Lab Manual CAMA

8/17/2019 Lab Manual CAMA

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Course outcomes

After the successful completion of this course, the student will be able to:

1. Model and solve simple problems in structural mechanics, thermal analysis and fluid

flow using a standard FEM software package

Part – A

Study of FEA Package and odeling Stress Analysis:

1. Bars of constant, tapered and stepped cross sections – one eercise each

!. "russes – two eercises

#. Beams – simply supported, cantilever, beams with $%&, beams with varying

loads, etc – si eercises

Part ! "

1. 'ectangular plate with a hole – one eercise

!. "hermal analysis – !% problem with conduction and convection boundary conditions

– two eercises

#. Fluid flow analysis – potential distribution in !% bodies – two eercises

(. %ynamic analysis – one eercise each in)

a. Fied*fied beam for natural fre+uency determination

b. Bar subected to forcing function

c# Fied*fied beam subected to forcing function

Department of Mechanical Engineering, NIEIT, Mysuru. Page


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Contents

Sl#

$o#$ame of the e%periment

Page

no#

1 Chapter &:-ntroduction – getting started with /00 #*2

! Chapter ':Bars of constant, tapered and stepped cross sections 3*14

#Chapter (:Beams – simply supported, cantilever, with $%& and

5arying &oads12*!3

( Chapter ):"russes !6*#7

7 Chapter *:'ectangular plate with a hole #4*#6

4

Chapter +:"hermal analysis –1% 8 !% problem with conduction and

convection boundary conditions (9*(6

2 Chapter : %ynamic nalysis 71*47



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Chapter &

-E../$- S.A0.E1 2/.3 A$S4S

(#& Performing a typical A$S4S analysis

"he /00 program has finite element analysis capabilities, ranging from asimple, linear, static analysis to a comple, non*linear, transient dynamic analysis. "he

analysis guide manual in the /00 documentation describes specific procedures for

performing analysis of different engineering disciplines.

n /00 analysis has maorly three distinct steps)

1. Build the model

!. pply loads and obtain the solution

#. 'eview the results.

"uilding a model:

Building a finite element model re+uires more of an /00 users time than anyother part of the analysis. First, specify a :obname; and :nalysis; title. "hen, use the


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"he element type determines, among other things)

• "he degree*of*freedom ?%CF@ set ?which in turn implies the discipline –

structural, thermal, magnetic, electric, +uadrilateral, brick etc>

• =hether the element lies in two*dimensional ?!%@ or three dimensional ?#%@ space

Eample)* BEAM4 – has si structural %CF ?$D, $, $, 'C"D, 'C", 'C"@ and is a line

element, and can be modeled in #% space.

Plane77 – has a thermal %CF ?"EM


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Creating the odel -eometry:

fter defining material properties, the net step in an analysis is generating a finite

element model*nodes and elements* that ade+uately describe the model geometry.

"here are two methods to create the finite element model)0olid modeling and

%irecteneration.

Solid modeling:%escribe the geometric shape of the model, and then instruct the /00

program to automatically mesh the geometry with nodes and elements. -t is possible to

control the siGe and shape of the elements that the program creates.

1irect generation:%efine the location of each node and the connectivity of each element.

0everal convenience operations, such as copying patterns of eisting nodes and elements,

symmetry reflection, etc.., are available.

'# Apply loads and obtain the solution

-n this step, the 0C&$"-C/ processor is used to define the analysis type and

analysis options, apply loads, specify load step options, and initiate the finite element

solution. &oads can also be applied using the


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/nitiating the solution:!

"o initiate solution calculations, use either of the following)

Aommand?s@) 0C&5E

-8/: ain enu Solution Current S

=hen this command is used, the /00 program takes model and loading information

from the database and calculates the results. 'esults are written to the results file

?Hobname."0", Hobname.'", Hobname.'M or Hobname.'F&@ and also to the database.

"he only difference is that only one set of results can reside in the database at one time,

while one can write all sets of results ?for all substeps@ to the results file.

(# 0e;iew the 0esults:

Cnce the solution has been calculated, use the /00 postprocessors to review

the results.



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(#' -eneral steps for sol;ing problems using A$S4S!&)#*

Step &* Ansys 8tility enu

File *Alear and start new *%o not read file –ok

File *Ahange ob name *Enter new ob name * – Ck

File *Ahange title *Enter new title *yyy –ok

Step ' *Ansys ain enu – Preferences

Select !S.08C.80A –ok

Step (* Preprocessor

Element Type * 0elect type of element from the 5"able; and the re+uired options

Real constants * %etails such as "hickness, reas, Moment of -nertia etc>

Mate!al popet!es ! 'e+uired depending on the nature of the problem

Select "n!ts* Ahoose one of the following types of unit

$0E' * $ser*defined system ?default@

0- * -nternational system ?0- or MI0J m, kg, s, I@

A0 * A0 system ?cm, g, s, KA@

M follow these steps

nsys tool bar 0ave * %B * File * 0ave as GGG.db * ok

Step && *"o :@pen7 the solution geometry etc)

nsys tool bar – 'esume – %B*File *'esume from *GGG.db

ok


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1 50 mm

300 mm

1500 N

E = 2.1*105 N/mm2

2

Chapter )

"A0S: C@$S.A$., .APE0E1 A$1 S.EPPE1 C0@SS SEC./@$S

Problem&: "ar of Constant Circular Cross!Section

%etermine) * 1@ "he nodal displacement !@ 0tress in each element #@ 'eaction forces.

