1. Determine the Stress Through the Thickness of the Tube

Utility Menu > Select > Entities... Select Nodes > By Location > Y coordinates and type 45, 55 in the Min, Max box.

General Postproc> List Results > Nodal Solution > Stress > Components SCOMP 2. Plotting the Elements as Axisymmetric

Utility Menu >PlotCtrls> Style > Symmetry Expansion > 2-D Axi-symmetric...

RESULT:

Thus the analyze a cylindrical pressure vessel (axis symmetric) for hoop stress and longitudinal stress

CALCULATION:

P=1.5MN/m2

P=1.5 N/mm2

d =1000 mm t=15 mm

σ c= pd2 t

σ C= 1.5∗1000

2∗15

σ C = 50 N/mm2

MODAL ANALYSIS OF BEAMS (CANTILEVER, FIXED, SIMPLY SUPPORTED)

MODAL ANALYSIS OF BEAMS CANTILEVER BEAM

Ex. No.

Date:

AIM

To perform modal analysis of a (cantilever beam) using ANSYS

PROCEDURE :

1. The modules available in ANSYS are

i. Preferences

ii. Pre-Processor

iii. Solution

iv. General post processor module

2. In the pre-processor module the element type for the analysis of the cantilever beam is

Chosen

Pre-processor > element type > add > beam3 > 2D ELASTIC 3

3. The Density & Young Modulus are given.

Main menu > Pre-processor > Material properties > Enter the Density & Young Modulus

4. Create two key points and a line to connect the points.

Modeling > create > key points > inactive c.s

Modeling > create > lines > between key points.

5. Mesh the line using mesh tool.

Main menu > Pre-processor > Meshing > mesh tool > line > mesh.

6. The boundary condition is defined for the beam.

Solutions > load > structural > displacement > On Left Key point > all DOF.

7. Solution: Assigning loads and solving, Define analysis type

Solution > analysis type > new analysis > modal

8. Set options for analysis type:

Solution > Analysis type > Analysis options….

Enter 10 for no. of modes to extract and no. of modes to expand in subspace and

Expand mode shapes.

9. Reduced method is chosen.

10. Apply constrains

Solution > Define loads > Apply > Structural > Displacement < On

Key points.

11. Solve the system


12. Post processing: Viewing the results

General Postproc > Result Summary.

13. To view mode shapes

General Postproc > Read results > First set

General Postproc > Plot results > Deformed shape

Repeat the process for the next set to view the next mode shapes.

14. To Animate mode shapes

Utility menu > Plot ctrls > Animate > Mode shapes

RESULT :

Thus the cantilever beam is modeled and analyzed in ANSYS for Modal

Analysis. Various mode shapes and respective frequencies are obtained.

MODAL ANALYSIS OF BEAMS SIMPLY SUPPORTED BEAM

Ex. No.

Date:

AIM :

To perform modal analysis of a (simply supported) beam using ANSYS

PROCEDURE:


i. Preferences

ii. Pre-Processor

iii. Solution


2. In the pre-processor module the element type for the analysis of the Simply Supported beam is

chosen

Pre-processor > element type > add > beam3 > 2DELASTIC 3

3. The Young Modulus and Material Density are given.

Main menu > Pre-processor > Material Properties > Enter the Young Modulus & Density







Solutions > load > structural > displacement > on key points > UY

7. Solution: Assigning loads and solving

Define analysis type



Solution > Analysis type > Analysis options


Expand mode shapes.



Solution > Define loads > Apply > Structural > Displacement > on key points.




General post processing > Result Summary.


General post processing > Read results > First set

General post processing > Plot results > Deformed shape




RESULT :

Thus the simple supported beam is modeled and analyzed in ANSYS for

Modal analysis. Various mode shapes and respective frequencies are obtained.

MODAL ANALYSIS OF BEAMS FIXED BEAM

Ex. No.

Date:

AIM

To perform modal analysis of a (Fixed beam) using ANSYS

PROCEDURE


i. Preferences

ii. Pre-Processor

iii. Solution


2. In the pre-processor module the element type for the analysis of the Fixed beam is

Chosen

Pre-processor > element type > add > beam3 > 2D ELASTIC 3

3. The Young Modulus and Material Density are given.

Main menu > Pre-processor > Material Properties > Enter the Young Modulus & Density







Solutions > load > structural > displacement > on key points > All DOF

7. Solution: Assigning loads and solving

Define analysis type



Solution > Analysis type > Analysis options….


Expand mode shapes.



Solution > Define loads > Apply > Structural > Displacement > On key points.




General post processing > Result Summary.


