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• ME5180/6900FiniteElementAnalysis

Chapter15

ThermalStressByAustinScheyer

12/1/2016

• Overview

MotivationHeattransferreviewModeling Formulatethethermalstressproblem Derivetheforcematrix

Onedimensional barelement Twodimensionplanestressandplanestrainelements

Exampleproblem ANSYS

• Motivation

Thermalstressescanoccurinstructuresfortworeasons

Restrictedmovement Differentcoefficientofthermalexpansion

• HeatTransferReview

FouriersLaw Conductiveheatflux

" = )

Newtonlawofcooling Conductiveheatflux

" = + ,

• ThermalStrain

Thermalexpansion,.

Coefficientofthermalexpansion(1/C) T Uniformchangeintemperature(C) L Originallength(m)

Strain,.:

T TL =L

L .

TT TL

= =

1

• MechanicsofMaterialL .

F Restorativeforce(N) E ModulusofElasticity A Crosssectionalarea

Setthe

Solvefortheforcegives:

ThermalStress:

RFLAE

=

R T = ETL FL

A =

F EAT=

TF TA

E = =

L

• OneElementBar

IsotropicmaterialUniformtemperaturechange,T

Force

Where:

Thus:

{ } { }TT TL

= =

{ } [ ] [ ]{ }V

TT Tf B D dV=

[ ] 1 1BL L

= [ ] [ ]D E=

{ }TE

fETATA

=

• 1D ThermalStress

Stiffnessmatrix

1

B

L

2 3

1 2

{ }(1) Ef ETATA

=

{ }(2) Ef ETATA

=

{ }(1) 1 11 12EAkL

=

{ }(2) 1 11 12EAkL

=

• 1D ThermalStress

Step2:ConstructtheglobalstiffnessmatrixWeknowthat

Thus

ApplyingtheboundaryconditionsUsingtheactivestiffnesstosolvefortheremainingdisplacements,thus

1

B

L

2 3

1 2

0{ } [ ]{ }F K d=

1

2

3

1 1 00 1 2 1

20 1 1

E uEA

TAu

LE uTA

=

1 30, 0u u= =

2 0u =

• 1D ThermalStress1

B

L

2 3

1 2

1 1

2 2

3 3

1 1 01 2 1 0 0

20 1 1

x

x

x

F u E EEAF uL

F

TA TA

TAu E TE A

= =

{ } 0[ ]{ } { }F K d F=

Step3:Solveforactualnodalforces Backsubstitutethedisplacementsintotheglobalstiffnessmatrix

For: E=200GPA A=24cm2 L=1.2m =12.5x10-6 (mm/mm)/C

1

2

3

1800180

x

x

x

FF kNF

=

• ConstantStrainTriangle(CST)

Theareaofthetriangleis: = 3+ 5 + A7

Assumeatemperaturefield:3 + 5+ 7Where) isconstant

Theshapefunctionsaredefinedas:3 , =

3 5 , =

5 7 , =

7

Shapefunctionmatrix = 3 5 7

, = []357

35

7

3

3 5

57

7

Heatfluxvector

So

11

231

32

23

NN NT Tx x x x TT NN N

Ty y y y

=

[ ]B

{ }"

"

0"

0x xx

y yy

TKq xq

TKqy

= =

{ }" [ ][ ]{ }q D B T=

{ }T

[ ]D

3

3 5

57

7

• ConstantStrainTriangle(CST)

Applyingenergyprinciple

Theequivalentforcevector

[ ] [ ] [ ][ ] [ ] [ ]V

TT

S

TK B D B dV h N N dS= +

{ } [ ] [ ] " [ ]V S S

T T Tf N QdV N q dS N hT dS= + +

{ } Heat sourceQf =

" 2{ } Heat flux on surface Sqf =3{ } Convection off surface Shf =

{ } { }[ ]TK T f=

• Planestressandplanestrain

PlaneStress

PlaneStrain

{0

TT

} =

( ){ 10

TT

} = +

• ANSYS CircularPipe

Given: Aluminum1100pipe E=69GPa r1=0.2m r2=1.0m =24x10^-6 K=177W/m*K Thickness=0.1m

5

3

5

3

• ANSYS CircularPipe

32Elements

5

3

5

3

Mesh

128Elements 512Elements

• TemperatureDistribution

Specifyboundarycondition

• ThermalStress

Switchtheelementtypefromthermalsolidtostructuralsolid Redefineboundaryconditions Fromthermalanalysis

32Elements 128Elements 512Elements

• Refernces

Logan, Daryl L. A first course in the finite element method. Cengage Learning, 2011.

Chris Wilsons Notes Parsons,R.,etal"INVESTIGATIONOFTHEUSEOFTHEJAVAPROGRAMMINGLANGUAGEFORWEB-BASEDFINITEELEMENTMODELING"

• Questions

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