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TRANSCRIPT
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2.094—
Finite
Element
Analysis
of
Solids
and
Fluids
—Fall‘08—
MITOpenCourseWare
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Contents
1
Large
displacement
analysis
of
solids/structures
3
1.1 Project Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Large Displacement analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.1 Mathematical model/problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.2 Requirementstobefulfilledbysolutionattimet . . . . . . . . . . . . . . . . . . . 5
1.2.3 Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.4 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2
Finite
element
formulation
of
solids
and
structures
7
2.1 Principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3
Finite
element
formulation
for
solids
and
structures
10
4
Finite
element
formulation
for
solids
and
structures
14
5 F.E.displacement formulation,cont’d 19
6
Finite
element
formulation,
example,
convergence
23
6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.1 F.E. model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6.1.2 Higher-order elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
7
Isoparametric
elements
28
8
Convergence
of
displacement-based
FEM
33
9
u/ p
formulation
37
10
F.E.
large
deformation/general
nonlinear
analysis
41
11
Deformation,
strain
and
stress
tensors
45
12TotalLagrangian formulation 49
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MIT2.094 Contents
13
Total
Lagrangian
formulation,
cont’d
53
14
Total
Lagrangian
formulation,
cont’d
57
15
Field
problems
61
15.1 Heat transfer
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
15.1.1 Differential formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
15.1.2 Principle of virtual temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
15.2 Inviscid, incompressible, irrotational flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
16
F.E.
analysis
of
Navier-Stokes
fluids
65
17
Incompressible
fluid
flow
and
heat
transfer,
cont’d
71
17.1 Abstract body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.2 Actual 2D problem (channel flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
17.3 Basic equations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
17.4 Model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
17.5 FSI briefly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
18
Solution
of
F.E.
equations
76
18.1 Slender structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
19
Slender
structures
81
20
Beams,
plates,
and
shells
85
21
Plates
and
shells
90
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture1- Largedisplacementanalysisof solids/structures
Prof. K.J.Bathe MITOpenCourseWare
1.1 ProjectExample
Physicalproblem
Reading:
Ch. 1 in
thetext
“Simple”
mathematical
model
analyticalsolution•
F.E.solution(s)•
More
complex
mathematical
model
holesincluded•
largedisp./largestrains•
F.E.solution(s)• ⇒
Howmanyfiniteelements?•
Weneedagooderrormeasure(especiallyforFSI)
“Even
more
complex”
mathematical
model
The“complexmathematicalmodel”includes FluidStructureInteraction(FSI).
YouwilluseADINAinyourprojects(andhomework)forstructuresandfluidflow.
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MIT2.094 1.Largedisplacementanalysisof solids/structures
1.2 LargeDisplacementanalysis
Lagrangianformulations:
• TotalLagrangianformulation
•
Updated
Lagrangian
formulation
Reading:Ch. 6
1.2.1
Mathematical
model/problem
Given theoriginalconfigurationof thebody,
thesupportconditions,theappliedexternal loads,theassumedstress-strainlaw
Calculate thedeformations,strains,stressesof thebody.
Question Isthereauniquesolution? Yes,forinfinitesimalsmalldisplacement/strain. Notnecessarily
forlargedisplacement/strain.
Forexample:
Snap-through
problem
Thesame load. Twodifferentdeformedconfigurations.
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MIT2.094 1.Largedisplacementanalysisof solids/structures
Column problem,statics
NotphysicaltR is in“direction”of bendingmoment Not inequilibrium.
⇒
1.2.2
Requirements
to
be
fulfilled
by
solution
at
time
t
I. Equilibriumof stresses (Cauchystresses, forcesperunitarea in tV and on tS f )withtheappliedbodyforcestf B andsurfacetractionstf S f
II. Compatibility
III. Stress-strain law
1.2.3 FiniteElementMethod
I. Equilibriumconditionmeansnow
• equilibriumatthenodesof themesh
• equilibriumof eachfiniteelement
II. Compatibilitysatisfiedexactly
III. Stress-strain lawsatisfiedexactly
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MIT2.094 1.Largedisplacementanalysisof solids/structures
1.2.4 Notation
Cauchystresses(forceperunitareaattimet):
tτ ij
i,
j= 1,2,3 tτ ij
= tτ ji
(1.1)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture2- Finiteelementformulationof solidsandstructures
Prof. K.J.Bathe MITOpenCourseWare
Reading:
Ch. 1,Sec.AssumethatontS u
thedisplacementsarezero(andtS u
isconstant). Needtosatisfyattimet: 6.1-6.2
Equilibrium of Cauchystressestτ ij
withapplied loads•
tτ T = tτ 11
tτ 22
tτ 33
tτ 12
tτ 23
tτ 31
(2.1)
(Fori= 1,2,3)
tτ ij,j +tf i
B = 0 intV (sumover j) (2.2)
tτ ij
tnj
= tf iS f
on tS f
(sumover j) (2.3)
(e.g. tf iS f
= tτ i1
tn1
+ tτ i2
tn2
+ tτ i3
tn3
) (2.4)
And: tτ 11tn1 +
tτ 12tn2 =
tf 1S f
• Compatibility Thedisplacementstui needtobecontinuousandzeroontS u.
Stress-Strain
law •
tτ ij = functiontuj (2.5)
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MIT2.094 2.Finiteelementformulationof solidsandstructures
2.1 Principleof VirtualWork∗
t V
tτ ij
teij
d
tV =
tV
tf Bi
ui
d
tV +
t S f
tf S f
i uS f
i dtS f
(2.6)
where
teij
withui
tS u
=
=
1
2
∂ui
∂
txj
0
+
∂uj
∂
txi
(2.7)
(2.8)
2.2 Example
Assume“planesectionsremainplane”
Principle
of
Virtual
Work
tV
tτ 11 te11 dtV
=
tV
tf B1 u1 dtV
+
tS f
tP r uS f 1
d
tS f (2.9)
Derivation
of (2.9)
tτ 11,1 +tf B1
=0 by(2.2)�
tτ 11,1
+ tf B1
u1
=0
∗orPrincipleof VirtualDisplacements
(2.10)
(2.11)
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MIT2.094 2.Finiteelementformulationof solidsandstructures
Hence,
tτ 11,1 +tf 1
B u1 dtV =0 (2.12)
tV
tτ u1,1tτ 11
d V + u1 dtV =0 (2.13) 11u 1
t
t
S
S
u
f
− tV t
tV
tf 1
B
Sf tτ tS
te11 u1 11 f
where tτ 11|tS f
=tP r.
Thereforewehave
te11tτ 11
d
tV = u1
tf 1B
d
tV +u
S
1
f tP rtS f (2.14)
tV
tV
From(2.12)to(2.14)wesimplyusedmathematics. Hence,if (2.2)and(2.3)aresatisfied,then(2.14)musthold. If (2.14)holds,thenalso(2.2)and(2.3)hold!
Namely,from(2.14)
t
tS f
t B t S f u1,1
tτ 11d V = u1tτ 11
tS u− u1
tτ 11,1d V = u1tf 1
d V +u1tP r
tS f
(2.15)tV
tV
tV
or
B t
S f u1tτ 11,1 +
tf 1 d V +u1
tP r −tτ 11
tS f =0 (2.16)tV
Nowletu1
=x 1− txL
tτ 11,1
+tf 1
B ,where tL=lengthof bar.
Hencewemusthavefrom(2.16)
tτ 11,1 +tf 1
B =0 (2.17)
andthenalso
tP =tτ (2.18)r
11
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture3- Finiteelementformulationforsolidsandstructures
Prof. K.J.Bathe MITOpenCourseWare
Reading:
Sec. 6.1-6.2Weneedtosatisfyattimet:
Equilibrium •
t∂ τ ij +tf B = 0 (i= 1,2,3)intV (3.1)t i∂ xj
tτ ij nt
j = f t S f
i
(i=1,2,3)ontS f (3.2)
• Compatibility
• Stress-strain law(s)
Principleof virtualdisplacements
tV
tτ ij
teij
dV t =
tV
uitf Bi
dV t +
tS f
ui|tS f
f t S f
i
dS t f
(3.3)
teij = 2
1
∂
∂ t
u
xi
j
+∂
∂utx
j
i
(3.4)
If (3.3)holdsforanycontinuousvirtualdisplacement(zeroon tS ),then(3.1)and(3.2)holdand• uviceversa.
RefertoEx. 4.2 inthetextbook. •
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MIT2.094 3.Finiteelementformulationforsolidsandstructures
Major
steps
I. Take(3.1)andweighwithui:
tτ ij,j +tf i
B ui = 0. (3.5a)
II. Integrate(3.5a)overvolumetV :
tτ ij,j +tf i
B ui dtV =0 (3.5b)
tV
III. Usedivergencetheorem. Obtainaboundarytermof stressestimesvirtualdisplacementsontS =tS u ∪
tS f .
IV. But,ontS u
theui
=0andontS f
wehave(3.2)tosatisfy.
Result: (3.3).
Example
tτ te11
d V =
S tf
ui
f 1 dS f t
S f t (3.6)11
t
tV
One
element
solution:
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u(r) =
tu(r) =
u(r) =
1 1(1+r)u1
+2 21
(1+r) tu1
+21 1
(1
+
r)
u1 +2 2
(1−r)u2
1
2(1−r) tu2
(1
−
r)
u2
(3.7)
(3.8)
(3.9)
Supposeweknowtτ 11,tV ,tS f ,
tu... use(3.6).
Forelement1,
te11
=∂u
=B(1)u1
(3.10)∂
tx u2
T t for el. (1) B(1)T
tV te11t
τ 11d V −→ [u1 u2]
t V
tτ 11 dtV
(3.11)
=
tF (1)
for
el.
