intro.pptemweb.unl.edu/negahban/femfundamentals/introduction/intro.pdf · title: microsoft...
TRANSCRIPT
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ENGM 918
Fundamentals of Finite Elements
Mehrdad Negahbang
W311 Nebraska HallDepartment of Engineering Mechanics
University of Nebraska-Lincoln
Phone: 472-2397E il hb @ l dE-mail: [email protected]
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Course description
Objective:
• Solution of partial differential equations of engineering importance usingSolution of partial differential equations of engineering importance using FEM. • Emphasizing the similarity of the development.• Development of common mathematical and numerical tools.• Using consistent and robust methods.• Developing an object oriented programming structure for FEM.• Addressing some advanced topics. 4 Elements
7 7 9 9 N d El t7x7 Element: 16 Load steps, 127 Newton Iterations9x9 Element: 24 Load steps 163 Newton Iterations
AuBv−
7x7 or 9x9 Node Elements 9x9 Element: 24 Load steps, 163 Newton Iterations
F Fw
FA
B
uv
w
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Topics
Course
• Object oriented programming in FEMW i h d R id l M h d• Weighted Residual Methods
• Linear Problems (scalar and vector fields, multi physics)• Heat transfer, elasticity, and thermoelasticity
• Constraints• Constraints• Non-Linear Problems
• Nonlinear elasticity (large deformation)• Initial-Value ProblemsInitial Value Problems
• Transient thermal analysis• Vibration of beams
• Bars, Plates and Shells,• Kinematically defined continua
• Numerical Implementation
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Peanut Sunflower Eggs
ShellsPeanut Sunflower Eggs
Solanum cinereum germinatingSolanum cinereum germinatingCoconut
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ShellsAppleApple
Red blood cell
Cell structure: Procaryote
Coronary artery
Cell structure: Procaryote
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Plant cell wall
Shells
Plant cells Plant cell wallPoplar leaf cell
Chloroplast: site of photosynthesis in plant cellsTallow tree
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Typical FEM Formulation
Shell Kinematics
Constitutive Characteristics
Finite ElementApproximation
Finite ElementFormulation
Initial and BoundaryConditions
Numerical Solution
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Finite Element Formulation
Shell Kinematics
Variational FEM Formulation
Finite ElementApproximation
Formulation FEM Formulation
Constitutive Characteristics
Initial and BoundaryConditionsConditions
Numerical Solution
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Shell Kinematics
),(),,( 21321 ξξξξξζ dρ = Q )( ξξ
3ξ Director
ξ
21321 Q ),( 21 ξξdρ
Curvilinear 2ξ
Pt
Coordinate
)( ξξ),,( 321 ξξξx
1ξ
),( 21 ξξsP
Q
“Mid”-surfacetO
Reference Frame
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Shell Kinematics
Reissner/Mindlin Shell
⎧ ∂∂∂
+= ),(),,(),(),,( 2132121321
dsdsx
ςξξξξξςξξξξξ jiijg gg o=
⎪
⎪⎪⎨
⎧
∂
=∂∂
+∂∂
+∂∂
=∂∂
=3
2,1
i
iiii
ii