General post processing > Read results > First set

General post processing > Plot results > Deformed shape




RESULT :

Thus the Fixed beam is modeled and analyzed in ANSYS for

Modal analysis. Various mode shapes and respective frequencies are obtained.

Description:

Cross section= 0.125*0.125

Formula:

Frequency, fn= λ2

2 π l2 √ EIm

where, E = young’s modulus (N/m2)

I = moment of inertia

m = mass (kg)

λ = wavelength (m)

λ values for the raw mechanical vibration:

Mode Cantilever SSB Fixed

1 1.875 π 4.730

2 4.697 2π 7.853

3 7.853 3π 10.995

Calculation:

E = 2.1*105N/mm2 for alloy steel

I = bd3/12 = 0.125*0.1253/12 = 2.03*10-3 m4

Density, ρ = 7700 kg/m3

Mass = ρ*v = 7700*0.125*0.153

m = 120.8kg/m3

Cantilever beam:

Mode 1:

fn = λ2

2 π l2 √ EIm

= 1.8752

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 4.217 Hz

Mode 2:

fn = 4.6972

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 26.429 Hz

Mode 3:

fn = 7.8532

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 73.97 Hz

Simply Supported beam:

Mode 1:

fn = π 2

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 11.85 Hz

Mode 2:

fn= (2 π )2

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 44.35 Hz

Mode 3:

fn = (3 π )2

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 106.5 Hz

Fixed beam:

Mode 1:

fn = 4.7302

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 26.82 Hz

Mode 2:

fn = 7.8532

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 73.94 Hz

Mode 3:

fn = 10.9952

2 π∗52 √ 2.1∗104∗2.03∗105

120.8

= 144.81 Hz

Result:

Beam Analytical value Simulation valueCantilever Mode 1 4.217 4.2169Cantilever Mode 2 26.429 26.488Cantilever Mode 3 73.97 74.592

SSB Mode 1 11.85 11.862SSB Mode 2 44.35 47.710SSB Mode 3 106.5 108.33

Fixed beam Mode 1 26.82 26.905Fixed beam Mode 2 73.94 74.636Fixed beam Mode 3 144.81 147.75

Mode Frequency analysis of 2d component .

EX NO:

DATE:

Aim:

To perform a mode frequency analysis of the given component using ANSYS

Description:

A simply supported steel plate of dimension ( 1000 × 500 × 25mm ). A fixed circular plate with thickness 25mm and diameter 200mm

Diagram:

Formula used:

ω=√ Dρ [(mπ

a )2

+( nπb )

2]D= ∈h3

12 (1−γ2 )

∈ = 2.1 x 1011 N/m2

γ = 0.28 (Poisson Ratio)

ρ = 7700 kg/m3

h = thickness of plate for rectangular lamina

m = number of parts of modal value

n = number of parts value in horizontal direct

Circular lamina:

λ2 = ω a2 √ ρD

500

1000

λ = Wavelength

a = radius

ω = 2πs

S = Number of parts modal value in radial direction

n = number of parts modal value in circular direction.

Procedure:

Rectangular plate :











Pre-processor → Material Properties → Material models → Structural → Density [ Enter the density value as 7700]


Pre-processor → Modelling → Create → Areas → Rectangle → By 2 Corners [Enter the following values: WPX – 0, WPY – 0, Width – 1, Height – .5]

7. Meshing




8. Analysis type

Solution → Analysis type → New analysis → Modal → ok

Analysis type → Analysis Option [ no of modes to extract – 10 ; Nmode number of modes to expand 10] → ok


Solution → define load → apply → structural → displacement → on lines [Select the left and right lines, select UY & click ok]

Solution → define load → apply → structural → displacement → on lines [Select the Bottom and top lines, select UX & click ok]




11. Mode Shape

General Postproc → Read Results → by pick [Select the frequency and click read] → close

Plot controls → animate → Mode shape [select DOF solution ; Deformed shape] → ok

12. Repeat step 11 for different frequency values.

Circular plate :










Pre-processor → Material Properties → Material models → Structural → Density [ Enter the density value as 7700]

5. Create the circular shape

Pre-processor → Modelling → Create → Areas → circle → solid circle [Enter the following values: WPX – 0, WPY – 0, Radius = .1]

7. Meshing




8. Analysis type

Solution → Analysis type → New analysis → Modal → ok

Analysis type → Analysis Option [ no of modes to extract – 10 ; Nmode number of modes to expand 10] → ok


Solution → define load → apply → structural → displacement → on lines [Select Circumference, select all Dof & click ok]




11. Mode Shape

General Postproc → Read Results → by pick [Select the frequency and click read] → close

Plot controls → animate → Mode shape [select DOF solution ; Deformed shape] → ok

12. Repeat step 11 for different frequency values. Result:

RESULT:

The mode frequency analysis of given 2D component has been successfully done using ANSYS.