(1)
tF
(1)−→
[u1
u2] (3.12)tF ̂(1)
=U 1 U 2 U 3 (3.13)
0
u2 u1
where
tF ̂1(1)
= tF 2(1)
(3.14)
tF ̂2(1)
= tF 1(1)
(3.15)
Forelement2,similarly,
0
=U 1
U 2 U 3 F (2)
(3.16)
t ˆ
u2 u1
R.H.S.
(unknownreactionatleft)
U 1
U 2
U 3
0 (3.17)
t S f tS f
U T
f
· 1
Nowapply,
U
T =
1 0 0
(3.18)
then,
U T
= 0 1 0 (3.19)
then,
U
T
= 0 0 1 (3.20)
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MIT2.094 3.Finiteelementformulationforsolidsandstructures
Thisgives,
tF̂
(1)
0
+
0tF̂
(2)
=
unknownreaction0
f
t
tS f 1
·tS f
(3.21)
We
write
that
as
tF = tR �
(3.22)tF
=fn tU 1,tU 2,
tU 3
(3.23)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture4- Finiteelementformulationforsolidsandstructures
Prof. K.J.Bathe MITOpenCourseWare
Weconsideredageneral3Dbody, Reading:Ch. 4
Theexactsolutionof themathematicalmodelmustsatisfytheconditions:
• Equilibrium withintV andontS f ,
Compatibility •
Stress-strain
law(s)
•
I. Differentialformulation
II. Variationalformulation(Principleof virtualdisplacements)(orweakformulation)
WedevelopedthegoverningF.E.equationsforasheetorbar
Weobtained
t
F
t
R =
(4.1)
wheretF isafunctionof displacements/stresses/materiallaw;andtR isafunctionof time.
Assume
for
now
linear
analysis: Equilibriumwithin0V andon0S f
,linearstress-strainlawandsmalldisplacementsyields
tF =K · tU (4.2)
Wewanttoestablish,
KU (t)=R (t) (4.3)
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MIT2.094 4.Finiteelementformulationforsolidsandstructures
Consider
U ˆ T =
U 1 V 1 W 1 U 2 W N
(N nodes) (4.4)· · ·
whereU ˆ T isadistinctnodalpointdisplacementvector.
Note: forthemoment“removeS u
”
Wealsosay
U ˆ T = U 1 U 2 U 3 U n
(n= 3N ) (4.5)· · ·
Wenowassume (m)u
(m) (m)u =H (m)U ˆ , u = v (4.6a)w
whereH (m) is3xnandU ˆ isnx1.
�(m) =B(m)U ˆ (4.6b)
whereB(m) is6xn,and
�(m)T
=
�xx �yy �zz γ xy γ yz γ zx
∂v
∂u
e.g. γ xy = +∂x
∂y
Wealsoassume
u(m) = H (m)U ˆ (4.6c)
�(m) = B(m)U ˆ (4.6d)
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MIT2.094 4.Finiteelementformulationforsolidsandstructures
Principleof VirtualWork:
T
�T τ dV = U ˆ f BdV V V
(4.7)canberewrittenas
(m)T
�(m)T
τ
(m)dV
(m) = U ˆ f B
(m)
dV
(m)
V (m) V (m)m m
Substitute(4.6a)to(4.6d).
T U ˆ B(m)
T
τ (m) dV (m) = V (m) m
U ˆT
H (m)T
f B(m)
dV
(m)
V
(m)m
τ
(m) =C (m)�(m) =C (m)B(m)U ˆ
Finally,
T U
̂� B(m)
T
C (m)B(m) dV (m) U ˆ =V (m)m
T
U ̂�
H (m)T
f B(m)
dV
(m)
V (m)m
with
�(m)T ˆ
T
B(m)T
=U
KU ˆ =R B
whereK isnxn,andR B isnx1.
Directstiffnessmethod:
K = K (m)
m
R B =
R B
(m)
m K (m) = B(m)
T C (m)B(m) dV
(m)
V (m)R B
(m)= H (m)
T f B
(m)
dV
(m)
V (m)
(4.7)
(4.8)
(4.9)
(4.10)
(4.11)
(4.12)
(4.13)
(4.14)
(4.15)
(4.16)
(4.17)
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MIT2.094 4.Finiteelementformulationforsolidsandstructures
Example
4.5 textbook
E =Young’sModulus
Mathematicalmodel Planesectionsremainplane:
F.E.
model
U 1U
= U 2 (4.18)U 3
Element1
U 1
x xu(1)(x) = 1−100 100 0
U 2
(4.19)
U 3H (1)
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MIT2.094 4.Finiteelementformulationforsolidsandstructures
�(1)xx(x)=
− 1100
1100
B(1)
0
U 1U 2U 3
(4.20)
Element2
u(2)(x)=
�(2)xx(x)=
0
0
1− x80x
80
H (2)
U
−180
1
80
B(2)
U
(4.21)
(4.22)
Then,
K =E
100
1−10
−110
000
+
13E
240
000
01
−1
0−11
(4.23)
where,
A
η=0
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture5- F.E.displacementformulation,cont’d
Prof. K.J.Bathe MITOpenCourseWare
Forthecontinuum Reading:Ch. 4
Differentialformulation•
Variationalformulation(Principleof VirtualDisplacements) •
Next,weassumedinfinitesimalsmalldisplacement,Hooke’sLaw, linearanalysis
KU =R (5.1a)
(m) =
H (m)U u (5.1b)
K = K (m) (5.1c)
m
R = R (m)
(5.1d)B
m
�(m) =B(m)U (5.1e)
U
T = U 1
U 2
U n
, (n=alld.o.f. of elementassemblage) (5.1f)
· · ·
K (m) = B(m)T
C (m) B(m) dV (m) (5.1g)V (m)
R B(m
)
=
H (m)T
f B(m)
dV
(m)
(5.1h)V (m)
Surface
loads
Recallthatintheprincipleof virtualdisplacements,
“surface”loads= U S f
T
f S f dS f (5.2)S f
u
S
(m)
=H S (m)
U (5.3)
H S (m)
=H (m) (5.4)evaluated at the surface
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MIT2.094 5.F.E.displacementformulation,cont’d
Substituteinto(5.2)
U
T H S
(m)T
f S (m)
dS (m) (5.5)S (m)
forelement(m)andonesurfaceof thatelement.
R (sm) = H S
(m)T
f S (m)
dS (m) (5.6)S (m)
Needtoaddcontributionsfromallsurfacesof allloadedexternalelements.
KU =R B
+R S
+R c
(5.7)
whereR c areconcentratednodalloads.
Assume
• (5.7)hasbeenestablishedwithoutanydisplacementboundaryconditions.
•
We,
however,
know
nodal
displacements
U b (rewriting
(5.7)).
KU =R ⇒
K aaK ba
K abK bb
U aU b
=
R aR b
(5.8)
SolveforU a:
K aaU a =R a −K abU b
whereU b
isknown!
Thenuse
(5.9)
K baU a +K bbU b =R b +R r (5.10)
whereR r areunknownreactions.
Example
4.6
textbook
τ xx E 1 ν 0 �xx
τ yy = ν 1 0 �yy (5.11)τ xy
1−ν 20 0 1−2
ν γ xy
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MIT2.094 5.F.E.displacementformulation,cont’d
u(x,y)v(x,y)
=H
u1u2
u3u4v1
v2
v3
v4
(5.12)
If wecansetthisrelationup,thenclearlywecangetH (1),H (2),H (3),H (4).
u(m) =H (m)U (5.13)
Alsowant�(m) =B(m)U . We want H . Wecouldproceedthisway
u(x,
y) =
a1 +
a2x
+
a3y
+
a4xy
(5.14)
v(x,y) =b1
+b2x+b3y+b4xy (5.15)
Expressa1. . .a4,b1. . .b4
intermsof thenodaldisplacementsu1. . .u4,v1. . .v4.
(e.g.) u(1,1)=a1
+a2
+a3
+a4
=u1.
h1(x,y) =14(1+x)(1+y) interpolation function
fornode1.
h2(x,y) =1
(1−x)(1+y)4
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MIT2.094 5.F.E.displacementformulation,cont’d
h3(x,y) =1(1−x)(1−y)4
h4(x,y) =1
(1+x)(1−y)4
u(x,y) =h1u1 +h2u2 +h3u3 +h4u4 (5.16)
v(x,y) =h1v1 +h2v2 +h3v3 +h4v4 (5.17)
u1
u2
u3u4v1v2v3v4
u(x,y)=
h1 h2 h3 h4 0 0 0 0 (5.18)v(x,y) 0 0 0 0 h1 h2 h3 h4
H (2x8)
Wealsowant,
u1u2u3u4v1v2
v3
v4
�xx h1,x h2,x h3,x h4,x 0 0 0 0 = 0 0 0 0 h1,y h2,y h3,y h4,y (5.19)�yy
γ xy
h1,y
h2,y
h3,y
h4,y
h1,x
h2,x
h3,x
h4,x
B (3x8)
�xx =∂u
∂x(5.20)
�yy =
∂v
∂y
(5.21)
γ xy =∂u
∂y+
∂v
∂x(5.22)
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�
2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture6- Finiteelementformulation,example,convergence
Prof. K.J.Bathe MITOpenCourseWare
6.1 Example
Reading:
Ex. 4.6 in
thetext
t= 0.1, E, ν planestress
(6.1)KU =R ; R =R B
+R s
+R c
+R r
= K (m); K (m) = B(m)T
C (m) B(m) d V (m) (6.2)K V (m)m
R B = R (m) (m)
; R = H (m)T
f B(m) d V (m) (6.3)B BV (m)m
6.1.1
F.E.
model
u1
u2
u3
u4
v1
v2
v3
v4↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
u1× × × × × ←(6.4)K
el.
(2)
=
...
...
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�
MIT2.094 6.Finiteelementformulation,example,convergence
Inpractice,
BT CBdV ; �=B
u1
...u4
v1
.
.
