FINITE ELEMENT : MATRIX FORMULATION

Transcription

1 FINITE ELEMENT : MATRIX FORMULATION Georges Cailletaud Ecole des Mines de Paris, Centre des Matériaux UMR CNRS 76 Contents /67

2 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

3 Continuous Discrete Continuous How much rain? Discrete versus continuous /67

4 Continuous Discrete Continuous Geometry discretization Discrete versus continuous 4/67

5 Continuous Discrete Continuous Unknown field discretization Discrete versus continuous 5/67

6 Continuous Discrete Continuous Use elements Discrete versus continuous 6/67

7 Finite Element Discretization Replace continuum formulation by a discrete representation for unknowns and geometry Unknown field: u e (M) = i N e i (M)q e i Geometry: x(m) = i N e i (M)x(P i ) Interpolation functions Ni e and shape functions Ni e M, Ni e (M) = and Ni e (P j ) = δ ij i such as: Isoparametric elements iff N e i N e i Discrete versus continuous 7/67

8 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

9 D-mapping Subparametric Superparametric Isoparametric element element element Geometry Unknown field Geometry Unknown field Geometry Unknown field more field nodes more geometrical nodes same number of than geometrical nodes than field nodes geom and field nodes Rigid body displacement not represented for superparametric element that has nonlinear edges! The location of the node at the middle of the edge is critical for quadratic edges Element 9/67

10 Shape function matrix, [N] Deformation matrix, [B] Field u, T, C Gradient ε, grad(t ),... Constitutive equations σ = Λ : ε, q = kgrad(t ) Conservation div(σ ) + f = 0,... First step: express the continuous field and its gradient wrt the discretized vector Element 0/67

11 Deformation matrix [B] () Knowing: u e (M) = i N e i (M)q e i Deformation can be obtained from the nodal displacements, for instance in D, small strain: ε xx = u x x = N (M) qx e + N (M) qx e +... x x ε yy = u y y = N (M) qy e + N (M) qy e +... y y ε xy = u x y + u y x = N (M) qx e + N (M) qy e +... y x Element /67

12 Deformation matrix [B] () Matrix form, 4-node quadrilateral ( ) {u} e = [N] T {q} e N 0 N = 0 N 0 N N 0 N 0 N 0 N 4 q e x q e y... q e 4y {ε} e = [B] T {q} e = N,x 0 N,x 0 N,x 0 N 4,x 0 0 N,y 0 N,y 0 N,y 0 N 4,y N,y N,x N,y N,x N,y N,x N 4,y N 4,x q e x q e y... q e 4y Shear term taken as γ = ε Element /67

13 Reference element η η y ξ ξ x Actual geometry Physical space (x, y) Reference element Parent space (ξ, η) f (x, y)dxdy = + + Ω f (ξ, η)jdξdη J is the determinant of the partial derivatives x/ ξ... matrix Element /67

14 Remarks on geometrical mapping The values on an edge depends only on the nodal values on the same edge (linear interpolation equal to zero on each side for -node lines, parabolic interpolation equal to zero for points for -node lines) Continuity... The mid node is used to allow non linear geometries Limits in the admissible mapping for avoiding singularities Element 4/67

15 Mapping of a -node line x 0 x =0 x =/ x = ξ Physical segment: x =- x = x Parent segment: ξ =- ξ = ξ =0 x = ξ + ξ x Element 5/67

16 Jacobian and inverse jacobian matrix ( ) x dx = ξ dy y ξ x η y η ( ) dξ = [J] dη ( ) dξ dη ( ) ξ dξ = x dη η x ξ ( ) ( ) y dx η = [J] dx dy dy y Since (x, y) are known from N i (ξ, η) and x i, [J] is computed from the known quantities in [J], using also: J = Det ([J]) = x ξ y η y ξ x η Element 6/67

17 Expression of the inverse jacobian matrix [J] = J y η x η y ξ x ξ For a rectangle [±a, ±b] in the real world, the mapping function is the same for any point inside the rectangle. The jacobian is a diagonal matrix, with x/ ξ = a, y/ η = b, and the determinant value is ab For any other shape, the mapping changes according to the location in the element For computing [B], one has to consider N i / x and N i / y: N i x = N i ξ ξ x + N i η η x N i y = N i ξ ξ y + N i η η y then ( ) ( ) Ni / x = [J] Ni / ξ N i / y N i / η Element 7/67