Calculation:

For Rectangular plate :

D= ∈h3

12 (1−γ2 ) = 2.1× 1011 ×0.0253

12 (1−0.28 )2 = 296698.67

ω1=√ Dρ [(mπ

a )2

+( nπb )

2] = √ 296698.67192.5 [( π

1 )2

+( π.5 )

2]f1 =

ω2 π

=308.34 Hz

f 2=√ 296698.67

192.5 [( 2 π1 )

2

+( π.5 )

2]2 π

= 493.3Hz

f 3=√ 296698.67

192.5 [( 3 π1 )

2

+( π.5 )

2]2 π

= 801.75Hz

f 4=√ 296698.67

192.5 [( π1 )

2

+( 2 π.5 )

2]2 π

= 1048.36Hz

f 5=√ 296698.67

192.5 [( 2 π1 )

2

+( 2 π.5 )

2]2 π

= 1233.36Hz

f 6=√ 296698.67

192.5 [( 4π1 )

2

+( π.5 )

2]2π

= 1233.36Hz

f 7=√ 296698.67

192.5 [( 3 π1 )

2

+( 2 π.5 )

2]2 π

= 1547.71Hz

f 8=√ 296698.67

192.5 [( 5 π1 )

2

+( π.5 )

2]2 π

= 1788.38Hz

f 9=√ 296698.67

192.5 [( 4π1 )

2

+( 2π.5 )

2]2 π

= 1973.38Hz

f 10=√ 296698.67

192.5 [( π1 )

2

+( 3 π.5 )

2]2 π

= 2281.72Hz

For circular plate :

D= 296698.67 a = 1m

ρ= 192.5 ω= 2πF

λ2 = ω a2 √ ρD

10.2158=2 π F1 ×12 √ 192.5296698.67

F1 = 63.83Hz

21.26=2 π F2×12 √ 192.5296698.67

F2 = 132.84Hz

34.83=2 π F4× 12√ 192.5296698.67

F4 = 217.63Hz

34.771=2 π F6 ×12 √ 192.5296698.67

F6 = 248.5Hz

51.04=2 π F7× 12√ 192.5296698.67

F7 = 318.92Hz

60.82=2 π F9 ×12 √ 192.5296698.67

F9 = 380.03Hz

RESULT :

Mode M n Mode shape Frequency Analytical Simulated

1 1 1 308.34 307.932 2 1 493.3 492.293 3 2 801.73 798.974 1 3 1048.36 1043.25 4 1 1233.36 1226.86 2 2 1233.36 1226.8

7 3 2 1541.71 1531.58 5 1 1788.4 1774.89 4 2 1968.38 1956.610 1 3 2281.72 2260.0

Mode M n Mode shape

Wavelengthλ

FrequencyAnalytical simulated

1 0 0 10.158 63.81 63.8122 0 1 21.26 132.71 132.773 0 1 21.26 132.71 132.774 0 2 34.83 217.63 217.635 0 2 34.83 217.63 217.636 1 0 34.73 248.5 248.247 0 3 51.04 318.92 318.378 0 3 51.04 318.07 318.459 1 1 60.82 380.07 379.4510 1 1 60.82 380.07 379.45

CONVECTIVE HEAT TRANSFER ANALYSIS OF A 2 D COMPONENT

EX NO:

DATE:

AIM:

To find the convective heat transfer analysis of a 2 D component

DIAGRAM:

PROCEDURE:

1) Preference > Thermal > ok

2) Preprocessor > Element type > Add/edit/delete > add > solid > Quad4node55 > ok

3) Material properties > material models > thermal > conductivity(Enter the value 50

w/mk)

4) Modeling > create > areas > rectangle > by 2 corners (Enter the value

w=0.012m;h=0.1m)

5) Meshing > mesh tool > area set > pick the area > ok > size element > edge

length(0.02m) > ok

6) Mesh tool > mesh > pick the area > ok

7) Loads > define loads > apply > thermal > convection > on lines > pick the inside

lines > ok(Enter the value of inside h and inside T) > on lines > pick the outside lines

> ok(Enter the value of outside h and outside T)

8) Solution > solve > current LS

9) See the result in General post processor > list results > nodal solution > DOF solution

> nodal temperature > ok

RESULT:

The temperature and heat flux are shown and results are completed.