.
v4
K (6.5)=el
V
whereK is8x8andB is3x8.
AssumewehaveK (8x8)forel. (2)
U 1 U 2 U 3 U 4 U 5 U 11 U 18· · · · · ·↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓× × × × × × × × ×
. . .. . .. . .
. . .. . .. . .
. . .. . .. . .
. . .. . .. . .
U 1←....
.
.
(6.6)
K =
assemblage
U 11←......
U 18←× × × × × × × × ×
Consider,
R S = H S T f S d S ; H S =H (6.7)
S
on surface
h1 h2 h3 h4 0 0 0 0 u(x,y)H =
0 0 0 0 h1 h2 h3 h4
←v(x,y)
(6.8)←
u1u2...
(6.9)u= u4v1...
v4
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MIT2.094 6.Finiteelementformulation,example,convergence
(6.10)H S =H y=+11
2
(1+x) 12
(1−x) 0 0 0 0 0 0=
(6.11)
0 0 0 0 1(1+x) 1(1−x) 0 02 2
From(6.7);
+1
=
=
1
(1+x) 021(1−x) 02
0
0(0.1) dx (6.12)R S
1(1+x) − p(x)2−1
10 (1−x) thickness2
0 0
− p0(0.1) − p0(0.1)
(6.13)R S
6.1.2 Higher-orderelements
Wanth1,h2,h3,h4,h5
u(x,y) =
i5=1hiui.
hi
=1atnodeiand0atallothernodes.
h5 =
1
(1
−
x2)(1
+
y)2
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MIT2.094 6.Finiteelementformulation,example,convergence
h1 =1
4(1+x)(1+y)−
1
2h5
h2 =1
4(1−x)(1+y)−
1
2h5
h3
=
1
4
(1
−
x)(1
−
y)
h1 =1
4(1+x)(1−y)
Note:
(6.14)
(6.15)
(6.16)
(6.17)
hi = 1
Wemusthave ihi =1tosatisfytherigidbodymodecondition.
u(x,y) = hiui
(6.18)i
Assumeallnodalpointdisplacements=u∗. Then,
u(x,y) = hiu∗
=u∗ hi =u∗ (6.19)i i
From(6.1),
K (m) U =R (6.20)m
B
(m)T
C
(m)
B
(m)
dV
(m)
U
=
R
(6.21)
V
(m)m
whereC (m)B(m)U =τ (m). (AssumewecalculatedU .)
B(m)T
τ
(m) d V
(m) =R (6.22)V
(m)m
F
(m) =R ; F (m) = B(m)
T τ
(m) d V
(m) (6.23)
V
(m)m
Twoproperties
I. Thesumof theF (m)’satanynode isequaltotheappliedexternalforces.
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MIT2.094 6.Finiteelementformulation,example,convergence
II. Everyelement isinequilibriumunderitsF (m)
Û
T
F
(m) =Û T
B(m)T
τ
(m) d
V
(m) (6.24)
V
(m)
=�(m)T
= �(m)T
τ
(m) d
V
(m) (6.25)
V
(m)
=0 (6.26)
whereU ˆ T
=virtualnodalpointdisplacement.
Applyrigidbodydisplacement.
If we move the element virtually in the rigid body modes, �(m) is zero. Therefore the virtual workobtainedduetovirtualmotionof theelementiszero. ThentheelementisinequilibriumunderitsF (m).
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture7- Isoparametricelements
Prof. K.J.Bathe MITOpenCourseWare
Reading:
Sec. 5.1-5.3
WewantK =V
BT CB
dV ,R B
=V
H T f B dV . Uniquecorrespondence(x,y)⇔(r,s)
(r,s)arenaturalcoordinatesystemorisoparametriccoordinatesystem.
4
x= hixi
(7.1)i=1
4
y= hiyi (7.2)i=1
where
1h1
= (1 +r)(1+s) (7.3)4 1
h2 =
4
(1−r)(1+s) (7.4)
. . .
4
u(r,s) = hiui (7.5)i=1
4
v(r,s) = hivi (7.6)i=1
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MIT2.094 7.Isoparametricelements
�=Bû ûT = u1
u2
v4
(7.7)· · ·
∂u ∂x
∂v =Bû (7.8)
∂y
∂u + ∂v∂y
∂x
∂ ∂x ∂y ∂ ∂r ∂r ∂r ∂x
= (7.9)∂ ∂x ∂y ∂
∂s ∂s ∂s ∂y
J
∂ ∂ ∂x ∂r
=
J −1
(7.10)∂ ∂
∂y ∂s
J mustbenon-singularwhichensuresthatthereisuniquecorrespondencebetween(x,y)and(r,s).
Hence, 1 1
K = BT CB tdet(J )drds (7.11)
−1 −1 d V
1
1
Also,R B
= H T f B tdet(J )drds (7.12)−1
−1
Numerical
integration
(Gauss
formulae)
(Ch.
5.5)
=t BT det(J ij)×(weighti, j) (7.13)i j
K
∼ijCBij
2x2Gaussintegration,
(i=1,2) (7.14)
( j=1,2) (weighti, j=1 inthiscase) (7.15)
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�
MIT2.094 7.Isoparametricelements
9-node
element
9
x= hixi
(7.16)i=1
9
y= hiyi (7.17)
i=1
9
u= hiui
(7.18)i=1
9
v= hivi (7.19)i=1
Use3x3Gaussintegration
Forrectangularelements,J =const
Considerthefollowingelement,
Note,
here
we
could
use
hi(x,
y)
directly.
3 = 6 0J
= 2 3
(7.20)0 2
Then,wecandeterminethenumberof appropriateintegrationpointsbyinvestigatingthemaximumorderof BT CB .
Forarectangularelement,3x3Gauss integrationgivesexactK matrix. If theelement isdistorted,aK matrixwhich isstillaccurateenoughwillbeobtained,(if highenoughintegrationisused).
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MIT2.094 7.Isoparametricelements
Convergence
Principle
of
virtual
work: Reading:
Sec. 5.5.5,
4.3�T C�
dV =R(u) (7.21)
V
Find u, solution, in V, vector space (any continuous function that satisfies boundary conditions),
satisfying
�T C�
dV = a(u,v) = (f ,v) forallv,anelementof V. (7.22)
V
bilinear
form R(v)
Example:
Finite
Element
problem Finduh
∈Vh,whereVh
isF.E.vectorspacesuchthat
a(uh,vh) = (f ,vh) ∀vh ∈Vh (7.23)
Sizeof Vh ⇒#of independentDOFs(hereit’s12).
Note:
a(w,w) >0forw∈V (w=0)
2x
(strain
energy
when
imposing
w)
Also,
a(wh,wh)>0forwh ∈Vh (Vh ⊂V,wh =0)
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MIT2.094 7.Isoparametricelements
Property
I Define: eh =u−uh.
From(7.22),a(u,vh) = (f ,vh) (7.24)
From(7.23),a(uh,vh) = (f ,vh) (7.25)
Hence,
a(u−uh,vh)=0 (7.26)
a(eh,vh)=0 (7.27)
(errorisorthogonalinthatsensetoallvh inF.E.space).
PropertyII
(7.28)a(uh,uh)≤a(u,u)
Proof:
a(u,u) =a(uh
+eh,uh
+eh) (7.29)
=a(uh,uh) +
0byProp. I2a(uh,eh) +a(eh,eh) (7.30)
≥0
∴
a(u,u)≥a(uh,uh) (7.31)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture8- Convergenceof displacement-basedFEM
Prof. K.J.Bathe MITOpenCourseWare
(A) Find
u∈Vsuchthata(u,v) = (f ,v) ∀v∈V(Mathematicalmodel) (8.1)
a(v,v)>0 ∀v∈V, v= 0. (8.2)
where(8.2)impliesthatstructuresaresupportedproperly. E.g.
(B)
F.E.
Problem Find
uh ∈Vh suchthata(uh,vh)=(f ,vh) ∀vh ∈Vh (8.3)
a(vh,vh)>0 ∀vh ∈Vh,
Propertieseh =u−uh
vh =0 (8.4)
(I)
a(eh,
vh)
=
0
∀vh ∈
Vh (8.5)
(II)a(uh,uh)≤a(u,u) (8.6)
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MIT2.094 8.Convergenceof displacement-based FEM
(C)
Assume Mesh “iscontainedin”Mesh
h1 h2
e.g. Mesh notcontainedinMesh
h1 h2
Weassume(C),butneedanotherproperty(independentof (C))
(III)
a(eh,eh)≤a(u−vh,u−vh) ∀vh
∈Vh
(8.7)
uh
minimizes! (Recalleh
=u−uh)
Proof: Pickwh ∈Vh.
a(eh +wh,eh +wh) =a(eh,eh) + 02a(eh,wh) ( )+a w wh, h
≥0
(8.8)
Equalityholdsfor(wh =0)
a(eh,eh)≤a(eh
+wh,eh
+wh) (8.9)
=a(u−uh
+wh,u−uh
+wh) (8.10)
Takewh =uh −vh.
a(eh,eh)≤a(u−vh,u−vh) (8.11)
Usingproperty(III)and(C),wecansaythatwewillconvergemonotonically,frombelow,toa(u,u):
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MIT2.094 8.Convergenceof displacement-based FEM
Pascal
triangle
(2D)
(Ch. 4.3)
error indisplacement∼C ·
4-nodeelementiscompletetok=1.⇒
9-node
element
is
complete
to
k
=
2.
⇒
hk+1 (8.12)
(C isaconstantdeterminedbytheexactsolution,materialproperty...)
errorinstresses∼C hk (8.13)·
errorinstrainenergy∼C h2k ( theseC aredifferent) (8.14)· ←
Hence,
E −E h
=C h2k (roughlyequalto) (8.15)·
Bytheory,
log(E −E h)=logC + 2klogh (8.16)
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MIT2.094 8.Convergenceof displacement-based FEM
Byexperiment,wecanevaluate log(E −E h)fordifferentmeshesandplotlog(E −E h)vs. logh
Weneedtousegradedmeshesif wehavehighstressgradients.