Fig (.1

Sol"t!on#

Step & ! Ansys 8tility enu

File * Alear and start new * %o not read file * ok

File * Ahange ob name * Enter new ob name * * ok

File * Ahange title * Enter new title – yyy * ok Step ' *Ansys ain enu ! Preferences

0elect * 0"'$A"$'& * ok.

Step ( * Preprocessor

Element "ype ! ddEdit%elete – dd * &ink !#% finit stn139– ok * close

'eal constants – dd – ok * real constant set no * 1 – rea * 17999 * ok.

Material properties* Material Model * Material Model /umber 1 – 0tructural –

&inear – Elastic – -sotropic – ED * !.1E7 *


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&oads *%efine&oads – pply – ForceMoment – Cn nodes – pply * pick node ! *

%irection of ForceMoment – FD * ForceMoment value * 1799?Ove value@ *

Aonstant 5alue * pply – ok.

Step + * Ansys ain enu Solution

0olve – 0olve current &0 –ok ?-f everything is ok – solution is done and displayed@* Alose.

Step ! Ansys ain enu-eneral Post processor

Element table – define table – add – "ype N0tressN in the user label item block*

results data item * By se+uence number – select &0, * change to &0, 1 * ok.

Element table – define table – add – "ype N0trainN in the user label item block*

results data item * By se+uence number – select &E


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500 N

2

600mm 500mm

1

3

E2 0.!"#05 N$mm2,%2 600 mm2

Problem ':SteppedCircular Cross!Section "ar

%etermine) * 1@ "he nodal displacement !@ 0tress in each element #@ 'eaction forces.

Fig (.!

Sol"t!on#

Step& – Ansys 8tility enu

File –Alear and start new –%o not read file –ok

File – Ahange ob name –Enter new ob name – –ok

File –Ahange title – Enter new title – yyy * ok.

Step ' – Ansys ain enu –Preferences

0elect – 0"'$A"$'&–ok.

Step ( – Preprocessor

Element "ype – ddEdit%elete – dd – &ink !#% finit stn139* ok – Alose

'eal constants –dd –ok – real constant) set no *1 rea – 699* apply, set no – ! –

cs area – 499 – ok* Alose

Material properties* Material Model – Material Model /umber 1)0tructural –

&inear –Elastic–-sotropic– !.1E7 –


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pick 18! – ok – Element attributes – Element type number) *1 &ink139 *Materiel

no – !*'eal const set no*! – ok* uto numbered – "hru node –pick !8#– ok.

Step * – Preprocessor

&oads –%efine &oads –pply – 0tructural %isplacement – Cn nodes –pick nodes

1* pply –ll %CF –ok.&oads * %efine &oads –pply –ForceMoment – Cn nodes –pick node # – pply*

%irection of ForceMoment –FD* ForceMoment value * 799?Ove value@ – ok.

Step + – Ansys ain enu –Solution

0olve – 0olve current &0 –ok ?-f everything is ok – 0olution is done is displayed@

*Alose.

Step –Ansys ain enu –-eneral post processor

Element table – define table – add – "ype N0tressN in the user label item block*

results data item * By se+uence number – select &0, * change to &0, 1 * ok

Element table – define table – add – "ype N0trainN in the user label item block*

results data item * By se+uence number – select &E


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0ES8.S



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1000 N

500 mm2

750 mm

1000 mm2

Problem (: .apered "ar

For the following tapered bar, find the nodal displacements:

"he cross sectional area decreases linearly from 1999 mm! to 799 mm!. $se two elements.

"ake E P ! L 197

M


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Step ) –Preprocessor

0ection – Beams – Aommon 0ections – 0ub type – 0elect 0olid Aircle.

-% – 1 – /ame – Big – ' – 49 – pply.

-% – 1 – /ame – Big – ' – #9 – Ck,

0ections – Beams – "aper 0ection – By D location – /ew "aper 0ection -% – # /ew 0ection name – "aper – Beginning 0ection -% – 1 – 9, 9, 9 – End 0ection -%

– ! 0mall – 1999, 9, 9.

Modeling –Areate –/odes – -n active A0 – /ode number – 1) D, , – ?9, 9, 9@ –

apply – /ode number – !) D, , – ?1999, 9, 9@ – ok.

Areate – &ines – &ines – 0traight line –


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16/75

1

2

600mm 500mm

3

E2 0.!"#05 N$mm2,%2 600 mm2

600mm

E# E& 2.0"#05 N$mm2, %# %& '00mm2

3

Assignment &:Stepped Cross!Section "ar

"he below stepped bar is subected to an aial force of 1999 / along positive *

ais. nd assume E1 P E#.

Fig (.(


1000 N


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5000 mm

#0 (N

# 2

Chapter (

"EAS: Cantile;er, Simply Supported, 2ith 81, 2ith Darying oads

Note#ssume rectangular cs area of !99mm #99mm, oung;s Modulus of!.1197

/mm! for all the problems on beams.

T!p#For solving the problems, all the values should be in the same units.

Problem &:– Aantilever beam subected to a concentrated load at free end.