d
ddsxg ς
ξς
ξς
ξξ
),(),,( 21321 ξξξξξζ dρ = Q ),( 21 ξξd
3ξ
ρ
⎪⎪⎩
=∂
∂ 33
ii dξςξ
2ξ
P
ρ
),( 21 ξξd
)( ξξdP
ξ
),( 21 ξξs),,( 321 ξξξx
),( 21 ξξd−
⎧ 1)1( ξξ 1ξ
O⎩⎨⎧
=−=−
1)1,,(1)1,,(
21
21
ξξςξξς
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Making the mappings into material mappings:
Shell Kinematics
1)( −=∇= oJJxF X
dInitial configuration
Current configuration
Making the mappings into material mappings:
D
g
ΩΓ
Γ
)(XJ ξ∇=o s )(xJ ξ∇=oΩ
oΓ
3ξS
ξ
ii egJ ⊗=o
2ξ
e
iio eGJ ⊗=
1ξ2e3e
1eO
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Shell Kinematics
1)( −=∇= oJJxF X Current configuration
Deformation Gradient: 1−= oJJFIFH −=Displacement Gradient:
D
dInitial configuration
)(1 HHε += T )(XJ ξ∇=o s
D
)(xJ ξ∇=
Strains:
)(21
)(2
IFFG −= TR
3ξS
)(ξ
)(212
IFFG −= TL
2ξ
2e3e
O1ξ2e
1e
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ξ
Isoparametric shell
),(),( 2121 ξξξξ ∑=nnpe
pp Nss 2ξ
3ξ
52
Current Configuration),(),,(),(),,( 2132121321 ξξξξξςξξξξξ dsx +=
),(),( 211
21
211
21
ξξξξ ∑
∑
=
=
=nnpe
pp
p
pp
Ndd1d
4
8 1
5
),()(),,( 211
3321 ξξξςξξξς ∑=
=nnpe
pp
p N1ξ 1s
4
3eIsoparametric 2-D
3ξO
2e1e
)1,1(2ξ
12 5
Isoparametric 2 DExtruded 3-D
2ξ
3e4 8 15
21ξ6 89
1ξ 2e1e
ξ
347)1,1( −−
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Isoparametric shell
I iti l C fi ti),(),,(),(),,( 2132121321 ξξξξξςξξξξξ DSX o+= Initial Configuration
3ξ3ξ
),(),( 2121 ξξξξ ∑=nnpe
pp NSS
2ξ
1D8 1
52
2ξ
1D8 1
52
),(),( 211
21
211
21
ξξξξ ∑
∑
=
=
=nnpe
pp
p
pp
NDD
Isoparametric 2-D 1ξ 1S
4
1ξ 1S
4),()(),,( 211
3321 ξξξςξξξς ∑=
=nnpe
pp
poo N
3ξ)1,1(2ξ
12 5
Isoparametric 2 DExtruded 3-D
1ξ S
2e3e
1e
1ξ S
2e3e
1e2e
3e
1e
2ξ
3e4 8 15
21ξ6 89 OO
1ξ 2e1e
ξ
347)1,1( −−
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)()()( ξξξξξξξξ ρs +
ii
exJ ⊗∂∂
=ξ i
io eXJ ⊗
∂∂
=ξJacobians
),(),,(),(),,( 2132121321 ξξξξξζξξξξξ dsx +=
),,(),(),,( 32121321 ξξξξξξξξ ρsx +=
ppp
ppp
=Δ
−=Δ
DddSss
ρ
),(),()(),(),,( 2121321321 ξξξξξζξξξξξ qq
pp
pp NNN dsx +=
p
ppp NN
Nξξξξ
ξξζ
ξξξ
ξ ∂∂
+∂
∂=
∂∂ ),(),()(),(
2121321 dsx
po
pp ζζζ −=Δ
−=Δ Dddxs
i
qqp
pq
q
i
pp
qpiii
NNN
Nξξξ
ξξξζξξξξξ
ξζ
ξξξξξξξ
∂
∂+
∂
∂+
∂∂∂),(
),()(),(),(
)(
),(),(
2121321
213
2121
dd Thickness unknowns
∑=
+=nif
ii
pi
po
p fa1
333 )()()( ξξζξζ
pi
po
i
pp
i
NN
NNN
ξξξξ
ξξξξξξζ
ξξξ
ξ∂∂
∂∂
+∂
∂=
∂∂
)()(
),(),()(),(2121
321 DSX Thickness basefunctions
i
qqp
poq
q
i
ppo
NNN
Nξξξ
ξξξζξξξξξ
ξζ∂
∂+
∂
∂+
),(),()(),(
),()( 21
2132121
3 DD33)( ξξζ =p
o
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3ξNumber of interpolation functions
Thickness interpolation
33)( ξξζ =po
∑+=nif
ii
pi
po
p fa1
333 )()()( ξξζξζ
3ξ1
)( 3ξif
Number of interpolation functions
=i 1
1−Legendre polynomials:
Material surface
⎪
⎪⎨
⎧ −=
+ oddisiPPf
oi
i
)()()(
331
3
ξξξ
⎪⎩ −+ evenisiPPi )()( 3131 ξξ
One term approximation:
)1)(0()()( 23131131 ξζξξζ −Δ==Δ ppp fa
Mid-surface
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R id l f FEM
FEM formulationNominal stress TFT 1−= J
∫∫∫∫ΩΩΩΓ
Ω−Ω+Ω∇−Γ= oooooToo
No ddddR aubuTutu X ooo ρρ:)()(
Residual for FEM: Nominal stress TFT = Jo
