MODELING & ANALYSIS LABORATORY

MODELING & ANALYSIS LABORATORY

Department of Mechanical Engineering, VSMSRKIT, Nipani 1

INTRODUCTION

Mechanical design is the design of a component for optimum size, shape, etc., against failure

under the application of operational loads. A good design should also minimize the cost of

material and cost of production. Failures that are commonly associated with mechanical

components are broadly classified as:

(a) Failure by breaking of brittle materials and fatigue failure (when subjected to repetitive loads)

of ductile materials.

(b) Failure by yielding of ductile materials, subjected to non-repetitive loads.

(c) Failure by elastic deformation.

The last two modes cause change of shape or size of the component rendering it useless and,

therefore, refer to functional or operational failure. Most of the design problems refer to one of

these two types of failures. Designing, thus, involves estimation of stresses and deformations of

the components at different critical points of a component for the specified loads and boundary

conditions, so as to satisfy operational constraints.

Design is associated with the calculation of dimensions of a component to withstand the applied

loads and perform the desired function.

Analysis is associated with the estimation of displacements or stresses in a component of

assumed dimensions so that adequacy of assumed dimensions is validated.

Optimum design is obtained by many iterations of modifying dimensions of the component

based on the calculated values of displacements and/or stresses vis-a-vis permitted values and re-

analysis.

An analytic method is applied to a model problem rather than to an actual physical problem.

Even many laboratory experiments use models. A geometric model for analysis can be devised

after the physical nature of the problem has been understood. A model excludes superfluous

details such as bolts, nuts, rivets, but includes all essential features, so that analysis of the model

is not unnecessarily complicated and yet provides results that describe the actual problem with

sufficient accuracy. A geometric model becomes a mathematical model when its behavior is

described or approximated by incorporating restrictions such as homogeneity, isotropy,

constancy of material properties and mathematical simplifications applicable for small

magnitudes of strains and rotations.



Several methods, such as method of joints for trusses, simple theory of bending, simple theory of

torsion, analyses of cylinders and spheres for axisymmetric pressure load etc., are available for

designing/analyzing simple\components of a structure. These methods try to obtain exact

solutions of second order partial differential equations and are based on several assumptions on

sizes of the components, loads, end conditions, material properties, likely deformation pattern

etc. Also, these methods are not amenable for generalization and effective utilization of the

computer for repetitive jobs.

Use of strength of materials approach for designing a component is, therefore, associated with

higher factor of safety. The individual member method was acceptable for civil structures, where

weight of the designed component is not a serious constraint. A more accurate analysis of

discrete structures with few members is carried out by the potential energy approach. Optimum

beam design is achieved by analyzing the entire structure which naturally considers finite

stiffness of the columns, based on their dimensions and material, as it ends. This approach is

followed in the Finite Element Method (FEM).



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:

ANSYS 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 solver

6. 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.



Ex.No:1 STRESS ANALYSIS OF A CONSTANT CROSS SECTION BAR

Aim: To determine the nodal displacement, stress in the element and reaction forces of

the bar shown in fig.

Procedure:

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

2. Preference > Structural > OK

3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Spar > OK.

4. Preprocessor > Real Constant > Add/Edit/Delete > Add > Real constant Set No.1 > C/S

area > Value[500] > OK

5. Preprocessor > Material Properties > Material Model > Structural > Linear > Elastic >

Isotropic > EX = 2E5, PRXY = 0 > OK

6. Preprocessor > Modeling > Create > Nodes > In active CS > Now enter the co-ordinates

of the nodes to be created > Apply

Node 1: X=0, Y=0, Z=0

Node 2: X=1000, Y= 0, Z=0

7. Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru Nodes > Pick

Node 1 and Node 2 > OK

8. Preprocessor > Loads > Define loads > Apply > Structural > Displacement > On Nodes >

Pick Node 1 > All DOF > Value [0] > OK

9. Preprocessor > Loads > Define Loads > Apply > Structural > Force/Moment > On Nodes

> Pick Node 2 > FX > Value[1000] > OK

10. Solution > Solve > Current LS > OK

E=2x105N/mm

2



11. General Postprocessor > Element table > Define table > Add by Sequence No >LS,1 >

OK > Close

12. General Postprocessor > Plot Results > Contour Plot > Line Element result > OK

13. General Postprocessor > List Results > Reaction solution > OK

14. General Postprocessor > List Results > DOF Solution > Vector Sum > OK



Ex.No:2 STRESS ANALYSIS OF A STEPPED BAR



Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Spar > OK.