Fig 7.1

Solution)

Step&! Ansys 8tility enu

File –Alear and start new –%o not read file –ok File – Ahange ob name –Enter new ob name –probl –ok

File –Ahange title –Enter new title –yyy– ok

Step ' –Ansys ain enu –Preferences

0elect – 0"'$A"$'& –ok

Step ( –Preprocessor

Element "ype–ddEdit%elete –dd –Beam * ! node 133– ok – Alose

Material properties* Material Model – Material Model /umber 1) 0tructural –

&inear – Elastic –-sotropic– !.1E7 –


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Step * –Preprocessor

&oads–%efine &oads –pply – 0tructural %isplacement – Cn Ieypoints –pick

Ieypoint– 1– pply –%CF to be constrained – ll %CF–ok.

&oads * %efine &oads –pply –ForceMoment – Cn Ieypoints –pick node ! – pply* %irection of ForceMoment –F ForceMoment value – *19999 ?*ve

value@*–ok.

Step + –Ansys ain enu –Solution

0olve – 0olve current &0 –ok ?-f everything is ok –solution is done is displayed@*

Alose.

? Note# Cnce 0olution is done is displayed, then only &ist of results will be

generates.@


For "ending oment:

Element table – define table – add – By se+uence number – select –0M-0A, *

change to 0M-0A, # – pply, By se+uence number – select –0M-0A, * change to

0M-0A, 14 * ok.

For Shear!Force:


change to 0M-0A, 4 – pply, By se+uence number – select –0M-0A, * change to

0M-0A, 16 – ok.

Step = – -eneral post processor


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&00 mm

)00 mm

#50 mm

#

2 &

)

)0 (N20 (N

Fig 7.!

Problem ': simply supported beam is subected to concentrated loads. Aompute the

shear force and bending moment for the beam shown and find the reactions at the

supports.

Fig 7.#

Solution:Step&! Ansys 8tility enu

File –Alear and start new –%o not read file –ok.

File – Ahange ob name –Enter new ob name –probl –ok.

File –Ahange title –Enter new title –yyy– ok.


0elect – 0"'$A"$'& –ok.


Element "ype–ddEdit%elete –dd –Beam * ! node 133* ok – Alose.




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Meshing – 0iGe cnrls– Manual siGe – lobal – 0iGe – /o of element divisions – 79

– Ck.

Mesh – &ines – 0elect Iey


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Assignment '# simply supported beam is subected to concentrated loads. Aompute the

shear force and bending moment for the beam shown and find the reactions at the

supports.

Fig 7.(



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2000mmM

)000 mm

#2 (N$m *UDL+

6000 mm

&000mmM

Problem ':0imply supported beam with uniformly distributed load.

Fig 7.7

Solution:




File –Ahange title –Enter new title –yyy *ok


0elect – 0"'$A"$'& –ok Step ( –Preprocessor

Element "ype–ddEdit%elete –dd –Beam * ! node 133* ok – Alose




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&oads * %efine &oads –pply –


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Fig 7.4



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Problem ( – Beam with angular loads, one end hinged and at the other end roller support.

?ll dimensions in mm@.

Fig 7.2

Solution:






0elect – 0"'$A"$'& –ok


Element "ype–ddEdit%elete –dd –Beam * ! node 133* ok – AloseMaterial properties* Material Model – Material Model /umber 1) 0tructural –



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&oads * %efine &oads –pply –ForceMoment – Cn nodes –pick node ! – pply*

%irection of ForceMoment –F– ForceMoment value * *199 ?*ve value@ – pply

* pick node # * pply – %irection of ForceMoment* FD * ForceMoment value *

*!99 ?*ve value@ * pick node ( * pply – %irection of ForceMoment* FD *

ForceMoment value* *!79 ?*ve value@ *ok Step + –Ansys ain enu –Solution


Alose.

? Note# Cnce 0olution is done is displayed, then only &ist of results will be

generates.@


For "ending oment:



0M-0A, 14 * ok

For Shear Force:



0M-0A, 16 – ok



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Fig 7.3

Problem ): Beam with applied moment and over*hanging.

Fig 7.6

Solution:



File – Ahange ob name –Enter new ob name –probl –ok



0elect – 0"'$A"$'& –ok


Element "ype–ddEdit%elete –dd –Beam * ! node 133* ok – Alose

Material properties* Material Model – Material Model /umber 1) 0tructural – &inear – Elastic –-sotropic– !.1E7 –


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&oads * %efine &oads –pply –ForceMoment – Cn Ieypoints –pick node ! –

pply* %irection of ForceMoment – M –ForceMoment value – 4999 ?Ove

value@– pply–pick node #– pply – %irection of ForceMoment * F *

ForceMoment value – *4999 ?*ve value@– pply–pick node 7– pply – %irection

of ForceMoment * F * ForceMoment value – *3999 ?*ve value@–ok Step + –Ansys ain enu –Solution


Alose.


For "ending oment:



0M-0A, 14 * ok.

For Shear Force:



0M-0A, 16 – ok



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&000 mm #500 mm #500 mm

0 (N

)0 (N$m

2 & )

Problem *:Beam subected to uniformly varying load.

