ME5180/6900FiniteElementAnalysis
Chapter15
ThermalStressByAustinScheyer
12/1/2016
Overview
MotivationHeattransferreviewModeling• Formulatethethermalstressproblem• Derivetheforcematrix
• Onedimensional barelement• Twodimensionplanestressandplanestrainelements
Exampleproblem• ANSYS
Motivation
• Thermalstressescanoccurinstructuresfortworeasons
• Restrictedmovement• Differentcoefficientofthermalexpansion
HeatTransferReview
Fourier’sLaw• Conductiveheatflux
𝑞" = −𝑘𝜕𝑇𝜕𝑥)
Newtonlawofcooling• Conductiveheatflux
𝑞" = ℎ 𝑇+ − 𝑇,
ThermalStrain
Thermalexpansion,𝛿.
• 𝛼 – Coefficientofthermalexpansion(1/°C)• T– Uniformchangeintemperature(°C)• L– Originallength(m)
Strain,𝜀.:
T TLδ α=L
L 𝛿.
TT T
Lα
δε = =
𝜀
𝑇
1
𝛼
MechanicsofMaterialL 𝛿.
• F– Restorativeforce(N)• E– ModulusofElasticity• A– Crosssectionalarea
Setthe
Solvefortheforcegives:
ThermalStress:
RFLAE
δ =
R Tδ δ=E
TL FLA
α =
F EATα=
TF TA
Eσ α= =
L
OneElementBar
IsotropicmaterialUniformtemperaturechange,T
• Force
Where:
Thus:
{ } { }TT T
Lδ
ε α= =
{ } [ ] [ ]{ }V
TT Tf B D dVε= ∫
[ ] 1 1BL L
⎡ ⎤= −⎢ ⎥⎣ ⎦[ ] [ ]D E=
{ }TE
fETATA
α
α⎧ ⎫
= ⎨ ⎬−⎩ ⎭
1D– ThermalStress
Step1:Determinetheelementalcomponents• Appliedthermalloading
• Stiffnessmatrix
1
B
L
2 3
1 2
{ }(1) Ef
ETATAα
α
−⎧ ⎫= ⎨ ⎬⎩ ⎭
{ }(2) Ef
ETATAα
α
−⎧ ⎫= ⎨ ⎬⎩ ⎭
{ }(1) 1 11 12
EAkL
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
{ }(2) 1 11 12
EAkL
−⎡ ⎤= ⎢ ⎥−⎣ ⎦
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
α
α
− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
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
α α
α α
− −⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= − − − =⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭
{ } 0[ ]{ } { }F K d F= −
Step3:Solveforactualnodalforces• Backsubstitutethedisplacementsintotheglobalstiffnessmatrix
• For:• E=200GPA• A=24cm2
• L=1.2m• 𝛼 =12.5x10-6 (mm/mm)/°C
1
2
3
1800180
x
x
x
FF kNF
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪
=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭
ConstantStrainTriangle(CST)
Theareaofthetriangleis:𝐴 = 𝐴3+ 𝐴5 + A7
Assumeatemperaturefield:𝛼3 + 𝛼5𝑥+ 𝛼7𝑦Where𝛼) isconstant
Theshapefunctionsaredefinedas:𝑁3 𝑥,𝑦 =
𝐴3𝐴 𝑁5 𝑥,𝑦 =
𝐴5𝐴 𝑁7 𝑥,𝑦 =
𝐴7𝐴
Shapefunctionmatrix𝑁 = 𝑁3 𝑁5 𝑁7
𝑇 𝑥,𝑦 = [𝑁]𝑇3𝑇5𝑇7
𝑦
𝑥
𝐴3𝐴5
𝐴7
𝑢3
𝑣3 𝑢5
𝑣5𝑢7
𝑣7
ConstantStrainTriangle(CST)Temperaturegradient
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
𝑦
𝑥
𝑢3
𝑣3 𝑢5
𝑣5𝑢7
𝑣7
ConstantStrainTriangle(CST)
Applyingenergyprinciple
Theequivalentforcevector
𝑦
𝑥
[ ] [ ] [ ][ ] [ ] [ ]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
𝑦
𝑥{0
TTε
α
α⎧ ⎫⎪ ⎪
} = ⎨ ⎬⎪ ⎪⎩ ⎭
( ){ 10
TTαε
α
ν⎧ ⎫⎪ ⎪
} = + ⎨ ⎬⎪ ⎪⎩ ⎭
ANSYS– CircularPipe
Given:• Aluminum1100pipe• E=69GPa• r1=0.2m• r2=1.0m• 𝛼 =24x10^-6• K=177W/m*K• Thickness=0.1m
𝑟5
𝑟3
𝑇5
𝑇3
ANSYS– CircularPipe
32Elements
𝑟5
𝑟3
𝑇5
𝑇3
Mesh
128Elements 512Elements
TemperatureDistribution
Procedure• DefineGeometry• Definematerialproperties• Defineelementtype(thermalsolidquad4node55)
• Specifyboundarycondition
ThermalStress
• Switchtheelementtypefromthermalsolidtostructuralsolid• Redefineboundaryconditions• Fromthermalanalysis
32Elements 128Elements 512Elements
Refernces
• Logan, Daryl L. A first course in the finite element method. Cengage Learning, 2011.
• Chris Wilson’s Notes• Parsons,R.,etal"INVESTIGATIONOFTHEUSEOFTHEJAVAPROGRAMMINGLANGUAGEFORWEB-BASEDFINITEELEMENTMODELING"
Questions
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