Isotropic > EX = 70E3, PRXY = 0





Node 1: X=0, Y=0, Z=0

Node 2: X=300, Y=0, Z=0

E=70x109N/m

2

E=200x109N/m

2



Node 3: X=700, Y=0, Z=0



10. Preprocessor > Modeling > Create > Elements > Element Attributes > Change Real

constant Real constant Set No.1 to 2 > Change Material N0.1 to 2 > Auto Numbered >

Thru Nodes > Pick Node 2 and Node 3 > OK


Pick Node 1 and Node 3 > All DOF > Value [0] > OK





OK > Close






Ex.No:3 STRESS ANALYSIS OF A TAPERED BAR



Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Spar > OK









Node 1: X=0, Y=0, Z=0

Node 2: X=375, Y=0, Z=0

Node 3: X=750, Y=0, Z=0

E=2x105N/mm

2






constant Real constant Set No.1 to 2 > Auto Numbered > Thru Nodes > Pick Node 2 and

Node 3 > OK






13. General Postprocessor > Element table > Define table > Add by Sequence No > LS,1 >

OK > Close






Ex.No: 4 STRESS ANALYSIS OF A TWO BAR TRUSS HAVING SAME ELEMENT

AREA


the truss shown in fig.

Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=-400, Y=300, Z=0

Node 3: X=-900, Y=300, Z=0

Take A1=A2=200mm2

E=2x105N/mm

2








Pick Node 1 and Node 3 > All DOF > Value [0] > OK


> Pick Node 2 > FY > Value [-12000] > OK



OK > Close






Ex.No: 5 STRESS ANALYSIS OF A TWO BAR TRUSS HAVING DIFFERENT

ELEMENT AREA


the truss shown in fig.

Procedure:









Isotropic > EX = 2E5, PRXY = 0.



Node 1: X=0, Y=0, Z=0

Node 2: X=750, Y=500, Z=0

Take

E=2x105N/mm

2



Node 3: X=0, Y=500, Z=0




constant Real constant Set No.1 to 2 > Auto Numbered > Thru Nodes > Pick Node 2 and

Node 3 > OK


Pick Node 1 and 3 > All DOF > Value [0] > OK


> Pick Node 2 > FX > Value [-50000] > OK


13. General Postprocessor > Element table > Define table > Add by Sequence No > LS,1 >

OK > Close






Ex.No: 6 SIMPLY SUPPORTED BEAM SUBJECTED TO POINT LOAD AT THE

CENTER

Aim: To determine SFD, BMD & Reaction forces at the supports for a rectangular cross

section beam having area 0.2mX0.3m and Young’s modulus =210GPa

Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Elastic 3 > OK


area > Value[0.2*0.3] > Moment of Inertia > Value[ ] > Height > Value [ ] > OK


Isotropic > EX = 2.1E11, PRXY = 0 > OK



Node 1: X=0, Y=0, Z=0

Node 2: X=2, Y= 0, Z=0

Node 3: X=4, Y= 0, Z=0







9. processor > Loads > Define loads > Apply > Structural > Displacement > On Nodes >

Pick Node 1 & 3 > UY > Value [0] > OK


> Pick Node 2 > FY > Value[-20000] > OK


12. General Postprocessor > Element table > Define table > Add by Sequence No >SMISC,2

> Apply > SMISC,6 > Apply > SMISC,8 > Apply > SMISC,12 > OK





Ex.No: 7 SIMPLY SUPPORTED BEAM SUBJECT TO UNIFORMLY DISTRIBUTED

LOAD



Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=4, Y= 0, Z=0

Node 3: X=6, Y= 0, Z=0









10. Preprocessor > Loads > Define Loads > Apply > Pressure > On Beams > Pick Element 1

> I=12000 > J=12000 > OK








Ex.No: 8 SIMPLY SUPPORTED BEAM SUBJECT TO POINT LOAD & COUPLE



Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=2, Y= 0, Z=0

Node 3: X=4, Y= 0, Z=0

Node 4: X=6, Y= 0, Z=0

Node 5: X=8, Y= 0, Z=0













12. Preprocessor > Loads > Define Loads > Apply > Force/Moment > On Nodes > Pick

Node 2 > MZ= 12000 > OK


Node 3 & 5 > FY = -6000 > OK








Ex.No: 9 SIMPLY SUPPORTED BEAM SUBJECT TO POINT LOAD & UDL



Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=1.5, Y= 0, Z=0

Node 3: X=2.5, Y= 0, Z=0

Node 4: X=4, Y= 0, Z=0












Node 2 > FY = -4000 > OK


> I=2000 > J=2000 > OK








Ex.No: 10 CANTILEVER BEAM SUBJECT TO CONCENTRATED POINT LOAD

AT FREE END



Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=5, Y= 0, Z=0






Node 2 > FY = -10000 > OK










Ex.No: 11 CANTILEVER BEAM SUBJECT TO UNIFORMLY DISTRIBUTED LOAD



Procedure:










Node 1: X=0, Y=0, Z=0

Node 2: X=4, Y= 0, Z=0








> I=3000 > J=0 > OK








Ex.No: 12 STRESS ANALYSIS OF A RECTANGULAR PLATE WITH A CIRCULAR

HOLE

Aim: In plate with a hole under plane stress, find deformed shape of hole and determine

maximum stress developed.