Fig 7.19

Solution:






0elect – 0"'$A"$'& –ok


Element "ype–ddEdit%elete –dd –Beam * ! node 133–ok – AloseMaterial properties* Material Model – Material Model /umber 1) 0tructural –



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&oads * %efine &oads –pply –


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1500 mm2

2000 mm22000 mm2

400 mm

400 mm400 mm

15 kN

Chapter )

A$A4S/S @F .08SSES

Problem &: For the three*bar truss, determine the nodal displacements, reactions at thesupports and the stress in each member. "ake modulus of elasticity as !99


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0elect –0"'$A"$'& –ok

Step ( – .o define element:Preprocessor

Element "ype–ddEdit%elete –dd – ink – (1!finit!stn!&=? * ok – Alose.

Step ) –.o define element 0eal constants:Preprocessor

'eal constants –dd – ok – 'eal constant set no – 1*cs area – 1799*ok.

'eal constants –dd – ok – 'eal constant set no – !*cs area – !999*ok.

Step* – .o define element aterial properties:Preprocessor

Material properties* Material Model – Material Model /umber 1 – 0tructural –

Alick – &inear – Alick – Elastic –-sotropic – Alick – !.1E7 – – .o apply oads:Preprocessor

&oads * %efine &oads –pply –ForceMoment – Cn nodes –pick node # – pply *

%irection of ForceMoment –F – ForceMoment value – 17,999?*ve value@ –ok.

Step &? – .o sol;e the problem

nsys Main Menu – 0olution*0olve – 0olve current &0 –ok ?-f everything is ok –

solution is done is displayed@ – Alose.



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Step && – .o interpret the results: -eneral post processor

.o plot the element stresses:

nsys Main Menu – eneral post processor * Element table – %efine table – dd – 'esults data item* By se+uence num * 0elect &0 – Enter*&0, 1 – ok BElement

diagram will be displayed

eneral post processor* Element table*


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Problem ': Aonsider the following (*bar truss and determine i@ Element stiffness matri

for each element ii@ $sing elimination approach, solve for the nodal displacements iii@

Aalculate stresses in each element.

For the given data, find 1@ 0tress in each element !@ 'eaction forces and #@ /odal

displacement. EP!6.7 194

units, P1 units for all elements.



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Q1

Q2

Q7

Q8 20,00

25,000 units

30 units

40 units

4 3

1 2

Q4

Q5

Q6

Fig 4.#

Assignment – "heoretical – Finite Element method calculations ?to be entered in record@



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{ } [ ] units9.99.9!!.!7197.47199.9!2.1!199.99.9T

)nswers

### T −−− −=

unitsunits

unitsunits

(142,!93,7

,339,!1,999,!9

(#

!1

==

−==

σ σ

σ σ






0elect – 0"'$A"$'& –ok

Step ( – .o define element –Preprocessor

Element "ype–ddEdit%elete –dd – ink !(1 finit!stn &=?* ok – Alose

Step )!.o define element 0eal constants – Preprocessor'eal constants –dd – ok – 'eal constant set no – 1*cs*1*ok

Step* – .o define element aterial properties ! Preprocessor

Material properties* Material Model – Material Model /umber 1 – 0tructural –

Alick – &inear – Alick – Elastic –-sotropic – Alick – !.1E7 –


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&oads –%efine &oads –pply – 0tructural %isplacement – Cn nodes –pick nodes 1

8 ( * pply – %CF to be constrained * ll %CF –ok

&oads –%efine &oads –pply – 0tructural %isplacement – Cn nodes ! – pply –

%CF to be constrained –$ –ok

Step > – .o apply oads ! Preprocessor&oads * %efine &oads –pply – ForceMoment – Cn nodes –pick node ! – pply*

%irection of ForceMoment –FD * ForceMoment value * O !9,999?Ove value@ *

pply

&oads * %efine &oads –pply – ForceMoment – Cn nodes – pick node # – pply*

%irection of ForceMoment –F* ForceMoment value –*!7,999?*ve value@ *ok

Step &? – .o sol;e the problem:

nsys Main Menu – 0olution*0olve – 0olve current &0 –ok ?-f everything is ok –

0olution is done is displayed@* Alose

Step && – .o interpret the results ! -eneral post processor

1. "o plot the element stress

nsys Main Menu –eneral post processor * Element table – %efine table – dd –

'esults data item* By se+uence num * 0elect &0 – Enter*&0, 1 – ok BElement

stress diagram will be displayed

nsys Main Menu –eneral post processor* Element table*


38/75

P0/$. 8 $@1A S@8./@$ PE0 $@1E

P@S.& $@1A 1E-0EE @F F0EE1@ /S./$-

@A1 S.EP G & S8"S.EP G & ./E G ???? @A1 CASE G ?

.3E FA@2/$- 1E-0EE @F F0EE1@ 0ES8.S A0E /$ .3E -@"A C@@01/$A.E S4S.E

$@1E 8

& ?#????

' ?#'&&>'E!?&

( ! ?#*+)>E!?'

) ?#????

A/8 A"[email protected] DA8ES

$@1E '

DA8E ?#'&&>*E!?&

Problem (: Aonsider the (*bar truss shown in figure. For the given data, find 1@ 0tress in

each element !@ 'eaction forces and #@ /odal displacement. iven EP !194 k/mm!,

P19 mm!.


P0/$. 8 $@1A S@8./@$ PE0 $@1E

P@S.& $@1A 1E-0EE @F F0EE1@ /S./$-

@A1 S.EPG& S8"S.EPG& ./EG???? @A1 CASEG ?

