l13. fem for a non-linear material mechanics problem v u
TRANSCRIPT
L13. FEM for a non-linear material mechanics problem
Weak formulation - Static virtual work
Major aim: Evaluate momentum balance in weak form via FE-approximations
àVΣ
t : l âV = àVf × u âV + à
G
t × u âG " u Î U
Ý Reduce to a 2D computational domain
V ® t W, âV ® t âW, G ® t S, âG ® t âS
Ý Plane strain (for the in-plane directions 8Α, Β< = 1, 2)
Σ ® ΣΑΒ eΑ Ä eΒ + Σ33 e3 Ä e3 ; Ε ® ÎΑΒ eΑ Ä eΒ
Þ Consider 2D problem
àW
Σt : Ε âW = à
W
f × u âW + àSt × u âS " u Î U
FE discretization
� Displacement and strain fields
Consider triangulation We, e = 1, ... NEL. On each We introduce interpolation
u = âI=1
NODE
NI@Ξ1, Ξ2D uI, x = âI=1
NODE
NI@Ξ1, Ξ2D xI, u = âI=1
NODE
NI@Ξ1, Ξ2D uI Þ
l = âI=1
NODE
uI Ä gI Þ Ε =1
2âI=1
NODE IuI Ä gI + gI Ä uIM8NI@Ξ1, Ξ2D<I=1,2,NODE = element interpolation functions
8gI<I=1,2,3 = gradients of NI; gI =¶NI
¶x
Restrict to CST elements:
N1 = 1 - HΞ1 + Ξ2L, N2 = Ξ1, N3 = Ξ2 Þ
gI =¶NI
¶x= â
i=1
2 ¶NI
¶Ξi
Gi with Gi =¶Ξi
¶x
Ý Basis vector 9Gi=i=1,2
may be assessed explicitly, cf. details in lecture notes L13!
:0.0
0.5
1.0
Ξ10.0
0.5
1.0
Ξ20.0
0.5
1.0
,
0.0
0.5
1.0Ξ1
0.0
0.5
1.0
Ξ20.0
0.5
1.0
,
0.0
0.5
1.0Ξ1
0.0
0.5
1.0
Ξ20.0
0.5
1.0 >
� Gradient of basis function - Convective basis
Assess the gradients gI Þ introduce isoparametric mapping
x = âI=1
3
NI@Ξ1, Ξ2D xI
xI = are element nodal positions in W.
Ý 8gI<I=1,2,3 specified in terms of the contra-variant basis vector 9Gi=i=1,2
gI =¶NI
¶Ξi
Gi with Gi = Gij Gj and Gi =¶x
¶Ξi
Ý Gij = IGijM-1and Gij = Gi × Gj = J G1 × G1
G2 × G1
G1 × G2
G2 × G2N
Assess the basis vector Gi via isoparametric mapping of geometry:
x = âI=1
3
NI@Ξ1, Ξ2D xI Þ dx = Gi dΞi with Gi = âI=1
3 ¶NI
¶Ξi
xI
Explicit expressions
x = H1 - HΞ1 + Ξ2LL x1 + Ξ1 x2 + Ξ2 x3
Þ G1 =¶x
¶Ξ1= x2 - x1 = Dx2; G2 =
¶x
¶Ξ2= x3 - x1 = Dx3Þ
Gij = Gji =Dx2 × Dx2
Dx3 × Dx2
Dx2 × Dx3
Dx3 × Dx3; Gij = IGijM-1
=1
D
Dx3 × Dx3
-Dx3 × Dx2
-Dx2 × Dx3
Dx2 × Dx2
Þ
gI =¶NI
¶Ξ1
G1 +¶NI
¶Ξ2
G2 , G1 = G11 G1 + G12 G2 , G2 = G21 G1 + G22 G2
Ý Area of the element
2 Ae = HG1 ´ G2L × Ez = IDx2 ´ Dx3M × Ez =.. = Dx2 × Dx3 - Dx3 × Dx2
Static virtual work - Voight matrix formulation
� Voight matrix formulation
The static virtual work is rewritten as
àW
Σ : Ε âW = àW
f × u âW + àSt × u âS " u Î U
2 L13.nb
Let us immediately reformulate in matrix form - named "Voight matrix formulation"!