Example Consideranalmost incompressiblematerial:
�V
= vol. strain (8.17)
or
∇·v→ verysmallorzero (8.18)
Wecan“see”difficulties:
p=−κ�V
κ= bulkmodulus (8.19)
Asthematerialbecomesincompressible(ν = 0.3 0.4999)→
κ
→ ∞
p finitenumber (8.20)�V
0→
→
(Smallerrorin�V resultsinhugeerroronpressureasκ→ ∞,theconstantC in(8.15)canbeverylargelocking)⇒
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture9- u/ pformulation
Prof. K.J.Bathe MITOpenCourseWare
Wewanttosolve Reading:Sec. 4.4.3
I. Equilibrium
τ ij,j +f iB
=0 inVolume(9.1)S f τ ijnj =f i
onS f
II. Compatibility
III. Stress-strain law
Usetheprincipleof virtualdisplacements
�T C�
dV =R (9.2)
V
Werecognizethatif ν 0.5→
�V
0 (�V
=�xx
+�yy
+�zz ) (9.3)→
E
κ=3(1−2ν )
→ ∞ (9.4)
p=−κ�V mustbeaccuratelycomputed (9.5)
Solution
τ ij
=κ�V
δ ij
+ 2G�ij
(9.6)
where
1 i= jδ ij = Kroneckerdelta = (9.7)
0 i= j
Deviatoricstrains:
�V �ij =�ij − 3
δ ij (9.8)
τ kkτ ij =− pδ ij + 2G�ij p=− 3
(9.9)
(9.2)becomes
�T C � dV + �V
κ�V
dV =R (9.10)
V V
�T C � dV − �T V pdV =R (9.11)
V V
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� =
MIT2.094 9.u/ pformulation
Weneedanotherequationbecausewenowhaveanotherunknown p.
p+κ�V
=0 (9.12)
p( p+κ�V
) dV =0 (9.13)
V −
V
p �V
+κ p
dV =0 (9.14)
Foranelement,
u=Hû (9.15)
� =BDû (9.16)
�V =BV û (9.17)
p=H p p̂ (9.18)
Planestrain(�zz4/1element
=0)Example
Reading:
Ex. 4.32 in
thetext
�V =�xx +�yy (9.19)
131
3
(�xx +�yy )�xx −
�yy − (�xx +�yy )
(9.20) γ xy(�xx +
�yy )− 13
Note:
�zz
=0but� =0!zz
p=H p p̂ =[1]{ p0} (9.21)
p(x,y) = p0
(9.22)
Weobtainfrom(9.11)and(9.14)
K uu K up û R
K pu K pp p̂ =
0(9.23)
BT
V
K uu
= DC BD
dV (9.24a)
K up
=− BV T H p
dV (9.24b) V
K pu
=− H pT
BV
dV (9.24c) V
1K pp =− H p
T
κH p dV (9.24d)
V
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�
MIT2.094 9.u/ pformulation
In practice, we use elements that use pressure interpolations per element, not continuous betweenelements. Forexample:
Then ,unlessν = 0.5(whereK pp =0),wecanusestaticcondensationonthepressuredof’s.
Use p̂ equationstoeliminate p̂ fromtheû equations.
K uu
−K upK −1
pp
K pu
û =R (9.25)
(Inpractice,ν canbe0.499999...)
The“bestelement”isthe9/3element. (9nodesfordisplacementand3pressuredof’s).
p(x,y)= p0
+ p1x+ p2y (9.26)
The
inf-sup
condition Reading:
Sec. 4.5
=�V
Volq h∇·vh dVol
inf sup ≥β >0 (9.27) q vh h qh∈Qh vh∈Vh
for
normalization
Qh: pressurespace.
If “this”holds,theelementisoptimalforthedisplacementassumptionused(ellipticitymustalsobesatisfied).
Note:
infimum=largest lowerbound
supremum=leastupperbound
Forexample,
inf {1,2,4}= 1
sup{1,2,4}= 4
inf {x∈R; 0< x
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MIT2.094 9.u/ pformulation
For (entry [3,1] inmatrix)assumethecircledentryistheminimum(inf)of .
Also,allentries inthematrixnotshownarezero.
Case
1
β h = 0
0 uh|i
=0 (fromthebottomequation)·
α uh
+ 0 ph
= Rh
(fromthetopequation)⇒
· |i
· |j
|i
=0
⇒noequationfor ph|j
spuriouspressure! (anypressuresatisfiesequation)⇒
Case2 β h =small=�
�·uh|i
= 0 ⇒ uh|i
= 0
∴�· ph|j
+uh|i
·α= Rh|i
i ph|j =
Rh| displ. = 0as�issmall⇒
�
⇒pressure large→
The behavior of given mesh when bulk modulus increases: locking, large pressures. See Example4.39textbook.
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture10- F.E.largedeformation/generalnonlinearanalysis
Prof. K.J.Bathe MITOpenCourseWare
Wedeveloped
t V
tτ ij
teij
d
V t = Rt
Reading:
Ch. 6
(10.1)
et ij =1
2
∂ui∂ xt j
+∂uj∂ xt i
(10.2)
t
V
tτ ij
δ teij
d
V t = Rt (10.3)
δ teij =1
2
∂ (δui)
∂ txj+
∂ (δuj)
∂ txi
(≡ teij) (10.4)
In
FEA:
tF = tR (10.5)
Inlinearanalysis
tF =K tU KU =R (10.6)⇒
Ingeneralnonlinearanalysis,weneedto iterate. Assumethesolutionisknown“attimet”
t
0
tx
= x+ u (10.7)
HencetF isknown. Thenweconsider
t+∆t t+∆tF
=
R
(10.8)
Considerthe loads(appliedexternalloads)tobedeformation-independent,e.g.
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•
MIT2.094 10.F.E.largedeformation/generalnonlinearanalysis
Thenwecanwrite
t+∆tF =tF +F (10.9)t+∆tU
=tU +U (10.10)
whereonlytF andtU areknown.
=tK ∆U , tK =tangentstiffnessmatrixattimet (10.11)F ∼
From(10.8),
tK ∆U =t+∆tR −tF (10.12)
WeusethistoobtainanapproximationtoU . WeobtainamoreaccuratesolutionforU (i.e. t+∆tF )using
t+∆tK (i−1)∆U (i) =t+∆tR −t+∆tF (i−1) (10.13)t+∆t
U
(i)
=
t+∆t
U
(i−
1)
+ ∆U (i)
(10.14)
Also,
t+∆tF
(0) =tF (10.15)
t+∆tK (0) =tK (10.16)
t+∆tU
(0) =tU (10.17)
Iteratefori= 1,2,3. . .untilconvergence. Convergenceisreachedwhen
∆U (i) < �D (10.18)2
t+∆tR
t+∆tF
(i−1) (10.19)
−
< �F 2
Note:
a2 = (ai)2
i
∆U (i) =U i=1,2,3...
∆U (1) in(10.13)is∆U in(10.12).
(10.13) isthe full Newton-Raphsoniteration.
Howwecould(inprinciple)calculatetK
Process
IncreasethedisplacementtU i
by�,withnoincrementforalltU j
, j=i
calculatet+�F •
• thei-thcolumnin tK = (t+�F −tF )/�=∂ tF
.t∂ U i
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MIT2.094 10.F.E.largedeformation/generalnonlinearanalysis
So, perform this process for i = 1,2,3, . . . , n, where n is the total number of degrees of freedom.Pictorially,
tK =
. .. .. .
. .. .. . · · ·
. .. .
. .
Ageneraldifficulty: wecannot“simply”incrementCauchystresses.
t+∆tτ
referred
to
area
at
time
t
+ ∆t• ij
• tτ ij referredtoareaattimet.
Wedefineanewstressmeasure,2nd Piola - Kirchhoff stress ,t+∆0
tS ij
,where0intheleadingsubscriptreferstooriginalconfiguration. Then,
t+∆t
0S ij
=0tS ij +0S ij (10.20)
Thestrainmeasureenergy-conjugatetothe2ndP-Kstress0tS ij istheGreen-Lagrangestrain 0
t�ij
Then,
0
t
0tS ij δ 0
t�ij d V = R (10.21)0V
Also,
t+∆t t+∆t 0 t+∆tS δ
� d V = R (10.22)0 ij 0 ij0V
Example
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MIT2.094 10.F.E.largedeformation/generalnonlinearanalysis
tF =tR (10.23)
t+∆t t+∆tF = R (10.24)
(everytimeit isinequilibrium)
(10.13)and(10.14)give:
i
=
1,
t+∆tK (0) ∆U (1) =t+∆tR t+∆tF (0) ≡fn(tU ) (10.25)−
t+∆tU
(1) =t+∆tU (0) + ∆U (1) (10.26)
i=2,
t+∆tK (1) ∆U (2) =t+∆tR −t+∆tF (1) (10.27)
t+∆tU
(2) =t+∆tU (1) + ∆U (2) (10.28)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture11- Deformation,strainandstresstensors
Prof. K.J.Bathe MITOpenCourseWare
Westatedthatweuse Reading:Ch. 6
tV
tτ ij δ eij d V t =
0V
tS
t� 0V =tR (11.1)t 0 ij δ 0 ij d
The
deformation
gradient Weuse txi
=0xi
+tui
∂
tx1 ∂ tx1 ∂
tx1∂ 0x1 ∂
0x2 ∂ 0x3
∂ tx2 ∂ t
x2 ∂ t
x2
tX 0
(11.2)= 0 0 0∂ ∂ ∂ x x x1 2 3
∂
x
t
3 ∂ xt
3 ∂ xt
3
∂
x0 1 ∂ x0
2 ∂ x0
3
td x1 dtx (11.3)td x= 2
td x30d x1
d0x 0 (11.4)
(11.5)
d x2
0d x3
Impliesthat
dtx=0
tX d0x
=
(0tX
is frequently denoted by0tF
or simply F , but we use F forforcevector)
WewillalsousetherightCauchy-Greendeformationtensor
tC tX T tX = (11.6)0 0 0
Some
applications
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�
�
�
MIT2.094 11.Deformation,strainandstresstensors
Thestretchof afiber(tλ): 2t t
t� 2
dxT dx d stλ = = (11.7)
d0xT d0x d0s
Thelengthof afiberis
0
0 0d s = d xT d x12 (11.8)
�
�
� 2 d0xT tX T tX d0xtλ
=0
0
0
0 ,
from(11.5) (11.9)d s d s·
Express
0 0 0d
x= d s n (11.10)
0
0n
= unitvector intodirectionof d x (11.11)
tλ
2 =
0n
T 0tC
0n
(11.12)
⇒
∴
tλ= 0nT 0
tC
0n
12 (11.13)
Also,�
d x̂tT
·�
dxt
=�
d ŝt
�
dst
costθ, (a·b=abcosθ) (11.14)
From(11.5),
costθ =
d
x̂0 T ˆtX T 0
�
tX 0 d x0
d ˆtsdst
=d
ŝ0 n̂0 T tC 0 n0 d
s0
d ˆts·dst
ˆtX 0 ≡tX 0
(11.15)
(11.16)
∴
0n̂T 0
tC 0ncostθ=
tλ̂tλ(11.17)
Also,
0ρtρ=det0
tX
(seeEx. 6.5) (11.18)
Example Reading:Ex. 6.6 in
thetext
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MIT2.094 11.Deformation,strainandstresstensors
1 0 0h1
=4
(1+ x1)(1+ x2) (11.19)
...