18 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

19 D solid elements Type shape interpol # of polynom of disp nodes terms CD tri lin, ξ, η CD4 quad lin 4, ξ, η, ξη CD6 tri quad 6, ξ, η, ξ, ξη, η CD8 quad quad 8, ξ, η, ξ, ξη, η, ξ η, ξη CD9 quad quad 9, ξ, η, ξ, ξη, η, ξ η, ξη, ξ η Element 9/67

20 D solid elements Type shape interpol # of polynom of disp nodes terms CD4 tetra lin 4, ξ, η, ζ CD6 tri prism lin 6, ξ, η, ζ, ξη, ηζ CD8 hexa lin 8, ξ, η, ζ, ξη, ηζ, ζξ, ξηζ CD0 tetra quad 0, ξ, η, ζ, ξ, ξη, η, ηζ, ζ, ζξ CD5 tri prism quad 5, ξ, η, ζ, ξη, ηζ, ξ ζ, ξηζ, η ζ, ζ, ξζ, ηζ, ξ ζ, ξηζ, η ζ CD0 hexa quad 0, ξ, η, ζ, ξ, ξη, η, ηζ, ζ, ζξ, ξ η, ξη, η ζ, ηζ, ξζ, ξ ζ, ξηζ, ξ ηζ, ξη ζ, ξηζ CD7 hexa quad 7 ξ i η j ζ k, (i, j, k) 0,, Element 0/67

21 Isoparametric representation Example: D plane stress elements with n nodes Element geometry n = N i x = i= n N i x i y = i= Displacement interpolation n u x = N i u xi u y = i= n N i y i i= n N i u y i i= Matrix form... x y u x = x x x... x n y y y... y n u x u x u x... u xn u y u y u y u y... u yn Element /67 N N N.. N n

22 The linear triangle IFEM Felippa Terms in, ξ, η Element /67

23 The bilinear quad IFEM Felippa Terms in, ξ, η, ξη Element /67

24 The quadratic triangle IFEM Felippa Terms in, ξ, η, ξ, ξη, η Element 4/67

25 The biquadratic quad IFEM Felippa Terms in, ξ, η, ξ, ξη, η, ξ η, ξη, ξ η Element 5/67

26 The 8-node quad IFEM Felippa Corner nodes: N i = 4 ( + ξξ i)( + ηη i )(ξξ i + ηη i ) Mid nodes, ξ i = 0: N i = ( ξ )( + ηη i ) Mid nodes, η i = 0: n I = ( η )( + ξξ i ) Terms in, ξ, η, ξ, ξη, η, ξ η, ξη Element 6/67

27 Approximated field Polynomial basis ξ η ξ ξη η ξ ξ η ξη η Examples : CD4 ( + ξ i ξ)( + η i η) CD8, corner 0.5( + ξ i ξ + η i η)( + ξ i ξ)( + η i η) CD8 middle 0.5(. ξ )(. + η i η) Element 7/67

28 The -node infinite element Displacement is assumed to be q at node and q = 0 at node x ξ x x Interpolation Geometry N = ξ N such as x = x + α + ξ ξ ξ =? Resulting displacement interpolation u(x) =?? N = + ξ N = 0 Element 8/67

29 The -node infinite element Displacement is assumed to be q at node and q = 0 at node x ξ x x Interpolation Geometry N = ξ N such as x = x + α + ξ ξ ξ = x x α x x + α Resulting displacement interpolation u(x) =? N = + ξ N = 0 Element 9/67

30 The -node infinite element Displacement is assumed to be q at node and q = 0 at node x ξ x x Interpolation Geometry N = ξ N such as x = x + α + ξ ξ ξ = x x α x x + α Resulting displacement interpolation N = + ξ u(x) = N (x) q = N (ξ(x)) q = N = 0 αq x x + α Element 0/67

31 Connecting element lin quad. Connection between a linear and a quadratic quad Quadratic interpolation with node number 8 in the middle of 7: u(m) = N q + N 8 q 8 + N 7 q 7 On edge 7, in the linear element, the displacement should verify: q 8 =? Overloaded shape function in nodes and 7 after suppressing node 8: u(m) =?? Element /67

32 Connecting element lin quad. Connection between a linear and a quadratic quad Quadratic interpolation with node number 8 in the middle of 7: u(m) = N q + N 8 q 8 + N 7 q 7 On edge 7, in the linear element, the displacement should verify: q 8 = (q + q 7 )/ Overloaded shape function in nodes and 7 after suppressing node 8: u(m) =?? Element /67