Procedure:

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

2. Preference > Structural > OK.

3. Preprocessor > Element Type > Add/Edit/Delete > Add > Solid > Quad 4 Node 42 > OK

> Options > Element Behavior > Plain Stress with thickness > Close

4. Preprocessor > Real Constant > Add/Edit/Delete > Add > Real constant Set No.1 >

Thickness > Value[10] > OK



6. Preprocessor > Modeling > Create > Areas > Rectangle > By dimension > OK

X1=0, X2=200

Y1=0, Y2=100

7. Preprocessor > Modeling > Create > Circle > Solid Circle > OK

X=100: Y=50: R=20



8. Preprocessor > Modeling > Operate > Booleans > Subtract > Areas > Select Rectangle >

OK > Select Circle > OK

9. Preprocessor > Meshing > Mesh Tool > Smart Size > Fine > Mesh > Select Rectangle >

OK > Refine > Select All Elements > 4 > OK


Select all Nodes of Left line of Rectangle > All DOF > Value [0] > OK

11. Preprocessor > Loads > Define Loads > Apply > Structural > Pressure > On Nodes >

Select all Nodes of Right line of Rectangle > Value [-200] > OK

12. Solution > Solve > Current LS > Ok.

13. General Postprocessor > Plot Results > Contour Plot > Nodal Solution > Stress >

Vonmises > OK



Ex.No: 13 CONDUCTION HEAT TRANSFER

Aim: To determine the nodal temperature for the composite wall shown in fig.

Procedure:


2. Preference >Thermal > OK

3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Conduction > OK


area > Value[0.1] > OK

5. Preprocessor > Material Properties > Material Model > Thermal > Conductivity >

Isotropic > KXX = 5 > OK

6. Preprocessor > Material Properties > Material > New Model > Thermal > Conductivity >






A=0.1m2



Node 1: X=0, Y=0, Z=0

Node 2: X=0.1, Y=0, Z=0

Node 3: X=0.2, Y=0, Z=0

Node 4: X=0.3, Y=0, Z=0



10. Preprocessor > Modeling > Create > Elements > Element Attributes > Change Material

No.1 to Material No.2 > Auto Numbered > Thru Nodes > Pick Node 2 and Node 3 > OK



12. Preprocessor > Loads > Define loads > Apply > Thermal > Temperature > On Nodes >

Pick Node 1 > Temp > Value [200] > OK

13. Preprocessor > Loads > Define Loads > Apply > Thermal > Temperature > On Nodes >

Pick Node 4 > Temp > Value[600] > OK


15. General Postprocessor > Plot Results > Contour Plot > Nodal Solution > DOF Solution >

Nodal Temperature > OK

16. General Postprocessor > List Results > Nodal Solution > DOF Solution > Temp > OK



Ex.No: 14 CONVECTION & CONDUCTION HEAT TRANSFER


Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Conduction > Apply >

Convection 34 > OK



5. Preprocessor > Material Properties > Material Model > Convection or Film Co-efficient

> HF = 25 > OK

6. Preprocessor > Material Properties > Material Model > New Model > Thermal >

Conductivity > Isotropic > KXX = 20 > OK





A=1m2





Node 1: X=0, Y=0, Z=0

Node 2: X=0.1, Y=0, Z=0

Node 3: X=0.4, Y=0, Z=0

Node 4: X=0.55, Y=0, Z=0

Node 5: X=0.7, Y=0, Z=0

10. Preprocessor > Modeling > Create > Elements > Element Attributes > Change Element

Link32 to Link34 > Auto Numbered > Thru Nodes > Pick Node 1 and Node 2 > OK


Link34 to Link32 > Change Material No.1 to Material No.2 > Auto Numbered > Thru

Nodes > Pick Node 2 and Node 3 > OK








Pick Node 5 > Temp > Value[20] > OK







Ex.No: 15 CONDUCTION & CONVECTION HEAT TRANSFER


Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > 2D Conduction > Apply >

Convection 34 > OK






Isotropic > KXX = 20 > OK > Convection or Film Co-efficient > HF = 25 > OK

7. Preprocessor > Material Properties > Material > New Model > Thermal > Convection or

Film Co-efficient > HF = 25 > OK



Node 1: X=0, Y=0, Z=0

A=1m2



Node 2: X=0.06, Y=0, Z=0

Node 3: X=0.08, Y=0, Z=0

Node 4: X=0.09, Y=0, Z=0






Link 32 to Link 34 > Change Material No.2 to Material No.3 > Auto Numbered > Thru