t
0
txi = xi + ui (11.20)
4
= hktxi
k, (i= 1,2) (11.21)
k=1
where txik
arethenodalpointcoordinatesattimet(tx11
=2, tx21
= 1.5)
Thenweobtain
tX =
1 5 +0x2 1 +0x1 (11.22)0 4 12
(1+0x2)12
(9+0x1)
At 0x1
= 0, 0x2
=0,
1 5 10tX
0x
=0x2=0=
41 9
(11.23)i
2 2
The
Green-Lagrange
Strain
1� 1� 0t� =
2 0tX T 0
tX −I =
2 0tC
−I (11.24)
t
0 t t∂ xi
∂ xi
+ ui
∂ ui
= =
δ ij
+
(11.25)
∂ 0x ∂ 0x ∂ 0xj j j
Wefindthat
1� t� = tu + tu + tu tu , sumoverk= 1,2,3 (11.26)0 ij 0 i,j 0 j,i 0 k,i0 k,j2where
t∂ u0tui,j = 0
i (11.27)
∂ xj
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MIT2.094 11.Deformation,strainandstresstensors
Polar
decomposition
of
tX
tR tU = 000 (11.28)
tX
isarotationmatrix,suchthat
0
where tR
tR T
0
tU
0
0
tR
isasymmetricmatrix(stretch)
0
=I (11.29)
and
Ex.
6.9
textbook
√ 3
41 0−√
32tX
=
Then, � 2tC
tX T tX
tU = = 0
t� tU =
0
0
2
0
0
0
0
(11.30)3 301 222
(11.31)
(11.32)1 2
I −
2
Thisshows,byanexample,thatthecomponentsof theGreen-Lagrangestrainareindependentof arigid-bodyrotation.
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture12- TotalLagrangianformulation
Prof. K.J.Bathe MITOpenCourseWare
Wediscussed:
t
tX =
∂
∂ x
xi
j
⇒ dtx= tX d0x, d0x=�
tX
−1
dtx (12.1)0 0 0 0
tC = tX T tX (12.2)0
0 0
d0x=0tX d
tx
where 0tX =�
0tX
−1
=∂
t
0xi (12.3)∂ xj
TheGreen-Lagrangestrain:
1
� 1
�
0
t� = 0
tX T 0
tX −I = 0
tC −I (12.4)
2 2
Polardecomposition:
1� 2
0tX
=0tR 0
tU ⇒0
t� =2 0
tU −I (12.5)
Wesee,physicallythat:
wheredt+∆tx anddtx arethesame lengthsthe components of the G-L strain do not⇒
change.
Note
in
FEA
0 0 kxi
= hk xi
k
foranelement (12.6)t t kui
= hk ui
k
t
0
t
xi =
xi +
ui →
for
any
particle
(12.7)
Hencefortheelement
t
0
k
t kxi = hk xi + hk ui (12.8)
k k
= hk0
xik
+tuik
(12.9)k
= hktxk
k (12.10)k
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MIT2.094 12.TotalLagrangianformulation
E.g.,k= 4
2ndPiola-Kirchhoff stress
0ρ
0
tS =
tρ
0t X
tτ
0tX
T → componentsalso independentof arigidbodyrotation (12.11)
Then
t 0 t ttS ijδ �ijd V =tτ ijδ eijd V = R (12.12)0 0 t
0V tV
Wecanusean incrementaldecompositionof stress/strain.
t+∆0
tS =0
tS +0S (12.13)
t+∆tS = tS +0S (12.14)0
ij
0 ij ij
t+∆t �= t� +0� (12.15)0 0
t+∆
0
t�ij =0t�ij +0�ij (12.16)
Assumethesolutioniskownattimet,calculatethesolutionattimet+ ∆t. Hence,weapply(12.12)attimet+ ∆t:
t+∆t t+∆t 0
t+∆tS δ
� d V = R (12.17)0
ij
0
ij0V
Lookatδ �0
t
ij
:
1� t t t t tδ � =δ u + u + u u (12.18a)0 ij 0 i,j 0 j,i 0 k,i0 k,j2t t1 ∂δui
∂δuj
∂δuk
∂ u ∂ u ∂δukt k k
δ � =
+
+
+
(12.18b)
0
ij 2
∂ 0x ∂
0x ∂ 0x ·
∂ 0x ∂ 0x ·
∂ 0xj i i j i j
1� δ �0
tij
=2
δ 0ui,j
+δ 0uj,i
+δ 0uk,i
0
tuk,j
+0
tuk,iδ 0uk,j
(12.18c)
Wehave
t+∆t�ij −t�ij =0�ij (12.19)0 0
0�ij
=0eij
+0ηij
(12.20)
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�
�
MIT2.094 12.TotalLagrangianformulation
where 0eij isthe linear incrementalstrain,0ηij isthenonlinearincrementalstrain,and
1 t t
0eij
= 0ui,j
+0uj,i
+0uk,i
0uk,j
+0uk,i
0uk,j
(12.21)2
initial displ. effect
1
0ηij =
0uk,i0uk,j (12.22)2
where
∂uk t+∆t t
0uk,j
= 0
,
uk
= uk
−
uk
(12.23)∂ xj
Note
t+∆t tδ � =δ � ∵δ � =0whenchangingtheconfigurationatt+ ∆t (12.24)ij 0 ij 0 ij
From(12.17):
� �
tS δ ij +δ η d V 0 ij +0S ij 0e 0 ij0
0V
= tS + + tS ijδ η + ijδ η d V (12.25)0 ijδ 0eij 0S ijδ 0eij 0 0 ij 0S 0 ij0
0V
=t+∆tR (12.26)
Linearization
tt K U K
0V
0S 0e +tS ijδ η
d V = R −0V
tS 0eijd V (12.27)
ijδ ij 0 0 ij0
t+∆t
0 ijδ 0
L
U 0 NL0 tF 0
Weuse,
0S ij �0C ijrs0ers (12.28)
Wearriveat,withthefiniteelementinterpolations,
0
tK L
+0
tK NL
U =t+∆tR − 0
tF (12.29)
whereU isthenodaldisplacementincrement.
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MIT2.094 12.TotalLagrangianformulation
Lefthandsideasbeforebutusing(k−1)andrighthandsideis
t+∆t t+∆t t+∆t (k−1) 0= R −0V
S δ �
d V (12.30)0 ij 0 ij
gives
t+∆tR −
t+∆0tF
(k−1)(12.31)
InthefullN-Riteration,weuse
t+∆t (k−1) t+∆t (k−1)∆U (k) t+∆t
t+∆t (k−1)0
K L
+0
K NL
= R −0
F (12.32)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture13- TotalLagrangianformulation,cont’d
Prof. K.J.Bathe MITOpenCourseWare
Exampletrusselement. Recall:
Principleof virtualdisplacementsappliedatsometimet+ ∆t:
t+∆t t+∆t t+∆tτ ijδ t+∆teijd V = R (13.1)t+∆tV
t+∆t t+∆t 0 t+∆tS δ
� δ V = R (13.2)0 ij 0 ij0V
t+∆0
tS ij
=0
tS ij
+0S ij
(13.3)
t+∆0t�ij =0
t�ij
+0�ij
(13.4)
0�ij =0eij +0ηij (13.5)
where0
tS ij
and 0
t�ij
areknown,but0S ij
and 0�ij
arenot.