1)( −=∇= oJJxF X
dInitial configuration
Current configuration
ΩΩΩΓ oooo
oTo
No NTNTt ==)(
DoΓN)(Not
3ξ
)(XJ ξ∇=o
Ss )(xJ ξ∇=oΩ
⎧ =Δ= jsU q 321
Nodal unknowns:
2ξ⎪⎩
⎪⎨
⎧
===Δ==Δ=
+
nifjaUjdUjsU
qjj
qjjq
jqj
,...,13,2,13,2,1
)6(
)3(
1ξ2e3e
1eO
⎩ + nifjaU jjq ,...,1)6(
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Nonlinear elasticity
D f ti di t ( l)
oldnew
UtU
t
Δ∂
+Δ∂
+
Δ+=FFFFF
FFF
&
&
Deformation gradient: (general)11
1−−
−
∂⊗∂
=∂∂
=∂∂
=∂∂
oqj
iio
qjqj
o
qj UUUUJegJJJJF
1⊗∂∂ i JgF ∂x
qjqj
oldqjqj
oldnew UU
tUU
Δ∂
+=Δ∂
+= FFF
Nominal stress: (nonlinear elastic)
1−⊗∂∂
=∂∂
oiqj
i
qj UUJegF
ii ξ∂
∂=
xg
qjqj
ooldooldonewo
oldonewo
UU
t
t
Δ∂∂
∂+=Δ∂+=
Δ+=FTFTFTFTFT
TFTFT
FF :)()(:)()()(
)()(
&
&
From shell kinematics
⎧ =Δ= jsU q 321qj
qjoldonewo U
UΔ
∂∂
+=FFTFT :)()( oE
M t i l ifi
⎪⎩
⎪⎨
⎧
===Δ==Δ=
+
+
nifjaUjdUjsU
qjjq
qjjq
jqj
,...,13,2,13,2,1
)6(
)3(
Material specific
)( oTF∂=oETangent modulus of nominal stress:
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Nonlinear elasticity
Two parameter nonlinear-elastic material model:
Nominal stress:
Tangent modulus:g
Material parameters: Shear and bulk modulus matched to linear response
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Examples from Nonlinear Elasticity
Cantilevered beam with distribute end load
4 Load steps41 Newton iterations 0.11.00.10
0.40.0102.1 max6
=====×=
bhLFE ν
Cantilevered beam with distribute end load
Ftipw
tipu−
w
u
vw
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Examples fromNonlinear Elasticity
Cantilevered beam with distribute end moment
Single (9x3)-node element16 loading steps
Cantilevered beam with distribute end moment
2360oM
M =
M16 loading steps175 Newton iterations
102.1 6×=E2
4360oM
M =
0.11.00.12
0.0
====
bhLν
oMM 36032
=
wTheoretical
oMM 360=tipu−
tipw
vw
u
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Examples from Nonlinear Elasticity
Cantilevered beam with distribute end momentCantilevered beam with distribute end moment
(a) 2-(4x3)-360o-16LINC-187NI (b) 3-(4x3)-360o-16LINC-147NI (c) 4-(4x3)-360o-16LINC-147NI( ) ( ) ( ) ( ) ( ) ( )
(f) 3-(5x3)-540o-24LINC-224NI(d) 2-(5x3)-360o-16LINC-150NI (e) 3-(5x3)-360o-16LINC-150NI
(g) 3-(6x3)-540o-24LINC-221NI (h) 3-(6x3)-720o-32LINC-309NI (i) 3-(7x3)-720o-32LINC-329NI
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Examples from Nonlinear Elasticity
Buckling in compression of a columnBuckling in compression of a column
u u
vw
tipu−
301002 11 =×=E ν
u
tipw
0045.0075.05.0
3.0100.2
===
=×=
hbL
E ν
L FL FCritical buckling load
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Examples from Nonlinear Elasticity
Transverse loading of an annular ringTransverse loading of an annular ring
Aw BwB
8.003.0
0.100.60.01021
max
6
==
===×=
qh
RRE
oi
νA
w
vu
A
B
qA
B
q
32 Load Steps243 Newton steps5 Elements(7x7)-Nodes per element
R
oR
A
B
q
iR
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Examples from Nonlinear Elasticity
Pinching of hemispherical shell with hole
4 Elements7x7 or 9x9 Node Elements
