Locking phenomena in Computational Mechanics: nearly incompressible materials and plate problems

Transcription

1 Locking phenomena in Computational Mechanics: nearly incompressible materials and plate problems C. Lovadina Dipartimento di Matematica Univ. di Pavia IMATI-CNR, Pavia Bologna, September, the 18th 2006

2 Aim of the talk Discuss the troubles of problems with small (or large) parameters. 1. Understand the roots of the difficulties. 2. Suggest possible cures. Examples: Finite Element approximation of nearly incompressible elasticity and plate problems.

3 THE LINEAR ELASTICITY PROBLEM Plane linear elasticity problem in the framework of the infinitesimal theory, and plane strain case. Isotropic and Homogeneous material. t Γ D f Γ N

4 Solve the problem: Find u such that div ( ) 2µε(u) + λ(divu)δ = f in u = u 0 on Γ D ( 2µε(u) + λ(divu)δ ) n = t on Γ N 1. u, f : R 2 are the displacement field and the loading term; 2. ε( ) is the symmetric gradient operator; 3. δ is the second-order identity tensor. µ and λ are the Lamé coefficients 0 < µ 0 µ µ 1 < + and 0 < λ 0 λ +

5 Equivalent formulation Find u which minimizes the elastic energy E(v) = µ ε(v) : ε(v) + λ 2 divv 2 f v Γ N t v over a suitable set of admissible functions V. Remark. For homogeneous Dirichlet boundary conditions, as in the sequel: V = { v H 1 () : v ΓD = 0 }

6 Find u V such that: Variational formulation 2µ ε(u) : ε(v) + λ divudivv = F(v) v V where F(v) := f v + Γ N t v The bilinear form 2µ ε(u) : ε(v) + λ divudivv is continuous, symmetric and coercive; F is linear and continuous. Consequence The problem has a unique solution and there is stability (continuous dependence on the data).

7 Standard Finite Elements Choose V h finite element space. Find u h V h such that: 2µ ε(u h ) : ε(u h ) + λ divu h divv h = F(v h ) v h V h U h

8 Test problem using the above mesh and data (given tractions on Neumann boundary) Rubber material (nearly incompressible): λ/µ >> 1 A good description of the deformation is expected!! HOWEVER...

9 Discrete solution analytical solution blue: the analytical solution. red: the discrete solution. The method HEAVILY underestimates the solution

10 WHY?? We need to recall the energy functional: E(v) = µ ε(v) : ε(v) + λ 2 which has to be minimized over V. divv 2 F(v) λ/µ >> 1 corresponds to µ 1, λ + Therefore, for λ/µ >> 1, the minimiser u V satisfies: divu 0

11 The limit problem It can be shown that u u 0 as λ +, where u 0 solves the the limit problem Find u 0 K which minimizes E 0 (v) = µ ε(v) : ε(v) F(v) in K K = {v V : divv = 0} Remark: The above problem is well-posed and reasonable. It is the elasticity problem for incompressible materials.

12 Finite Element & Minimization Standard Finite Elements corresponds to Find u h V h which minimizes E(v h ) = µ ε(v h ) : ε(v h ) + λ 2 divv h 2 F(v h ) in V h Remark: The SAME energy, but different admissible functions.

13 The Finite Element limit problem It can be shown that u h u 0 h as λ +, where u h 0 solves the the limit problem Find u h 0 K h which minimizes E(v h ) = µ ε(v h ) : ε(v h ) F(v h ) in K h K h = {v h V h : divv h = 0} = K V h Question: Is it a good limit problem?

14 { } v h K h = v h is continuous, piecewise linear, and divv h = 0 P v h P v h Therefore, it holds: v h (P) = 0

15 Repeating the argument: v h K h = v h 0 Therefore, K h = (0)

16 The Finite Element limit problem revisited Find u 0 h K h which minimizes E 0 (v h ) = µ ε(v h ) : ε(v h ) F(v h ) in K h K h = (0) The divergence-free constraint is too severe for finite element functions: Volumetric Locking!

17 A possible cure The enemy: the term λ 2 divv h 2 for λ + Idea Take λ 2 P h (divv h ) 2 P h suitable reduction operator (typically a projection operator)

18 New Finite Element problem Find u h V h which minimizes E h (v h ) = µ ε(v h ) : ε(v h ) + λ 2 P h (divv h ) 2 F(v h ) in V h Find u 0 h K h which minimizes Limit Problem Eh(v 0 h ) = µ ε(v h ) : ε(v h ) F(v h ) in K h K h := {v h V h : P h (divv h ) = 0 } Remark: P h (divv h ) = 0 may be weaker than divv h = 0.

19 Example: pw. quadratic functions (not recommended) C x U h P h : Projection operator on piecewise constant functions v h K h = divv h (C) = 0

20 Classical Approaches Galerkin Least Squares Methods: Hughes Franca ( ) Enhanced Strain Methods: Simo Rifai (1990), Pantuso Bathe (1995)... MINI element: Arnold, Brezzi, Fortin (1984). Taylor Hood Elements (1973). Non-conforming methods: Crouziex-Raviart Element (1973). Remark: Most of them are Mixed Methods.