1� 0eij = 2 0
ui,j +0uj,i +0tuk,i0uk,j +0
tuk,j0uk,i (13.6)
1� 0ηij = 0uk,i0uk,j (13.7)2
Substituteinto(13.2)andlinearizetoobtain
0 0 t+∆t 0δ 0e 0C 0ers + ijδ ηijd V = δ 0eij ijd V 0V
ij ijrs d V 0V
0tS 0 R −
0V 0tS (13.8)
F.E.
discretization
gives
0
tK L
+0
tK NL
∆U =t+∆tR −0
tF (13.9)
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MIT2.094 13.TotalLagrangianformulation,cont’d
tK
tBT tB
0
0 L = 0 L 0C 0 Ld V (13.10)0V
T
0tK = tB tS tBNL d V (13.11)0 NL 0 NL 0 0
0V
matrix
T 0tF = tB tS ˆ d V (13.12)0 0
L
00V
vector
Theiteration(fullNewton-Raphson)is
t+∆tK
(i−1)
+t+∆t
K (i−1)
∆U (i) =t+∆tR t+∆t
F
(i−1)
(13.13)0 L 0 NL − 0
t+∆tU (i) =t+∆tU (i−1) + ∆U (i) (13.14)
Truss
element
example (p. 545)
Herewehavetoonlydealwith0
tS 11, 0e11, 0η11
t
0e11
=∂u1
+∂ uk
∂uk(13.15)
∂ 0x1
∂ 0x1
·∂ 0x1
1 ∂uk
∂uk
0η11
=2 ∂ 0x1
·∂ 0x1
(13.16)
Weareafter
1u1
u1
tB 2 tB0e11 =0 L
2
=0
L
û (13.17)u12u2
2
ui = hkuki (13.18)
k=1
2
t t kui = hk ui (13.19)k=1
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�
�
MIT2.094 13.TotalLagrangianformulation,cont’d
t t
=∂u1
+∂ u1 ∂u1 +
∂ u2 ∂u2 (13.20a)0e11 ∂ 0x1
∂ 0x1
∂ 0x1
∂ 0x1
∂ 0x1
2 0L cosθ−0L (13.20b)t + ∆L=u1
2
0Lt + ∆L sinθ (13.20c)=u2
1−1 0 1 0 û0e11
=0L
0L+ ∆L 1ûcosθ−1 −1 0 1 0+ ·
0L
0L
∂
tu1
0∂ x1
0L+ ∆L 1 û
=0tBLû (13.20e)
Hence,
0
−1 0 1sin
θ
(13.20d)
+
·0L
0L
∂
tu2
∂ x01
0e11 =0L+ ∆L
(0L)2−cosθ −sinθ cosθ sinθ û (13.20f)
where the boxed quantity above equals 0
tBL. In small strain but large rotation analysis we assume∆L0L,
1 θcos− −sinθ cosθ sinθ û (13.20g)0e11
=0L
1 ∂u1
∂u1
∂u2
∂u2+ (13.21a)0 0 0 0x x
=0η11 2 ∂ ∂ ∂ ∂ x x1 1 1 1
1δ η =0
11 2 1
∂δu1
∂u1
∂u1
∂δu1
∂δu2
∂u2
∂u2
∂δu2+ + + (13.21b)0 0
0 0
0 0
0 0x x∂
∂
∂
∂
∂
∂
∂
∂ x x x x x x1
1
1
1
1
1
1
∂δu1
∂u1
∂δu2
∂u2
= 0 0 + 0 0 (13.21c)∂ x1 ∂ x1 ∂ x1 ∂ x1
∂u1tS 00
0∂δu1 ∂δu 2 ∂ 11tS x1 (13.21d)11δ η =0 11 0 0 0 0
tS
∂u2∂ ∂ x x1 1011
∂ x1
tS 0
∂u10 1 −1 0 1 0
0∂
x1 = û (13.21e)∂u2 0L −1 0 1∂
0x1
BNL
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MIT2.094 13.TotalLagrangianformulation,cont’d
0C =E (13.22)
tS ˆ = tS (13.23)0 0 11
Assumesmallstrains
EAtK =0 0L
cos2θ cosθsinθ −cos2θ −cosθsinθsin2θ −sinθcosθ −sin2θ
cos2θ sinθcosθ
=
sym sin2θ
tK (13.24)0 L
1 0 −1 00tP
0
L
0 1 −1
0 + −1 0 10 −1 0 1
tK 0 NL
Whenθ=0,0tK L doesn’tgivestiffnesscorrespondingtou2
2,but0tK NL does.
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�
2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture14- TotalLagrangianformulation,cont’d
Prof. K.J.Bathe MITOpenCourseWare
Trusselement. 2Dand3Dsolids.
t+∆t t+∆t t+∆tτ ij
δ t+∆teijd V = R (14.1)t+∆tV
t+∆t t+∆t 0
t+∆t
0S ijδ 0�ijδ V = R (14.2)0V
linearization
δ ij
δ V + tS ij
δ ηij
δ V = tS ij
δ V (14.3)0V
0C ijrs
0ers 0e 0
0V 0
00 t+∆tR −
0V 0
ij
δ 0e 0
Note:
tδ =δ �0eij
0
ij
varyingwithrespecttotheconfigurationattimet.
F.E.
discretization
0 0
k
t t k
t+∆t t+∆t kxi = hk xi xi = hk xi xi
= hk xi
(14.4a)k k k
t t k t+∆t t+∆t k kui = hk ui ui = hk ui
ui = hkui (14.4b)k k
k
(14.4)
into
(14.3)
gives
0tK L +0
tK NL U =t+∆tR − 0
tF (14.5)
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MIT2.094 14.TotalLagrangianformulation,cont’d
Truss
∆L 1smallstrainassumption:0L
0E AtK
=
0 0L
cos2θ cosθsinθ −cos2θ −cosθsinθcosθsinθ sin2θ −sinθcosθ −sin2θ
cos2θ sinθcosθ
=−cos2θ
−cosθsinθ
−cosθsinθ−sin2θ
(14.6)sinθcosθ sin2θ
1 0 −1 00tP
0L
0 1 −10
+−1 0 1
0 −1 0 1
(noticethatthebothmatricesaresymmetric)
cosθ sinθu1 u1 (14.7)=v1
−sinθ cosθ v1
Correspondingtotheuandv displacementswehave:
0E AtK = (14.8)0L
1 0 −1 00
1 0 −1 00
+ tP 0L
0 0 0 0 1 −10 (14.9)= −1 0 1 00 0
−1 0 100 0 −1 0 1
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�
MIT2.094 14.TotalLagrangianformulation,cont’d
tP 0 tP Q L = ∆ Q= ∆ (14.10)0L
· ⇒ ·
tP where the boxed term is the stiffness. In axial direction, 0L is not very important because usually
E 0A tP tP 0L
0L
. But, inverticaldirection, 0L
is important.
−cosθ
tF tP
−sinθ
0
=
cosθ
(14.11)
sinθ
2D/3D
(e.g.
Table
6.5)
2D:
1� 2 � 20�11
= 0u1,1 +0tu1,1 0u 1,1 +0tu2,1 0u2,1
+2 0
u1,1
+ 0u2,1
(14.12)
0e11 0η11
0�22
= (14.13)· · ·
0�12
= (14.14)· · ·
(Axisymmetric)
0�33 = ? (14.15)
1� 2 t� =2
tU −I (14.16)0 0
× × 0× × 0
0
tU 2 = 0 0 × (14.17)
(tλ
↑
)2
t
0 td s 2π x1
+ u1tλ= =
0 0d s 2π x1
(14.18)
t
= 1 +u1
0x1
2 0
t�33
=2
11 + 0
tu
x1 −1 1
(14.19) 2tu1 1 tu1= +0x1
2 0x1
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MIT2.094 14.TotalLagrangianformulation,cont’d
2tu +u1 1 tu +u1t+∆t 1 1� = + (14.20)0 33 0
0x1
2 x1
2u1 tu u1 1
u10�33 =
t+∆0
t�33
−0t�33 = 0x1
+ 0x1
1
· 0x1+
2 0x1 (14.21)
How
do
we
assess
the
accuracy
of
an
analysis? Reading:
Sec. 4.3.6
• Mathematicalmodel∼u
• F.E.solution∼uh
Findu−uhandτ −τ h.
References
[1] T.SussmanandK.J.Bathe.“Studiesof FiniteElementProcedures- onMeshSelection.”Computers
&
Structures ,
21:257–264,
1985.
[2] T. Sussman and K. J. Bathe. “Studies of Finite Element Procedures - Stress Band Plots and theEvaluationof FiniteElementMeshes.”Journal of Engineering Computations ,3:178–191,1986.
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture15- Fieldproblems
Prof. K.J.Bathe MITOpenCourseWare
Heattransfer,incompressible/inviscid/irrotationalflow,seepageflow,etc. Reading:Sec. 7.2-7.3
Differentialformulation•
Variationalformulation•
Incrementalformulation•
F.E.discretization•
15.1 Heattransfer
AssumeV constantfornow:
S
=
S θ ∪
S q s
θ(x,y,z,t)isunkownexceptθ|S θ
=θ pr. Inaddition, q |S q isalsoprescribed.
15.1.1
Differential
formulation
I. HeatflowequilibriuminV andonS q.
II. Constitutive lawsq x =−k∂θ .∂x
∂θ
q y
=−k (15.1)∂y
∂θ
q z =−k (15.2)∂z
III. Compatibility: temperaturesneedtobecontinuousandsatisfytheboundaryconditions.
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| |
MIT2.094 15.Fieldproblems
Heatflowequilibriumgives
∂ ∂θ
∂ ∂θ
∂ ∂θ Bk
+ k + k =−q (15.3)∂x
∂x
∂y
∂y
∂z
∂z
whereq B istheheatgeneratedperunitvolume. Recall1Dcase:
unitcross-section
dV =dx (1) (15.4)·
q x
−q x+dx
+q Bdx=0 (15.5)
∂q xq |x − q |x +
∂xdx
+q Bdx=0 (15.6)
∂
∂θ B
−∂x
−k∂x
dx+q dx =0 (15.7)
∂
∂θ Bk =−q (15.8)
∂x
∂x
We
also
need
to
satisfy
k
∂θ
=q S (15.9)∂n
onS q.