.3E F@@2/$- ,4,H S@8./@$S A0E /$ .3E -@"A C@@01/$A.E S4S.E

$@1E F F4

& !')&+ !(&'*#?

' !'&=*#

) )&++# ?#????

A/8 DA8ES

DA8E !'???? !'*???


39/75

2500 N4000 N

1000 mm

4 3

1 2

800 mm

50000 N

Fig 4.(

ssignment – "heoretical – Finite element method calculations ?to be entered in record@

Alass work – Aarry out the /00 software steps to solve and enter the same in the

record

Problem number ): Aonsider the (*bar truss shown in figure. For the given data, find 1@

0tress in each element !@ 'eaction forces and #@ /odal displacement. iven EP !19 4

k/mm!, P19 mm!.



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200

#000000

Fig 4.7

Assignment – "heoretical – Finite element method calculations.

Chapter *

0EC.A$-8A0 PA.E 2/.3 A 3@E BS.0ESS A$A4S/S

Problem &: plate of !99mm 199mm siGe and 19mm thick has a centrally drilled hole

of (9mm diameter subected to a pressure of !999/mm ! under plane stress condition,

find deformed shape of the hole and plot maimum stress distribution. "ake EP !E7,UP9.#

Fig 2.1

Solution#


!999

/mm!

V (9 mm


41/75





Step ' – Ansys ain enu – Preferences0elect – 0"'$A"$'& –ok

Step ( – Preprocessor

Element "ype–ddEdit%elete – dd – 0olid*+uad (node*13!* ok – Cptions –

Element

behavior I# – 0elect –


42/75

ist results

1. /odal solution –%CF solution*&& %CFs –ok* 5alue displayed – /odal solution

– 0tresses* 5on*mises stress*ok*values displayed

Note# T$e es"lts o%ta!ne& t$o"'$ Ansys s$o"l& %e (al!&ate& )!t$ t$e analyt!cal Met$o&*



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Problem ':"he corner angle bracket is shown below. "he upper left hand pin*hole is

constrained around its entire circumference and a tapered pressure load is applied to the

bottom of lower right hand pin*hole. Aompute Maimum displacement, 5on*Mises stress.

Fig 2.!


File –Alear and start new –%o not read file –ok File – Ahange ob name –Enter new ob name – –ok


Step ' ! Ansys ain enu – Preferences

0elect – 0"'$A"$'& – ok

Step ( ! Preprocessor

Element type – ddEdit%elete – dd – 0olid – Tuad 3 node – 13! – ok – Cption

– Element behavior I# –


44/75

Areate – rea – Aircle – 0olid circle – D, radius – 9, 1!.7, 7 – pply – D,,

radius, *32.7 *27, 7 – ok.

Cperate – Booleans – 0ubtract– reas – pick area which is not to be deleted

?bracket@ – pply – pick areas which is to be deleted ?pick two circles@ – ok

Meshing – Mesh tools – Mesh areas – +uad – free – mesh – pick all – ok. Meshtools – 'efine – pick all – level of refinement – # – ok

Step * ! Preprocessor

&oads– %efine loads – pply – 0tructural – %isplacement – Cn lines – 0elect the

inner lines of the upper circle – pply – %CFs to be constrained – && %CF – ok

&oads– %efine loads – pply – 0tructural –


45/75

# 2 & )

0.# m 0.# m 0.# m

t# t2 t& t)

Chapter +

.3E0A A$A4S/S

'1 Problem 2ith Conduction and Con;ection "oundary Conditions

eneal Steps to %e +ollo)e& )$!le sol(!n' a $eat tans+e po%lem - "s!n' .EM#

1. %iscretiGe and select the element type

!. Ahoose a temperature function.

#. %efine the temperature gradienttemperature and heat flutemperature gradient

relationships.

(. %erive the element conduction matri and e+uations by using either variation

approach or by usingalerkin;s approach

7. ssemble the element e+uations to obtain the global e+uations and introduce

boundary conditions.4. 0olve for the nodal temperatures

2. 0olve for the element temperature gradients and heat flues

Problem For the composite wall idealiGed by the 1*% model shown in figure below,

determine the interface temperatures. For element 1, let I 1 P 7 =m9A, for element !, I !

P 19 =m 9A and for element #, I # P 17 =m9A. "he left end has a constant temperature

of !99 9A and the right end has a constant temperature of 499 9A.

Fig 3.1

Solution:

. "heoretical approach * Finite element method

iven)

Finite Element model)?For "heoretical Method refer any FEM tet book@

Department of Mechanical Engineering, NIEIT, Mysuru. Page"heoretical results

t1 P !99 9A t( P 499 9A P 9.1 m!

