Post on 05-Oct-2020
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: mnegahban@unl.edu
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
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
Peanut Sunflower Eggs
ShellsPeanut Sunflower Eggs
Solanum cinereum germinatingSolanum cinereum germinatingCoconut
ShellsAppleApple
Red blood cell
Cell structure: Procaryote
Coronary artery
Cell structure: Procaryote
Plant cell wall
Shells
Plant cells Plant cell wallPoplar leaf cell
Chloroplast: site of photosynthesis in plant cellsTallow tree
Typical FEM Formulation
Shell Kinematics
Constitutive Characteristics
Finite ElementApproximation
Finite ElementFormulation
Initial and BoundaryConditions
Numerical Solution
Finite Element Formulation
Shell Kinematics
Variational FEM Formulation
Finite ElementApproximation
Formulation FEM Formulation
Constitutive Characteristics
Initial and BoundaryConditionsConditions
Numerical Solution
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
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
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
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
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( −−
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( −−
)()()( 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
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
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(
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:
Nonlinear elasticity
Two parameter nonlinear-elastic material model:
Nominal stress:
Tangent modulus:g
Material parameters: Shear and bulk modulus matched to linear response
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Multi-scale analysis
Each integration point can have an RVE
NaBRO
Questions?Q