15.1.2
Principle
of
virtual
temperatures
θ
∂ k
∂θ+ +q B =0 (15.10)
∂x ∂x· · ·
(θ =0andθ tobecontinuous.)S θ
θ
∂ k
∂θ+ +q B dV =0 (15.11)
V
∂x ∂x· · ·
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�
MIT2.094 15.Fieldproblems
Transformusingdivergencetheorem(seeEx4.2,7.1)
S qθ
T kθ dV = θq BdV + θ q S dS q
(15.12)V V S q
heat flow
∂θ ∂x
θ = ∂θ (15.13)
∂y
∂θ
∂z
k 0 0
k= 0 k 0 (15.14)
0 0 k
Convection
boundary
condition
q
S =h θe −θS (15.15)
where
θe is
the
given
environmental
temperature.
Radiation
q
S =κ∗
(θr)4
−�
θS 4
(15.16)
=κ∗
(θr)2 +�
θS 2
�
θr +θS �
θr −θS
(15.17)
=κ θr −θS (15.18)
whereκ=κ(θS )andθr isgiventemperatureof source. Attimet+ ∆t
θT t+∆tkt+∆tθdV = θ t+∆tq BdV + θ
S t+∆tq S dS q (15.19)V V S q
Let t+∆tθ =tθ+θ (15.20)
or t+∆tθ(i) =t+∆tθ(i−1) + ∆θ(i) (15.21)
with t+∆tθ(0) =tθ (15.22)
From(15.19)
θT t+∆t
(i−1)∆θ(i)k dV V
= θt+∆tq BdV − θT t+∆tk(i−1)t+∆tθ(i−1)dV (15.23)
V V
S (i−
1)
∆θS (i)
θ
S t+∆th(i−1) t+∆t e t+∆t+
θ θ
+
dS q−S q
wherethe∆θS (i)
termwouldbemovedtothe left-handside.
Weconsideredtheconvectionconditions
θS t+∆th
�
t+∆tθe −t+∆tθS
dS q (15.24)S q
Theradiationconditionswouldbeincludedsimilarly.
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MIT2.094 15.Fieldproblems
F.E.
discretization
t+∆t t+∆tθ =H 1x4
· θ̂4x1 for4-node2Dplanarelement
t+∆tθ t+∆tθ̂2x1 =B2x4 · 4x1t+∆tθS =H S t+∆tθ̂·
For(15.23)
t+∆t (i−1)∆θ(i) gives t+∆t (i−1)θ
T k
dV = BT k B dV ∆θ̂(i)⇒
V V
4x2 2x2 2x4 4x1
θt+∆tq
BdV
H T t+∆tq
BdV ⇒
V V
θT t+∆tk(i−1)t+∆tθ(i−1)dV
BT t +∆tk(i−1)BdV
t+∆tθ̂(i−1)⇒ V V known
S q
θS T t+∆th(i−1) t+∆tθe − t+∆tθS
(i−1)
+ ∆θS (i)
dS q =⇒
H S T t+∆th(i−1) H S t+∆tθ̂e t+∆tθ̂(i−1)+∆θ̂(i)dS q
−
S q 4x1
1x4 4x1
4x1
4x1
15.2
Inviscid,
incompressible,
irrotational
flow
2Dcase: vx,vy
arevelocitiesinxandy directions.
∇·v= 0
or∂vx
+∂vy
=0 (incompressible)∂x
∂y
∂vx
∂vy
∂y
−∂x
=0 (irrotational)
Usethepotentialφ(x,y),
∂φ
∂φ
vx =
vy =
∂x
∂y
∂ 2φ ∂ 2φ+ = 0 inV ⇒
∂x2 ∂y 2
(Sameastheheattransferequationwithk=1,q B =0)
64
(15.25)
(15.26)
(15.27)
(15.28)
(15.29)
(15.30)
(15.31)
Reading:
Sec. 7.3.2
(15.32)
(15.33)
(15.34)
(15.35)
(15.36)
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture16- F.E.analysisof Navier-Stokesfluids
Prof. K.J.Bathe MITOpenCourseWare
Incompressibleflowwithheattransfer
Reading:
Sec.Werecallheattransferforasolid: 7.1-7.4,
Table7.3
Governing
differential
equations
(kθ,i),i +q B = 0 inV (16.1)
isprescribed,k =q S ∂θ
(16.2)θ∂nS θ
S q S q
Sθ ∪
Sq = S
Sθ ∩
Sq =
∅
(16.3)
Principle
of
virtual
temperatures
θ,ikθ,idV = θq B
dV + θ
S q
S dS q
(16.4)V
V S q
forarbitrarycontinuousθ(x1, x2, x3)zeroonSθ
Forafluid,weusetheEulerianformulation.
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MIT2.094 16.F.E.analysisof Navier-Stokesfluids
ρc pvθ|x
−
ρc pvθ|x
+∂
∂x
(ρc pvθ)dx
+ conduction + etc (16.5)
Ingeneral3D,wehaveanadditionaltermforthe lefthandsideof (16.1):
−∇
·
(ρc pvθ)=−ρc p∇·(vθ)=−ρc p( ∇·v)θ−ρc p(v·∇)θ
term (A)
(16.6)
where∇·v=0inthe incompressiblecase.
∇·v=vi,i
=div(v)=0 (16.7)
So(16.1)becomes
(kθ,i),i +q B =ρc pθ,ivi
⇒(kθ,i),i +
�
q
B −ρc pθ,ivi
=0 (16.8)
Principle
of
virtual
temperatures
is
now
(use
(16.4))
V
θ,ikθ,idV +
V
θ(ρc pθ,ivi)dV =
V
θq BdV +
S q
θS
q S dS q
(16.9)
Navier-Stokes
equations
• Differential form
τ ij,j +f Bi =ρvi,jvj (16.10)
withρvi,jvj liketerm(A) in(16.6)=ρ(v·∇)v inV .
τ ij
=− pδ ij
+2µeij
eij
=1
2
∂vi∂xj
+∂vj∂xi
(16.11)
• Boundary conditions (needbemodifiedforvariousflowconditions)
τ ijnj =f S f i
onS f (16.12)
Mostly used as f n = τ nninflowconditions).
= prescribed, f t = unknown with possibly∂vn
∂n= ∂vt
∂n= 0 (outflow or
Andvi prescribedonS v,andSv ∪Sf = SandSv ∩Sf =∅.
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MIT2.094 16.F.E.analysisof Navier-Stokesfluids
Variational
form •
viρvi,j
vj
dV + eij
τ ij
dV = vif i
BdV + vi
S f f i
S f dS f
(16.13)V
V V S f
p∇·vdV =0 (16.14)V
F.E.
solution •
Weinterpolate(x1, x2, x3),vi,vi,θ,θ, p, p. Goodelementsare
×: linearpressure: biquadraticvelocities◦
(Q2, P 1),9/3element
9/4celement
Bothsatisfytheinf-supcondition.
Soingeneral,
Example:
ForS f e.g.
τ nn = 0,∂vt
=0; (16.15)∂n
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MIT2.094 16.F.E.analysisof Navier-Stokesfluids
and ∂vn issolvedfor. Actually,wefrequently justset p=0.∂t
Frequentlyused isthe4-nodeelementwithconstantpressure
Itdoesnotstrictlysatisfythe inf-supcondition. Oruse Reading:Sec. 7.4
3-nodeelementwithabubblenode.Satisfies inf-supcondition
1D
case
of
heat
transfer
with
fluid
flow, v= constant Reading:
Sec. 7.4.3
Re
=
vL
Pe
=
vL
α
=
k
(16.16)
ν
α
ρc p
Differential
equations •
kθ =ρc pθv (16.17)
θ|x=0 =θL θ|x=L =θR (16.18)
Innon-dimensionalform Reading:p. 683
1(nowθ andθ arenon-dimensional) (16.19)θ =θ
Pe
exp Pexθ−θL L
−1⇒
θR −
θL
=
exp
(Pe)
−
1
(16.20)
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MIT2.094 16.F.E.analysisof Navier-Stokesfluids
• F.E.discretization
θ =Peθ (16.21)
1
0
θθdx+Pe
1
0
θθdx=0+{effectof boundaryconditions =0here}
Using2-nodeelementsgives
(16.22)
1 Pe
(h∗)2(θi+1
−2θi
+θi−1) =2h∗
(θi+1
−θi−1) (16.23)
vL
Pe= (16.24)α
Define
h
vh
Pee =Pe = (16.25)·L α
Pee Pee−1−
2θi−1 + 2θi +
2−1 θi+1
=0 (16.26)
whatishappeningwhenPee islarge? Assumetwo2-nodeelementsonly.
θi−1 =0 (16.27)
θi+1 =1 (16.28)
1 Peeθi =
21−
2(16.29)
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MIT2.094 16.F.E.analysisof Navier-Stokesfluids
1 Peeθi
=2
1−2
(16.30)
ForPee >2,wehavenegative θi
(unreasonable).