33 Connecting element lin quad. Connection between a linear and a quadratic quad Quadratic interpolation with node number 8 in the middle of 7: u(m) = N q + N 8 q 8 + N 7 q 7 On edge 7, in the linear element, the displacement should verify: q 8 = (q + q 7 )/ Overloaded shape function in nodes and 7 after suppressing node 8: ( u(m) = N + N ) ( 8 q + N 7 + N ) 8 q 7 Element /67

34 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

35 Thermal conduction Strong form: GIVEN r : Ω R, a volumetric flux, φ d : Γ f R, a surface flux, T d : Γ u R, a prescribed temperature, FIND T : Ω R, the temperature, such as: in Ω φ i,i = r on Γ u T = T d on Γ F φ i n i = Φ d Constitutive equation (Fourier, flux (W /m ) proportional to the temperature gradient) φ i = κ ij T, j conductivity matrix: [κ] (W /m.k) Global problem 5/67

36 Thermal conduction () Weak form: S, trial solution space, such as T = T d on Γ u V, variation space, such as δt = 0 on Γ u GIVEN r : Ω R, a volumetric flux, Φ d : Γ f R, a surface flux, T d : Γ u R, a prescribed temperature, FIND T S such as δt V φ i δt, i dω = δtrdω + δt Φ d dγ Ω Ω Γ F For any temperature variation compatible with prescribed temperature field around a state which respects equilibrium, the internal power variation is equal to the external power variation: δt, i φ i is in W /m T is present in φ i = κ ij T, j Global problem 6/67

37 Elastostatic Strong form: volume Ω with prescribed volume forces f d : σ ij,j + f i = 0 surface Γ F with prescribed forces F d : Fi d = σ ij n j surface Γ u with prescribed displacements u d : u i = ui d Constitutive equation: σ ij = Λ ijkl ε kl = Λ ijkl u k,l So that: Λ ijkl u k,lj + f i = 0 Global problem 7/67

38 Principle of virtual power Weak form: volume V with prescribed volume forces : f d surface Γ F with prescribed forces : F d surface Γ u with prescribed displacements : u d Virtual displacement rate u kinematically admissible ( u = u d on Γ u ) The variation u is such as: u = 0 on Γ u. Galerkin form writes, u: σ : ε dω = f d u dω + u ds Ω Ω Γ F F d Global problem 8/67

39 Discrete form of virtual power Application of Galerkin approach for continuum mechanics: virtual displacement rate u w h ; σ u h, x { u e }, nodal displacements allow us to compute u and ε : u = [N]{ u e } ; ε = [B]{ u e } Galerkin form writes, { u e }: ( ) {σ({u e }).[B].{ u e } dω = elt Ω elt ( f d.[n].{ u e } dω Ω + F d.[n].{ u e } ds) Γ F Global problem 9/67

40 Internal and external forces In each element e: Internal forces: {F e int} = External forces: {Fext} e = {σ({u e }).[B] dω = [B] T {σ({u e }) dω Ω Ω f d.[n]dω + F d.[n]ds Γ F Ω The solution of the problem: {F e int ({ue })} = {F e ext} with Newton iterative algorithm will use the jacobian matrix : [K e ] = {F int e } {u e } = [B] T. {σ} Ω {ε}. {ε} {u e } dω = [B] T. {σ}.[b] dω {ε} Ω Global problem 40/67

41 Linear and non linear behavior Applying the principle of virtual power Stationnary point of Potential Energy For elastic behavior [K e ] = [B] T.[Λ ].[B] dω Ω is symmetric, positive definite (true since σ and ε are conjugated) [ ] {σ} For non linear behavior, one has to examine [L c ] =. Note {ε} that [L c ] can be approached (quasi-newton). {F e ext} may depend on {u e } (large displacements). Global problem 4/67

42 Elastostatic, strong and weak form, a summary BCu Displacements u Body forces f Kinematics Equilibrium Strains ε Constitutive equations Stresses σ BCs STRONG BCu: u = u d on Γ u Kinematics: ε = [B] u in Ω Constitutive equation: σ = Λε Equilibrium: [B] σ + f = 0 BCs: σn = F on Γ F WEAK BCu: u h = u d on Γ u Kinematics: ε = [B] u h in Ω Constitutive equation: σ = Λε Equilibrium: δπ = 0 BCs: δπ = 0 Global problem 4/67