I 1P 7 =m9A I !P 19 =m

9A I #P 17 =m9A


46/75

{ } [ ]A499A7!2.#A(13.!A!99t 9999" =

Analysis using A$S4S &)#*Step &! Ansys8tility enu

File *Alear and start new *%o not read file *ok

File *Ahange ob name *Enter new ob name *probl –ok

File *Ahange title *Enter new title *yyy –ok

Step ' !Ansysain enu !Preferences

0elect *"E'M& *ok

Step (!Preprocessor

Element "ype*ddEdit%elete *dd* &-/I *#% AC/%$A"-C/##

* ok * Alose'eal constants* ddEdit%elete*dd* "ype 1* &-/I * 'eal constant set no 1 –

A0 area – Enter *9.1*ok* Alose

Material properties– Material models – Material model number 1* "hermal –

conductivity – -sotropic * Aonductivity for material * "hermal conductivity ?I @ –

7 *ok * Material – /ew model – %efine material -%*!*ok * "hermal – conductivity –

-sotropic * Aonductivity for material * "hermal conductivity ?I @ –19 –ok *

Material – /ew model – %efine material -%*#*ok * "hermal – conductivity –

-sotropic * Aonductivity for material * "hermal conductivity ?I @ –17 –ok* Alose

Step ) ! Preprocessor

Modeling *Areate */odes * -n active A0 * ?9, 9, 9@ ?,y,G@ ?,y value w.r.t first

node@ * pply ?First node is created@* 9.1,9,9 * pply ?0econd node is created@ *

?9.!,9,9@ – pply ?"hird node is created@*?9.#,9,9@ * pply ?Fourth node is

created@ *ok

Areate – Elements* Element attributes* 0elect material number*1* 'eal constant

set number*1* uto numbered – "hru nodes – pick 1, ! ?Element 1 is created@*ok

* Element attributes* 0elect material number*!* 'eal constant set number *1*

uto numbered – "hru nodes – pick !, # ?Element ! is created@*ok * Element

attributes* 0elect material number*#* 'eal constant set number*1* uto numbered –

"hru nodes – pick #, ( ?Element # is created@*ok Step * –Preprocessor

&oads * %efine &oads * pply – "hermal – 0elect temperature – Cn nodes*pick

node 1* pply* 0elect–"EM


47/75

ist of results

1. /odal solution * %CF solution * "emp *ok B.emperature at all the nodes will

be displayed

!.


48/75

(# (&

0.& m 0.#5 m 0.#5 m

h,

T0 20 0-

K1 = 20 W/m 0C , K2 = 30 W/m 0C, K3 = 50 W/m 0C,

= 25 W/m20C , = 800 0C

h ,

# 2 & )

0.& m 0.#5 m 0.#5 m

t#t2 t& t)

Element # Element 2 Element &

Problem '. composite wall consists of three materials as shown. "he outer temperature

is "9 P !99A. Aonvection heat transfer takes place on the inner surface of the wall with

A399" 9=∞

and h P !7 = m!9A. %etermine the temperature distribution in the wall.

Fig 3.!

Solution:

. "heoretical approach * Finite Element model

"aking P 1 m!, the element conduction matrices are,

For element 1, we must consider the convection from the free end ?left end@

{ } [ ]A!9 A1(.27A116.97A#9(.24t 9999" =

Alass work) Aarry out the /00 software steps to solve and enter the same in the record


.heoretical results


49/75

t

W

L

235 0C

h ' $m20-, 20 0-

/ #0 cm 0.# m

t 0.# cm # #01& m

# m

t)t&t2t#

h ' $m20-, 20 0-

2&5 0-

# 2 & )

0

Problem (: ?@ne dimensional finite element formulation of fin@

metallic fin, with thermal conductivity k P #49 =m9A, 9.1 cm thick, and 19 cm long,

etends from a plane wall whose temperature is !#7 9A. %etermine the temperature

distribution and amount of heat transferred from the fin to the air at !9 9A with h P 6

=m!9

A "ake the width of fin to be 1 m.

Fig 3.#

Solution:

. "heoretical calculations and Finite element model)?For detailed steps refer any FEM

tet book@

Fig 3.(

"heoretical results

A169.#6tA,167.12tA,!96.22tA?given@,!#7t 9(9

#

9

!

9

1 ====

"he total heat loss in the fin can be calculated as,

∑=

= #

1eelementtotal

TT

nd

∞−= t t AhQ

avg selement


L


50/75

"herefore,

[ ] [ ]

=1!1.(#1!9!

!96.22!#71

#9

1!6

t!

tt=!&htthT !11avgs1element

=

−+

=

−+

=−= ∞∞

=(3!.961!9!

167.12!96.221

#9

1!6T

!element =

−

+

=

=443.19#!9!

169.#6167.121

#9

1!6T

#element =

−

+

=

=##(.73119#.443196.(3!1!1.(#1TT#

1eelementtotal

=++==∑=


Note# T$e es"lts o%ta!ne& t$o"'$ Ansys s$o"l& %e (al!&ate& )!t$ t$e analyt!cal

Met$o&*



51/75

T #00 0-

L = 0.4 m 5000 $m2

T #00 0-

/ 0.# m

5000 $ m2

/ 0.# m / 0.# m / 0.# m

t# t2 t& t) t5

Problem )# "he fin shown in figure is insulated on the perimeter. "he left end has a

constant temperature of 199 9A. positive heat flu +LP 7999 =m! acts on the right end.

&et I P 4 =m9A and A0 area P9.1 m!. %etermine the temperatures at

L

4

L

2

3 L

4 and

&, where & P 9.( m.