CALCULATION:

T – Temperature of water (℃)

T a – Atmospheric temperature (℃)

T i - Inner wall temperature (℃)

T 0 - Outer wall temperature (℃)

u – Overall heat transfer coefficient

K – Thermal conductivity

ho - Heat transfer coefficient at outer

surface(w/m2 k)

hi - Heat transfer coefficient at inner

surface(w/m2 k)

1u= 1

hi

+ 1k+ 1

h0

= 1

2850+ 0.012

50 +

110

u = 9.94 w/m2 k

q =u (T 0−T a)

= 9.94(393-293)

q = 795.2 w/m2

q = hi(T 0−T i)

795.2= 2850 (373-T i¿

T i = 372.72 K

T i = 99.72 ℃

q = ho(T0−T a)

795.2= 10 (T 0−293)

T 0 = 372.52 K

T 0 = 99.52 ℃

CONDUCTIVE HEAT TRANSFER ANALYSIS OF A 2-D ELEMENT

Ex No :

Date :

AIM:

To Conduct heat transfer analysis of a 2-D element for the given example and determine the temperature at the specified points

DIAGRAM:

FORMULAE USED:

Q = ∆Toverall/R

h1 = l 1

k 1a1 +

l 2k 2a 2

+ l 3

k 3a 3

T1, T2, T3, T4 = Temperatures of wall section

R = Resistance

k1, k2, k3 = Thermal Conductivity (W/mK)

A1, A2, A3 = Area of wall

q = heat flux

PROCEDURE:

Preprocessing: Defining the Problem

1. Give example a Title

2. Open pre-processor menu ANSYS Main Menu > Preprocessor

3. Define the Type of Element Preprocessor > Element Type > Add/Edit/Delete> click 'Add' > Select Link 3D conduction 33

4. Define Real ConstantsPreprocessor >Real Constants>Add>Select the element>Define Cross-sectional area as 1

5. Element Material Properties Preprocessor > Material Props > Material Models > Thermal > Conductivity > Isotropic > KXX1 = 1.05 (Thermal conductivity) > Select New Model > KXX2 = 0.15> Select new model> KXX3 = 0.85

6. Create geometry Preprocessor > Modeling > Create >Key points>On active CS >By dimensions> X=0, Y=0, Z=0 , X=0.25, X=0.37, X=0.57Preprocessor > Modeling > Create >lines>Straight lines

7. Mesh Preprocessor > Meshing > Mesh Attributes>Picked Lines > Pick the line>OK> Define the Material Number

Preprocessor > Meshing > Mesh Tool> Pick Lines> OK> Edge Length > .05> OK> Mesh

8. Define Loads

Preprocessor >Loads > Define Loads > Apply > Thermal > Temperature > On Keypoints> Define The Temp at T1=850 and T2= 65


1. Solve the System Solution > Solve > Current LS


1. Results Using ANSYS Plot Temperature General Postproc> Plot Results > Contour Plot > Nodal Solution > DOF solution, Temperature

General Postproc> List Results > Nodal Solution > Report for Node 7 and 9

RESULT :

Thus the temperatures at the specified points were found throughconductive heat transfer analysis successfully.

CALCULATIONS :

Q = ∆Toverall/R

h1 = l 1

k 1a1 +

l 2k 2a 2

+ l 3

k 3a 3

q = QA

= 850

0.251.005

+0.120.15

+0.2

0.85

q = 616.46 W/m2

Temperature at the surface, clay fire and insulation brick

To find T2

q = T 1−T 4

k =

T 1−T 2k

= T 2−T 3

k =

T 2−T 4k

q = T 1−T 2

kA =

T 1−T 2h 1k 1

616.46 = 850−T 20.25/1.05

T2 = 703.22 K

Temperature at the surface of insulation brick and red brick

q = QA

= T 2−T 3h 2/k 2

616.46 = 703.22−T 30.12/0.15

T3 = 210.056 K

THERMAL STRESS ANALYSIS OF 2D COMPONENT

Ex No :

Date :

AIM:

To perform thermal stress analysis of a given 2D component using solid works simulation software.

DIAGRAM

PROBLEM DESCRIPTION:

A rectangular plate of dimensions 50mm×25mm×2.5mm in subjected to a uniform temperature from 0oC to 38oC. Determine the maximum displacement and normal stress in X direction. The properties of the plate are, Young’s modulus (E) =2.1×10ˆ5 N/mm2, Poisson’s ratio (ν) =.28 and thermal coefficient of expansion (α) =1.3×10ˆ-5 K-1

FORMULA USED:

Normal stress in X direction = EαΔΤ1. Displacement in Y direction = αΔTb(1+ ν)

Where, E → Young’s modulus

α → Thermal co-efficient

ΔΤ → Temperature difference

b → Breadth

ν → Poisson’s ratio

PROCEDURE:


2. Give Title name.

File → Change title → Enter new title → ok.