Nodes > Pick Node 3 and Node 4 > OK




Pick Node 4 > Temp > Value[-5] > OK







Ex.No: 16 2D CONDUCTVE HEAT TRANSFER

Aim: To determine the temperature distribution for the body shown in fig.

Procedure:



3. Preprocessor > Element Type > Add/Edit/Delete > Add > Solid > Quad 4 Node 55 > OK

4. Preprocessor > Real Constant > No Real Constant


Isotropic > KXX = 1.7307 > OK



7. Preprocessor > Modeling > Create > Elements > Auto Numbered > Thru Nodes > Pick >

OK


Pick Inner Surface > Temp > Value [40] > OK


Pick Outer Surface > Temp > Value [-20] > OK

10. Preprocessor > Loads > Define Loads > Apply > Thermal > Heat Flux > On Nodes >

Pick Top and Bottom Surface > Heat Flux > Value[0] > OK









Ex.No: 17 DYNAMIC ANALYSIS OF FIXED- FIXED BEAM FOR NATURAL

FREQUENCY DETERMINATION

Aim: To determine the natural frequency of a fixed fixed beam shown in figure by modal

analysis. Take E=2x1011

N/mm2 and Density=7830Kg/m

3

Procedure:


2. ANSYS Main Menu > Solution > Analysis type > New Analysis > Modal

3. Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D Elastic > OK

4. Preprocessor > Real Constant > Add/Edit/Delete > Add > Enter the value of C/S area,

Moment of Inertia > Height > OK

5. Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > 2E11

> Density > 7830 > OK > Close

6. Modeling > Create > Nodes > In active CS > Now Enter the Co-ordinates of nodes > OK

7. Modeling > Create > Element > Auto numbered > Thru nodes > Pick node 1 and 3 >

Apply > Similarly pick the next two nodes > Apply

8. Loads > Define loads > Apply > Structural > Displacement > On nodes > Pick node 1

and 2 > All DOF > OK

9. Solution > Analysis type > Analysis option > Select > Subspace > Enter 5 in the number

of modes to extract > Enter 5 in number of nodes to Expand > OK > OK


11. General Post Processor > Results Summary

12. Read Result > First Set

13. Plot result > Deformed shape > Def + Undeformed > OK

14. Read Result > Next set



15. Plot result > Deformed Shape > Def + Undeformed > OK

16. Repeat the above steps for the remaining node shapes.



Ex.No: 18 DYNAMIC ANALYSIS OF FIXED- FIXED BEAM FOR NATURAL

SUBJECTED TO FORCING FUNCTION

Aim: To determine the simple harmonic analysis of a fixed-fixed beam subjected to a cyclic

load as shown in fig. Take E=2.068x1011

N/mm2 and Density=7830Kg/m

3

Procedure:


2. ANSYS Main Menu > Solution > Analysis type > New Analysis > Modal

3. Preprocessor > Element Type > Add/Edit/Delete > Add > Beam > 2D Elastic > OK

4. Preprocessor > Real Constant > Add/Edit/Delete > Add > Enter the value of C/S area,

Moment of Inertia > Height > OK

5. Material Properties > Material Model > Structural > Linear > Elastic > Isotropic > 2E11

> Density > 7830 > OK > Close

6. Modeling > Create > Nodes > In active CS > Now Enter the Co-ordinates of nodes > OK

7. Modeling > Create > Element > Auto numbered > Thru nodes > Pick node 1 and 2 >

Apply > Similarly pick the next two nodes > Apply

8. Loads > Define loads > Apply > Structural > Displacement > On nodes > Pick node 1

and 3 > All DOF > OK


Node 2 > FY = -100 > OK

10. Solution > Load step option > Time or Frequency > Frequency & Sub steps > Harmonic

Frequency Range > [0-300] > No. of Sub steps > [300] > Stepped > OK


12. ANSYS Main Menu > TimeHist Postpro > Variable Viewer > ADD > DOF Solution >

Y-Component > Pick node 2 > OK > Click graph button

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MODELING & ANALYSIS LABORATORY