Fig 3.7

Solution:

. "heoretical calculations or Finite Element model

Fig 3.4

.heoretical results) "he temperatures are,

{ } [ ]A(##.!62A#(6.64A!44.4#A13#.#A199t 99999" =



-i;en: t& G &??? C, J %% G + 2 < m

? C, A G ?#& m'

& G ' G ( G ) G ?#& m


52/75

Problem *:%etermine the temperature distribution and the rate of heat flow N+N per meter

of the height for a tall chimney whose cross section is shown below, #*% view a P ( m and

b P ! m. ssume that the inside gas temp is "g P #11 I, the inside convection co*efficient

is hi, the surrounding air temp is "a P !77 I and the outside convection coefficient is ho.

iven * Element type *"hermal solid element *


53/75

Step & ! Ansys8tility enu

File *Alear and start new *%o not read file *ok

File *Ahange ob name *Enter new ob name *probl –ok File *Ahange title *Enter new title *yyy –ok

Step ' !Ansysain enu !Preferences

0elect *"E'M& *ok

Step ( !Preprocessor

Element "ype*ddEdit%elete *dd* 0C&-% *T$% ( /C%E *77* ok * Alose

'eal constants*/o real constants

Material properties– Material models – "hermal – Aonductivity – -sotropic *

Aonductivity for material * "hermal conductivity ?I @ –1.2#92 *ok* Alose

Step ) ! Preprocessor

Modeling *Areate */odes * -n active A0 *,y,G location in A0*1,9 ?,y value

w.r.t first node@ * pply ?First node is created@* 1.7,9 * pply ?0econd node is

created@ *!,9 * pply ?"hird node is created@ *1,9.7 * pply ?Fourth node is

created@*1.7,9.7 * pply ?Fifth node is created@*!, 9.7 * pply?0ith node is

created@* 1,1 * pply ?0eventh node is created@ *1.7, 1 * pply ?Eighth node is

created@ *! , 1 * pply ?/inth node is created@* 1.7,1.7 * pply ?"enth node is

created@ * !, 1.7 * pply ?Eleventh node is created@*!, ! * pply ?"welfth node is

created@ –ok

Areate – Elements*uto numbered – "hru nodes – pick 1, !,7 8 ( ?nticlockwise,

element 1@* pply* pick !, #, 4 8 7 ?Element !@ * pply* pick (, 7, 3 8 2 ?Element

#@* pply*pick 7, 4, 6 8 3 ?Element (@ * pply * pick 2, 3 8 19 ?Element 7@*

pply*pick 3, 6, 11 819 ?Element 4@* pply *pick 19, 11 8 1! ?Element 2@*ok

?total seven elements ate created through nodes@

Step * –Preprocessor

&oads * %efine &oads * pply – "hermal *Aonvection * Cn nodes *pick inner

surface by bo option * pply* Film co*efficient ?-nner@ *43.1( * "emperature ?t

inner surface@* #11 ?value@*ok

Aonvection * Cn nodes *pick outer surface by bo option * pply* Film coefficient

?outer@ *12.9(* "emperature ?t inner surface@ – !77 ?value@ * Ck Step + !Ansys ain enu –Solution

0olve * 0olve current &0 *ok ?-f everything is ok, *solution is done and displayed@

– Alose

Step !Ansys ain enu ! -eneral Post Processor


54/75



55/75

0esults:



56/75

).0 cm ).0 cm

5 cm40 0C !20 0C

1 2 3 4 5

6 7 8 " 10

11 12 13 14 15

0.& m 0.2 m 0.& m

0.) m&00 o-

&00 o-&00 o-0.2 m

Problem +# For the body shown in figure, determine the temperature distribution. "he

body is insulated along the top and bottom edges, I P I yy P 1.2#92 =m9A. /o internal

heat generation is present.

Fig 3.3


Problem # For the two*dimensional stainless*steel body shown below, determine the

temperature distribution. the left and right sides are insulated. the top surface is subected

to heat transfer by convection. the bottom and internal portion surfaces are maintained at

#99 oA. "he thermal conductivity of stainless steel P 14 =moA.

Fig 3.6



T ∞=40 oA

P !o


57/75

0.5 m 0.5 m

100 mm250 mm2

#$t%

Chapter

1ynamic Analysis:


58/75

11 1

! !

!1 !

! #

/ow writing the global %CF numbers for the row and column for both the elemental

stiffness and mass matrices ?E+s ?1@ and ?!@@ and assembling, we get global stiffness and

mass matri as

1 ! # 1 ! #

#

!

1

19!19!9

19!19419(

919(19(

22

222

22

−

−−

−

=

x x

x x x

x x

K

and

#

!

1

947!7.99#!4#.99

9#!4#.916727.9947!7.9

9947!7.91#97.9

= M

But we have the characteristic e+uation as

9=− M K λ

, where

!ω λ =

0ubstitution for K and M yields,

9

947!7.99#!4#.99

9#!4#.916727.9947!7.9

9947!7.91#97.9

19!19!9

19!19419(

919(19(

22

222

22

=

−

−

−−

−

λ

x x

x x x

x x

9

947!7.919!9#!4#.919!9

9#!4#.919!16727.9194947!7.919(

9947!7.919(1#97.919(

22

222

22

=

−−−

−−−−−

−−−

λ λ

λ λ λ

λ λ

x x

x x x

x x

pplication of Boundary condition?s@) 0ince node 1 is fied, we have X 1 P 9. "herefore,

modifying the above e+uation using elimination method i.e. eliminate 1st

row and 1st

column of the above e+uation we get,

9947!7.919!9#!4#.919!

9#!4#.919!16727.919422

22

=−−−

−−−

λ λ

λ λ

x x

x x

***** ?@

"herefore,

9@9#!4#.919!?@947!7.919!@?16727.9194? !222 =−−−−− λ λ λ x x x

Cn solving,



59/75

91967.41927.4143!