21 THE PLATE PROBLEM We consider a plate subjected to a transversal load (with respect to its mid-plane) The 2D Reissner-Mindlin model Undeformed plate: made up by fibers, which are rectilinear and perpendicular to the mid-plane. Due to deformation, the fibers remain rectilinear, but they are not perpendicular to the mid-plane anymore.

22 θ w Problem unknowns The vectorial field θ = θ(x, y) (fiber rotations). The scalar field w = w(x, y) (vertical displacements).

23 Find (θ, w) s.t. The equations (clamped plate) divcε(θ) λt 2 ( w θ) = 0 in, div ( λt 2 ( w θ) ) = g in, θ = 0, w = 0 on. C and λ: material parameters; t: plate thickness (t << diam()); g: transversal load.

24 Find (θ, w) which minimizes Energy minimization E(η, v) = 1 2 Cε(η) : ε(η) + λt 2 2 v η 2 gv over the admissible space Θ W Θ = (H 1 0()) 2, W = H 1 0()

25 Find (θ, w) Θ W such that Variational formulation Cε(θ) : ε(η) + λt 2 ( w θ) ( v η) gv for every (η, v) Θ W. The bilinear form Cε(θ) : ε(η) + λt 2 ( w θ) ( v η) is continuous, symmetric and coercive; gv is linear and continuous. Consequence The problem has a unique solution and there is stability (continuous dependence on the data).

26 Standard Finite Elements Choose Θ h Θ and W h W finite element spaces. Find (θ h, w h ) Θ h W h s.t. Cε(θ h ) : ε(η h ) + λt 2 ( w h θ h ) ( v h η h ) gv h for every (η h, v h ) Θ h W h. θ w

27 Test problem using the above mesh and uniform constant load Thin plate: diam()/t >> 1 A good description of the deformation is expected!! HOWEVER...

28 Discrete solution analytical solution blue profile: the analytical solution. red profile: the discrete solution. The method HEAVILY underestimates the solution

29 Energy minimization The minimization problem Θ W for: E(η, v) = 1 2 Cε(η) : ε(η) + λt 2 2 v η 2 gv converges to the limit problem: Find (θ 0, w 0 ) K which minimizes E 0 (η, v) = 1 2 Cε(η) : ε(η) gv (η, v) K K = {(η, v) Θ W : v = η} Remark It is a coercive problem in K.

30 Finite Elements The minimization problem in Θ h W h for: E(η h, v h ) = 1 2 Cε(η h ) : ε(η h ) + λt 2 2 v h η h 2 gv h converges to the limit problem: Find (θ 0 h, w 0 h ) K h which minimizes E 0 (η h, v h ) = 1 2 Cε(η h ) : ε(η h ) gv h (η h, v h ) K h K h = {(η h, v h ) Θ h W h : v h = η h }

31 Structure of K h (η h, v h ) K h = v h = η h C 0 () = v h C 1 () But { vh C 1 () and piecewise linear } = v h is globally linear It follows {v h is globally linear and v h = 0 on the boundary} = v h 0 = η h = 0 Hence K h = (0) Shear Locking!!

32 A possible cure The enemy: the term λt 2 2 v h η h 2 for t 0 Idea Take λ t 2 2 R h ( v h η h ) 2 R h suitable reduction operator

33 Example: pw. quadratic functions θ w R h : Projection operator on piecewise constant functions

34 Computed vertical displacements w κ e = C 10 5 x 10 4 x

35 Modified Finite Elements & Minimization Minimization problem in Θ h W h for the new energy: E h (η h, v h ) = 1 2 Cε(η h ) : ε(η h )+ λ t 2 2 R h ( v h η h ) 2 gv h If R h ( v h ) v h we risk. It may happen: v h 0 but R h ( v h ) = 0 The energy on (0, α v h ): E h (0, α v h ) = λ t 2 2 α R h ( v h ) 2 α gv h = α gv h LINEAR functional along the direction v h!!!

36 Finite Elements for plates We need a reduction operator R h If R h reduces too much : Spurious modes. If R h do not reduce enough : Shear Locking Remarks Balancing R h is not trivial. Other difficulties arise: boundary layer effects.

37 Arnold Falk Element (1989). Possible Approaches M IT C Elements: Brezzi, Bathe, Fortin, Stenberg... ( 80 90). Linking Technique: Auricchio, Taylor, L.... ( 90). Stabilized Elements: Chapelle, Hughes, Stenberg... ( 90). Non-conforming and DG Elements: Arnold, Brezzi, L., Marini ( 04 ) Remark: Most of them are Mixed Methods.

38 Conclusions In Computational Mechanics: often problems with small or large parameter. Finite element discretization of such problems requires care. Different situations may arise,with different peculiarities. Shell Problems fall into this structure, but MUCH MORE DIFFICULT.