Þ Consider the Voight matrix form of Σt and Ε via Σt : Ε ® Σt
Ε Þ
Σt : Ε = Σij Εij ® Ε
tΣ =
HΕ11, Ε22, Ε33 = 0, Γ12, Γ13 = 0, Γ23 = 0LΣ11
Σ22
Σ33 ¹ 0Σ12
Σ13 = 0Σ23 = 0
where the situation at plane strain was introduced. In this case we may thus reduce the involved Voight arrays as
Εt
= HΕ11, Ε22, 0, Γ12L with Γ12 = Ε12 + Ε21,
Σt = HΣ11 Σ22 Σ33 Σ12L
� The B-matrix relation
B-matrix relation Ε = Be ue in terms of components of the gradients 8gI<I=1,2,3 as
Ε11
Ε22
0
Γ12
=
gx1 0 gx
2 0 gx3 0
0 gy1 0 gy
2 0 gy3
0 0 0 0 0 0
gy1 gx
1 gy2 gx
2 gy3 gx
3
ux1
uy1
ux2
uy2
ux3
uy3
The motivation for the planar CST element considered simply comes from
Ε =1
2âI=1
3 IuI Ä gI + gI Ä uIM =1
2
Igx1 + gx
1M ux1 ux
1 gy1 + uy
1 gx1
uy1 gx
1 + ux1 gy
1 Igy1 + gy
1M uy1
+
1
2
Igx2 + gx
2M ux2 ux
2 gy2 + uy
2 gx2
uy2 gx
2 + ux2 gy
2 Igy2 + gy
2M uy2
+1
2
Igx3 + gx
3M ux3 ux
3 gy3 + uy
3 gx3
uy3 gx
3 + ux3 gy
3 Igy3 + gy
3M uy3
We may further note the properties of the CST approximation in terms of the volumetric strain written as
Εv = 1 : Ε =1
2âI=1
3 I1 : IuI Ä gIM + 1 : IgI Ä uIMM = âI=1
3
uI × gI
In particular, let us consider the situation when Εv = 0 corresponding to pure deviatoric deformation as depicted in the
Figure below.
L13.nb 3
� Identity and tangent stiffness tensor representations
Fourth order: obtained via Ε : Ε = Ε : HI : ΕL = Ε : H1 Ä� 1L : Ε = Εt I Ε Þ
I =
1 0 0 00 1 0 00 0 1 0
0 0 0 1
2
Second order: via Εv = 1 : Ε ® Εv = Εt 1 Þ 1 =
1110
Deviatoric projection: Εdev = II -1
31 1tM Ε = Idev ΕÞ
Idev =
2
3-
1
3-
1
30
-1
3
2
3-
1
30
-1
3-
1
3
2
30
0 0 0 1
2
Algorithmic tangent: Ea ® Ea = Ee - 2 G J 2 G
3 G+H
nΚ+Σy
Σetr Ν
tr HΝtrLt
+ Μ3 G
Σetr IdevN, Νtr =
3
2
Σdevtr
Σetr
where it appears that the elastic stiffness modulus tensor Ee = 2 G Idev + K 1 1t can be represented as
Ee = 2 G
2
3-
1
3-
1
30
-1
3
2
3-
1
30
-1
3-
1
3
2
30
0 0 0 1
2
+ K
1 1 11 1 11 1 10 0 0
0000
In view of the elastic stiffness properties
G =E
2 H1 + ΝL , K =E
3 H1 - 2 ΝLwe thus retrieve the well known matrix representation of the isotropic elastic behavior at plane strain
Σ11
Σ22
Σ12
=E
H1 + ΝL H1 - 2 ΝL1 - Ν Ν 0
Ν 1 - Ν 0
0 0 1
2H1 - 2 ΝL
Ε11
Ε22
Γ12
where the out-of-plane component was omitted in this represenation.
� Static virtual work - Nodal FE-forces
Internal virtual work Wi:
Wi = àW
Εt
Σ âW = utA
e=1
NEL àWe
Bet
Σ âW = ut b with b = A
e=1
NELbe and be = à
We
Bet
Σ âW
8be<e=1,NEL =internal nodal forces
External virtual work We:
4 L13.nb
We = àW
f × u âW + àSt × u âS = ut
A
e=1
NELfe = ut f with f = A
e=1
NELfe
8fe<e=1,NEL=applied element nodal forces.
Equilibrium, i.e. Wi = We Þ g = b@DuD - f = 0
Ý Assembly operator Ae defines topology between ue and u, e.g. ue = Ae u.
� Linearized Static virtual work - Stiffness relationships
Consider the linearization
dg = A
e=1
NEL àWe
Bet d Σ âW = 8d Σ = Ea d Εe = Ea Be due< = A
e=1
NELA
e=1
NELKe du = K du
K = ¶b � ¶u =tangent stiffness; Ke =element stiffness
Ke = àWe
Bet Ea Be âW
Ý Also the algorithmic stiffness tensor Ea is formulated in Voight form.
� Solution procedure - Newton-Raphson's method
Solve for the balance between internal and external finite element nodal forces:
g@uD = b@uD - f@tD = 0
Þ Newton iteration procedure: given uHiL; Compute improved solution uHi+1L:
uHi+1L = uHiL + Ξ
Ξ is the iterative improvement computed from
ut Hg + K ΞL = 0 Þ Ξ = -K-1 g
L13.nb 5