=+− x xλ is the characteristic e+uation.

"he roots of the above characteristic e+uation give the Eigenvalues. "herefore,

Eigenvalues are

31 19997.1 ×=λ

and

3! 193.4 ×=λ

"herefore, the first natural fre+uency is

199!(19997.1 31 =×=ω radsec or

7.1767!1

1 ==π

ω

G.

0econd natural fre+uency is

3.4924.!193.4 3! =×=ω radsec or

#.(179!

!! ==

π

ω

G.

Eigen vectors or mode shapes)

fter modification for the boundary condition, E+. ?@ can be written as

9947!7.919!9#!4#.919!

9#!4#.919!16727.9194

#

!22

22

=

−−−

−−−i

i

ii

ii

X

X

x x

x x

λ λ

λ λ

*** ?#@First Mode 0hape)

"ake any one e+uation in the above matri and substitute for

31 19997.1 ×== λ λ i

=e get first mode shape ?i!1".

[email protected]#!4#[email protected]?4 1#321

!32 =+−××−× X x x x X

1

#

1

! ##.!9##.( X X

= or

722.91

#

1! =

X

X

/ormaliGation of the above mode is obtained by fiing the value of

1! X

P 1 ?a unity@

"hen, first mode shape will be,

{ } { }1722.91#1! =T

X X

0econd Mode 0hape)



60/75

X

x

# 2 &

0.5!!#

X

x#

2

&0.5!!

#

#000mm

0.0# m

0.0# m

0ubstitution of

3! 193.4 ×== λ λ i

into any one e+uation of E+. ?@ yields second

mode shape.

[email protected]#!4#[email protected]?4 !

#

32!

!

32 =+−××−× X x x x X

"he above e+uation gives second mode shape as

{ } { }1722.9!#!! −=T

X X

First Mode

0econd Mode

Problem '#Fied*fied beam for natural fre+uency determination

Fig 3.1!

iven) Modulus of elasticity, E P !.9431911 /m!


61/75

Step ' ! 1efine Element .ypes


62/75

Fig 3.1#

0elect the Subspace method.

Enter 7 in the :/o. of modes to etract;

Aheck the bo beside :Epand modes shapes;

Enter 7 in the :/o. of modes to epand;

Alick :CI;. "he following window will then appear

Fig 3.1(

Ieep default options in the above window and click on :CI;.

Step ' ! 0olution * %efine &oads * pply * 0tructural * %isplacement * Cn Iey points –

pick 18! – %CF to be constrained – ll %CFStep ( ! 0olution * 0olve * Aurrent &0



63/75

Post processing

Step &! eneral


64/75

The irst mode shape is no$ appear in the graphic $indo$#

Fig 3.12

"o view the net mode shapes, select

eneral


65/75

#.0 m

0.0# m

0.0# m

0.5 m3*t+

Problem (# armonic nalysis of a Fied*Fied Beam

Fig 3.16

iven) Modulus of elasticity, E P !.9431911 /m!

.

%or this problem& the 'EAM( )'eam *+ elastic" element is used#

Step ( ! 1efine Element .ypes


66/75

-n the window that appears, enter the following material properties

oung;sModulus, ED * !.943e11


67/75

Fig 3.!9

4 0elect the Full solution method, the 'eal O imaginary %CF printout format

4 %o not use lumped mass approimation.

4 Alick :CI;

"he following window will then appear

Fig 3.!1

Ieep default options in the above window and click on :CI;.



68/75

!. pply Aonstraints

0olution * %efine &oads * pply – 0tructural – %isplacement * Cn /odes

4 Fi all %CFs constraints on Iey point 1 and Iey point !

pply &oad

0olution * %efine &oads * pply* 0tructural * ForceMoment * Cn /odes4 0elect the node at mid*point of the beam ?i.e. at P9.7@

4 "hen the following window will appear.

4 Fill it as shown to apply a load with real value of 199 and an imaginary

value of 9 in the positive :y; direction.

Fig 3.!

(. 0et the Fre+uency 'ange

0olution * &oad 0tep Cpts * "imeFre+uency *Fre+ and *0ubstps . . .

s shown in the window below, specify a fre+uency range of 9 – #99 G, #99 sub

steps and stepped b.c.



69/75

Fig 3.!#

7. 0olve the 0ystem

0olution * 0olve * Aurrent &0

Post processing

@pen the .ime3ist Processing BP@S.'+ enu:'# 1efine Dariables:

"imeist


70/75

Fig 3.!(

0elect dd ?the green :O; sign in the upper left corner@ from this window and thefollowing window will appear.



71/75

Fig 3.!7

4 =e are interested in the /odal solution * %CF 0olution – *Aomponent.

4 Alick CI.

4 raphically select node ! when prompted and click CI.

Fig 3.!4

(# ist Stored Dariables:

-n the :"ime istory 5ariable; window, click the :&ist; button ?#rd buttons to the left

of :dd; button@.



72/75

Fig 3.!2

)# Plot 84 ;s# FreLuency

-n the :"ime 5ariable; window click the :


73/75

"o get a better view of the response, view the log scale of $.

0elect $tility Menu –


74/75

Fig 3.!6

Fig 3.#9.



75/75

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