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture17- Incompressiblefluidflowandheattransfer,cont’d
Prof. K.J.Bathe MITOpenCourseWare
17.1 Abstractbody
Reading:
Sec. 7.4
Fluid
Flow
Heat
transfer
Sv, Sf
Sθ, Sq
Sv
∪Sf
Sθ
∪Sq
=S =S Sv ∩Sf = 0 Sθ ∩Sq = 0
17.2 Actual2Dproblem(channelflow)
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MIT2.094 17.Incompressiblefluidflowandheattransfer,cont’d
17.3 Basicequations
P.V.
velocities
viρvi,jvjdV + τ ijeijdV = vif iB
dV + viS f f i
S f dS f (17.1)V V V S f
Continuity
pvi,idV =0 (17.2)V
P.V.
temperature
θρc pθ,ividV + θ,ikθ,idV = θq B
dV + θ
S q
S dS
(17.3)V
V V S q
F.E.
solution
xi
= hkxki
(17.4)
vi
= hkvik (17.5)
θ= hkθk (17.6)
p= h̃k pk (17.7)
v
F (u) =R u=
p
nodalvariables (17.8)⇒
θ
17.4 Modelproblem
1Dequation,
dθ
d2θ
ρc pv =k (17.9)dx
dx2
(v isgiven,unitcrosssection)
Non-dimensionalform(Section7.4)
dθ
d2θPe = (17.10)
dx
dx2
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�
MIT2.094 17.Incompressiblefluidflowandheattransfer,cont’d
vL
kPe= , α= (17.11)
α
ρc p
θ∗ isnon-dimensional
θ
−
θL
exp
PeLx
−
1
θR
−θL
=exp(Pe)−1
(17.12)
(17.10) inF.E.analysisbecomes
dθ dθ dθθPe dV + dV =0 (17.13)
dx dx dxV V
Discretizedby linearelements:
hh∗ =
L
ξ ξ
θ(ξ ) = 1− θi−1
+ θi
(17.14)h∗ h∗
Fornodei:
Pee Pee −θi−1 −
2θi−1 + 2θi −θi+1 +
2θi+1 =0 (17.15)
where
vh
hPee = =Pe (17.16)
α L
Thisresultisthesameasobtainedbyfinitedifferences
1(θi+1 −2θi +θi−1) (17.17)θ =
i (h∗)2
θ =θi+1 −θi−1
(17.18)i 2h∗
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�
MIT2.094 17.Incompressiblefluidflowandheattransfer,cont’d
Consideredθi+1 =1,θi−1 =0. Then
θi
= 1−(Pee/2)
(17.19)2
PhysicallyunrealisticsolutionwhenPee >2. Forthisnottohappen,weshouldrefinethemesh—averyfinemeshwouldberequired. Weuse“upwinding”
dθ=
θi −θi−1(17.20)
h∗dx i
Theresultis
(−1−Pee)θi−1 +(2+Pee)θi −θi+1 =0 (17.21)
Verystable,e.g.
θi−1
=0=
1(17.22)
θi+1
= 1⇒θi
2+Pee
Unfortunately
it
is
not
that
accurate.
To
obtain
better
accuracy
in
the
interpolation
for
θ,
use
the
function
exp PexL
−1
exp(Pe)−1(17.23)
TheresultisPee dependent:
This impliesflow-conditionbasedinterpolation. Weusesuch interpolationfunctions—seereferences.
References
[1] K.J. Bathe and H. Zhang. “A Flow-Condition-Based Interpolation Finite Element Procedure for
Incompressible
Fluid
Flows.”
Computers
&
Structures ,
80:1267–1277,
2002.
[2] H. Kohno and K.J. Bathe. “A Flow-Condition-Based Interpolation Finite Element Procedure forTriangularGrids.”International Journal for Numerical Methods in Fluids ,51:673–699,2006.
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MIT2.094 17.Incompressiblefluidflowandheattransfer,cont’d
17.5 FSIbriefly
Lagrangianformulationforthestructure/solid
Arbitrary
Lagrangian-Eulerian
(ALE)
formulation Letf beavariableof aparticle(e.g. f =θ).
Consider1D
f ̇∂f
∂f
= + vparticle
∂t
∂x
(17.24)
where
v
is
the
particle
velocity.
For
a
mesh
point,
f ∗ =∂f
∂f
+ vm
(17.25)mesh point ∂t ∂x
wherevm
f ̇
isthemeshpointvelocity. Hence,
∂f =f ∗ +
∂x(v−vm) (17.26)
particle mesh point
Use(17.26) inthemomentumandenergyequationsanduse forceequilibriumandcompatibilityattheFSIboundarytosetupthegoverningF.E.equations.
References
[1] K.J.Bathe,H.ZhangandM.H.Wang.“FiniteElementAnalysisof IncompressibleandCompressibleFluidFlowswithFreeSurfacesandStructuralInteractions.” Computers & Structures ,56:193–213,1995.
[2] K.J. Bathe, H. Zhang and S. Ji. “Finite Element Analysis of Fluid Flows Fully Coupled withStructuralInteractions.”Computers & Structures ,72:1–16,1999.
[3] K.J.BatheandH.Zhang.“FiniteElementDevelopmentsforGeneralFluidFlowswithStructuralInteractions.”International Journal for Numerical Methods in Engineering ,60:213–232,2004.
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture18- Solutionof F.E.equations
Prof. K.J.Bathe MITOpenCourseWare
Instructures, Reading:Sec. 8.4
F (u, p)=R . (18.1)
Inheattransfer,
F (θ) =Q (18.2)
Influidflow,
F (v, p,θ) =R (18.3)
Instructures/solids
F
(m)
t (m)T tˆ(m) 0 (m)F = = 0BL
0S d V (18.4)0V (m)
m m
Elasticmaterials
Example p. 590textbook
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MIT2.094 18.Solutionof F.E.equations
Plasticity
• yieldcriterion
flowrule•
• hardeningrule
ttτ
=t−∆tτ + dτ (18.16)t−∆t
Solution
of
(18.1)
(similarly
(18.2)
and
(18.3))
Newton-Raphson FindU ∗ asthezeroof f (U ∗)
t+∆t t+∆tR f (U ∗) = (18.17)F −
·t+∆t (i−1)U
U ∗ −t+∆tU (i−1) +H.O.T. (18.18)∂ f t+∆t
(i−1) +=f U
∂ U
where t+∆tU (i−1) isthevaluewe justcalculatedandanapproximationtoU ∗.
Assume t+∆tR isindependentof thedisplacements.
·∆U (i) (18.19)∂ t+∆tF t+∆tR
t+∆tF
(i−1)0= − −
∂ U t+∆t (i−1)U
Weobtain
t+∆tK (i−1)∆U (i) =t+∆tR −t+∆tF (i−1) (18.20)
(18.21)∂
t+∆tF
∂ F t+∆tK (i−1) = =∂ U
t+∆t (i−1) ∂ U t+∆t (i−1)U U
Physically
∆t+∆t
F 1(i−1)
t+∆t (i−1)K 11 = (18.22)∆u
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MIT2.094 18.Solutionof F.E.equations
Pictorially
for
a
single
degree
of
freedom
system
i=1; tK ∆u(1) =t+∆tR−tF (18.23)
i=2; t+∆tK (1)∆u(2) =t+∆tR t+∆tF (1) (18.24)−
Convergence Use
∆U (i)2
< � (18.25)
a2 = (ai)
2
(18.26)
i
But,if incrementaldisplacementsaresmall ineveryiteration,needtoalsouse
t+∆tR −t+∆tF (i−1)2 < �R (18.27)
18.1 Slenderstructures
(beams,plates,shells)
t
1 (18.28)Li
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MIT2.094 18.Solutionof F.E.equations
Beam
t 1e.g.L
= 100
(4-nodeel.) The element does not have curvature we have aspuriousshearstrain
→
(9-node
el.)
Wedonothaveashear(better)→But, still for thin structures, it has problems→
likeill-conditioning.
We need to use beam elements. For curved structures also spurious membrane strain can be⇒present.
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2.094—FiniteElementAnalysisof SolidsandFluids Fall‘08
Lecture19- Slenderstructures
Prof. K.J.Bathe MITOpenCourseWare
Beamanalysis, tL
1(e.g. tL
= 11100
,1000
,· · ·) Reading:Sec. 5.4,
6.5
(plane
stress)
2L 0
(19.1)J =0
1
t
2
(1−r) (1+s) (19.2)h2 =41
(1−r)(1−s) (19.3)h3
=4
∗ ∗=v v32
3
2
Beamtheoryassumptions(Timoshenkobeamtheory):
u∗
u∗
(19.4)
(19.5)
(19.6)
=v2t
=u2
+ θ22t
=
u2
−
2
θ2
∗ ∗ ∗ ∗u v u v3322
1 2L
0 − 14 (1−s)2L
0
2(1 )−rt
(1+s)− 4
0
etc (19.7)B∗ =1
42(1 )−rt
(1+s)
1
42(1 )−rt
2(1−s)
0 −
1 2L
− 142(1 )−rt
81
− 14− 4 L
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MIT2.094 19.Slenderstructures
u2
v2
θ2 ∂u
∂x1 t0− s
∼
L 2L
Bbeam =
· · ·0
(19.8)0 0 0 ∂v
∂y
0 −1 −1(1−r)∂y
∂x
L
2
∂u + ∂v
1v(r)=
2(1−r)v2
1 st u(r)=
2(1−r)u2
−4
(1−r)θ2
atr=−1,
v(−1)=v2
st
u(−1)=−2
θ2
+u2
Kinematics is
1u(r)=
2(1−r)u2
resultsinto�xx
∂u 2 1
→
�xx
=∂r
·
L
=−L
st
u(r,s) =−4
(1−r)θ2
resultsinto�xx,γ xy
st
�xx =→2L
∂u 2 1
γ xy =∂s
·t
=−2
(1−r)
1v(r)=
2(1−r)v2
resultsintoγ xy
1→γ xy
=−L
(19.9)
(19.10)
(19.11)
(19.12)
(19.13)
(19.14)
(19.15)
(19.16)
(19.17)
(19.18)
(19.19)
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MIT2.094 19.Slenderstructures
Forapurebendingmoment,wewant
1 1 −
Lv2 −
2(1−r)θ2 =0 (19.20)
forallr! Impossible (exceptforv2 =θ2 =0) So,theelementhasaspuriousshearstrain!⇒ ⇒
Beam
kinematics (Timoshenko,Reissner-Mindlin)
dwγ
=dx
−β (19.21)
1I = bt3 (19.22)
12
Principle
of
virtual
work
EI
L
0
dβ
dx
dβ
dx
dx+AS
G
L
0
dw
dx
−β
dw
dx
−β
dx=
L
0
pwdx (19.23)
As
=kA=kbt (19.24)
Tocalculatek Reading:
A
1
2G(τ a)
2dA=
AS
1
2G
V
As
2
dAs (19.25)
p. 400
where
τ a is
the
actual
shear
stress:
τ a
=3
2·
V
A