7x7 Element: 16 Load steps, 127 Newton Iterations9x9 Element: 24 Load steps, 163 Newton Iterations
Pinching of hemispherical shell with hole
AuBv−
ric
Symm
etric
Hole
z
F F
FA
B
uv
wSym
met
r c
F
F
R = 10h 0 04
A
B y
FuF h = 0.04E = 6.825x107
ν = 0.3Hole = 18o
x
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Examples from Nonlinear Elasticity
Pinching of cylindrical shell with free endsPinching of cylindrical shell with free ends
v
40 Load Steps174 Newton steps4 Elements(9x9) Nodes per element
FF
Bv−Cv−Aw(9x9)-Nodes per element
A
Cv−
Bv−
FFww
R
2L
B
C
400003125010510 6 ==×= FE νvu
vu 094.0953.435.10
400003125.0105.10 max
=====×=
hRLFE ν
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Examples from Nonlinear Elasticity
Point load on a hinged cylindrical roof
Hinged
FreeF
Hinged
FreeF
3.075.31021.02542540
=====
νθ
EradLR
Point load on a hinged cylindrical roof
Hinged
H
Free
R
LHinged
H
Free
R
L
Rθ
Rθ
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Four parameter elastic model
⎤⎡ ⎞⎛ 2
τ ⎟⎠⎞
⎜⎝⎛ −⎥⎥⎥⎥⎤
⎢⎢⎢⎢⎡
+⎟⎟⎠
⎞⎜⎜⎝
⎛−
+−= ∞
∗
IBBIT )(31
)3(tanh
)1( 3/52
12
2
3/5 trJG
G
IG
JGJ
o
o
oo
τκ
G
⎠⎝⎥⎥
⎦⎢⎢
⎣−∗
3)3( 12
JIGJ
o
o
τ
∞G1 )det(F=J
)(tr Coτ
( )GG κτ ,,, FFC T
32
*1
)(
J
trI C=
oGγ1
( )ooo GG κτ ,,, ∞ FFC T=
γ1
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Large deformation elastic-plastic model
E
⎟⎟⎠
⎞⎜⎜⎝
⎛−+−= IBBIT
3)(1)1(
35
ee
e
e tr
JGJκ )21(3 ν
κ−
=E
EG
Cauchy stress:
⎠⎝3J)1(2 ν+
=G
be1e TTFFΔT −= − 0Tb =Over-stress:
22f ΔSΔS IΔTΔTΔS )(tr
Yield function:
2
3: yf σ−= ΔSΔS IΔTΔS
3)(
−=
Power-law hardening rule:
nyoy a )1( ξσσ +=
pLΔT :Tξ& pLΔT :=ξ
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Two models in tension
3500
4000 Four Parameter Model
Large Deformation Elastic-Plastic Model
2500
3000
4 parameter Elastic-plastic
2000
2500
Stre
ss
000,45000,90
==
Go
oκ000,45000,90
==
Gκ
4-parameter Elastic plastic
1000
1500
700520,1
,
==
∞Go
o
τ6.0
700,1
=
=
ayoσ
500115.0=n
00 0.05 0.1 0.15 0.2 0.25 0.3 0.35
Strain
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Beam in compression
12000
Four Parameter Brick
8000
10000
on)
Four Parameter Brick
Four Parameter Shell
Large Deformation Elastic-Plastic Brick
6000
8000
Com
pres
sio
u
4000Loa
d (C
2000L=10, b=1, h=1
0-7 -6 -5 -4 -3 -2 -1 0
Displacement
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Beam in tension
3000
3500
2500
3000
1500
2000
Loa
d
u
1000
1500Four Parameter Brick
Four Parameter Shell
500Large Deformation Elastic-Plastic Brick
L=10, b=1, h=1
00.0 0.4 0.8 1.2 1.6
Displacement
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Cantilevered beam
1200Four Parameter BrickFour parameter brickLarge Deformation Elastic-Plastic BrickL D f ti El ti Pl ti B i k
uv
800
1000Large Deformation Elastic-Plastic BrickFour Parameter ShellFour Parameter Shell
v
600
Loa
d
L=10, b=1, h=1
uv
400
0
200
0-12 -10 -8 -6 -4 -2 0
Displacement
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Multi-scale analysis
Each integration point can have an RVE
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NaBRO
Questions?Q