43 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

44 Matrix vectors formulation of the weak form of the problem [K] {q} = {F } Thermal conduction: [K] = [B] T [κ] [B] dω {F } = Ω Elasticity: [K] = [B] T [Λ] [B] dω {F } = Ω Internal forces: {F e int} = External forces: {Fext} e = Ω In each element e: Ω Ω Ω [N] rdω + [N] Φ d dγ Ω [N] f d dω + [N] F d dγ Ω {σ({u e }).[B] dω = [B] T {σ({u e }) dω f d.[n]dω + F d.[n]ds Γ F Global problem Ω 44/67

45 The stiffness matrix Example of a 4-node quad and of a 0-node hexahedron () [B] T [D] [B] [K] (6) (6) 8 (60) 8 (60) 8 8 (60) (6) (6) (60) = The element stiffness matrix is a square matrix, symmetric, with no zero inside. Its size is equal to the number of dof of the element. Global problem 45/67

46 Nodal forces () {Fext} e = [N] T F d ds Γ F η 7 F t F n η ξ ξ F x ds = F t dx F n dy F y ds = F n dx + F t dy with ( ) ( ) dx dξ = [J] dy dη Global problem 46/67

47 Nodal forces () Integration on edge 5 7: dx = x ξ dξ y dy = ξ dξ Components 9, 0, for the nodes 5;, for nodes 6;, 4 for nodes 7 ( Fext(i e x ) = e N i F t ξ F n y ξ ) dξ ( ) Fext(i) e x = e N i F n ξ + F y t dξ ξ Example, for a pressure F n = p, and no shear (F t = 0) on the 5 7 edge of a 8-node rectangle a x a y = b represented by ξ η = x ξ = a y ξ = 0 N 5 = ξ( + ξ)/ N 6 = ξ N 7 = ξ( ξ)/ Global problem 47/67

48 Nodal forces () (4) () () F 0 = F 5y = e F = F 6y = e ξ( + ξ)padξ = ap ( ξ )padξ = 4ap The nodal forces at the middle node are 4 times the nodal forces at corner nodes for an uniform pressure (distribution... after adding the contribution of each element) Global problem 48/67

49 Nodal forces (4) Axisymmetric 8-node quad z () () Face of a 0-node hexahedron (4) ( ) (4) ( ) (4) (4) ( ) ( ) Global problem 49/67

50 Nodal forces (5) Face of a 7-node hexahedron who knows? Face of a 5-node hexahedron () () () Global problem 50/67

51 Assembling the global matrix C B A Local versus global numbering Global problem 5/67

52 Assembling the global matrix 0 F = F A F = F A +F B F = +F B F 4 = F A +F4 B +F C BF 5 = +F B +F C F 6 = +F4 C A F 7 = +F C A q = q A q = q A = q B q = = q B q 4 = q A = q4 B = q C Bq 5 = = q B = q C q 6 = = q4 C A q 7 = = q C C B Global problem 5/67

53 Assembling the global matrix C B A Global problem 5/67

54 Assembling the global matrix C B A Global problem 54/67

55 Assembling the global matrix 6 4 C A B Global problem 55/67

56 Assembling the global matrix 6 4 C A B Global problem 56/67

57 Assembling the global matrix 6 4 C A B Global problem 57/67

58 Assembling the global matrix 6 4 C A B Global problem 58/67

59 Assembling the global matrix C A B Global problem 59/67

60 Assembling the global matrix C A B Global problem 60/67

61 Assembling the global matrix C A B Global problem 6/67

62 Assembling the global matrix C B A Global problem 6/67

63 Assembling the global matrix C B A Global problem 6/67

64 Contents Discrete versus continuous Element Interpolation Element list Global problem Formulation Matrix formulation Algorithm

65 Global algorithm For each loading increment, do while {R} iter > EPSI : iter = 0; iter < ITERMAX ; iter + + Update displacements: {u} iter+ = {u} iter + δ{u} iter Compute {ε} = [B]. {u} iter+ then ε for each Gauss point Integrate the constitutive equation: ε σ, α I, σ ε 4 Compute int and ext forces: {F int ({u} t + {u} iter+ )}, {F e } 5 Compute the residual force: {R} iter+ = {F int } {F e } 6 New displacement increment: δ{u} iter+ = [K].{R} iter+ Global problem 65/67

66 Convergence Value of the residual forces < R ɛ, e.g. {R} n = Relative values: ( i ) /n Ri n ; {R} = max R i i Displacements {R} i {R} e {R} e < ɛ {U}k+ {U} k n < U ɛ Energy [ {U}k+ {U} k ] T. {R}k < W ɛ Global problem 66/67