PREFERENCE:


Preferences → structural & thermal→ ok

PREPROCESSOR:


Pre-processor → Element Type → Add/Edit/Delete → Add → Solid → Quad

4 node 182 → ok → options → plane strs w/thk → ok → close.


Preprocessor → Real Constants → Add/Edit/Delete → Add → ok → THK =

0.0025m → ok → close


Pre-processor → Material Properties → Material models → Structural →

linear → Elastic → Isotropic → EX = 2.1×10ˆ11 & PRXY = 0.24 → ok → Thermal

expansion → Secant coefficient →Isotropic → ALPX = 1.3×10ˆ-5 → ok → close.


Pre-processor → Modeling → Create → Areas → Rectangle → By 2 Corners

→ WP X – 0, WP Y = 0, Width = .05, Height .025 → ok.

8. Meshing

Pre-processor → mesh → mesh tool → Areas [set] → select the area → apply

→ enter element size = .001 → ok → mesh → select the area → ok

9. Define temperature

Solution → define loads → apply → structural →temperature → on lines →

Pick line → ok → VAL1 temperature = 273 → apply → Pick line → ok → VAL1

temperature = 311 → ok.

SOLUTION:

10.Solving the System

Solution → Solve → Current LS.

GENERAL POSTPROCESSOR:

11.Stress

General Postproc → Plot Results → Contour Plot → Nodal Solution → Nodal

Solution → X component of stress → ok.

12.Stress values

General postproc → list result → Nodal Solution → Nodal Solution → X

component of stress → ok → Note values from the table.

13.Displacement values

Note the value of DMX

RESULT:

Thus the thermal stress analysis of 2D component has been done and results are obtained.

CALCULATION:

Normal stress in X- direction,

σx = EαΔΤ

σx = 2.1×10ˆ5×1.3×10ˆ-5×38

= 103.74 N/mm 2

Displacement in Y-direction

y = αΔTb(1+ ν)

= 1.3×10ˆ-5×38×25×(1+0.25)

= 1.524×10ˆ -2 mm

Note: Restrain geometry along X direction and rotation on all directions.

HARMONIC ANALYSIS OF 2D COMPONENT

Ex No:

Date :

AIM:

To calculate the peak vs displacement and bending stress at the centre of square plate of side 10m and thickness 0.05m subjected to a steady state harmonic pressure

All dimensions are in mm

PROBLEM DESCRIPTION:

To calculate and perform harmonic analysis of a given 2D component with the given dimensions of side 10m and thickness 0.05m

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 82 > options > plane stress with thickness > close.

4. Preprocessor > Real Constant > Add/Edit/Delete > thickness = 0.05 > Ok

5. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > EX = 200 E9, PRXY = 0.3 & Density = 8000.

6. Preprocessor>Modeling>create>Areas>Rectangle> By dimensions>Enter the Value

7. Preprocessor > meshing > mesh tool > size control > Areas >Enter the Element edge length > Ok > mesh > Areas > free> pick all.

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

9. Solution > define loads > apply > structural > force/moment > on nodes > click right corner > FY real value> Enter the Value & Imaginary value = 0 > Ok.

10. Solve > current L.S > ok.

11. Load step option > time frequency > frequency & sub steps > Enter the Value > stepped > Ok.

12. Time history postprocessor > variable viewer > add > nodal solution > DOF solution > Y-component of displacement > click right corner > ok > graph data > Ok.

13. Utility Menu > plot controls > style > graphs > modify axis ( change the Y-axis scale to logarithmic)

14. Utility menu > plot > replot.

Model Damping Rayleigh DampingPeak U = displacement at Center (mm)

7.831mm at 2.088Hz 7.0833mm at 2.087Hz

Peak Bending Stress at Center 5.4N/mm2 at 2.088Hz 5.4N/mm2 at 2.087Hz

RESULT:

Thus the thermal stress analysis of 2D component has been done and results are obtained.

CALCULATION:

Density = 8000 kg / m³

E=200 x109 N / M ²

Harmonic option = 0.10 Hz

Model damping: Damping ratio = 0.2

Poisson Ratio = 0.3

Rayleigh damping: α= 0.299

β = 0.00139

.

Laboratory Manual

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