MIXED FINITE ELEMENTS FOR PLATES. Ricardo G. Durán Universidad de Buenos Aires

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1 MIXED FINITE ELEMENTS FOR PLATES Ricardo G. Durán Universidad de Buenos Aires - Necessity of 2D models. - Reissner-Mindlin Equations. - Finite Element Approximations. - Locking. - Mixed interpolation or reduced integration. - General Error Analysis. - L 2 Error estimates - Examples. 1

2 2 Why do we need to use 2D models if 3D elasticity can be solved by FE? 3D ELASTICITY EQUATIONS D IR 3 Initial configuration of elastic solid. u = (u 1, u 2, u 3 ) Displacement. ε ij (u) = 2( 1 ui x + u ) j j x Strain tensor. i 2µ D ε(u) : ε(v)+λ D div udiv v = v V H 1 (D) 3 D f v+ g v Γ Coercivity of this bilinear form follows from Korn s inequality : u 1 K ε(u) 0 Under appropriate boundary conditions.

3 3 Remark 1. : The constant K depends on the geometry of D. RECALL CEA S LEMMA: If u is the exact solution and u h the FE approximation then, u u h 1 M α u v h 1 v h V h where M and α are the continuity and coercivity constants of the bilinear form. In our problem α depends on the Korn s constant K. If K is too large then α is too small and the constant in Cea s Lemma is large.

4 Consider a PLATE of thickness t: 4 D = Ω ( t/2, t/2) where Ω IR 2 and t > 0, with t small in comparison with the dimensions of Ω. In this case: CONSEQUENCE: K = O(t 1 ) THE METHOD IS NOT EFFICIENT: VERY SMALL MESH SIZE WILL BE NEEDED! This is a serious computational drawback specially in 3D. SOLUTION: USE 2D MODELS!

5 REISSNER-MINDLIN EQUATIONS 5 Ω IR 2 D = Ω ( t/2, t/2) Displacements are approximated by u 1 (x, y, z) zθ 1 (x, y) u 2 (x, y, z) zθ 2 (x, y) u 3 (x, y, z) w(x, y) θ 1, θ 2 rotations, w transverse displacement. Assuming a transverse load of the form t 3 f(x, y) and a that the plate is clamped, θ = (θ 1, θ 2 ) and w satisfy the system of equations:

6 6 a(θ, η) + κt 2 ( w θ, v η) = (f, v) η H 1 0 (Ω)2, v H 1 0 (Ω). E a(θ, η) := 12(1 ν 2 [(1 ν)ε(θ)ε(η) ) Ω +νdiv θdiv η], E Young modulus, ν Poisson ratio, κ := Ek/2(1 + ν) shear modulus, k correction factor usually taken as 5/6. We change notation and keep only the parameter t. So, our equations are a(θ, η) + t 2 ( w θ, v η) = (f, v) For our purposes, the only important fact about a is that it is coercive in H0 1 (which follows from the 2D Korn inequality). We will not make other use of the explicit form o a.

7 The deformation energy is given by 1 t 2 a(θ, θ) + w θ 2 fw 2 2 Ω Ω It can be shown that the second term remains bounded when t 0. In particular, 7 t 0 w θ 0 For the limit problem: w = θ Kirchkoff constraint THIS IS A PROBLEM FOR THE NUMER- ICAL SOLUTION! If t is small the problem is close to a constrained minimization problem. FOR EXAMPLE: If we use standard P 1 finite elements for θ and w, the restriction of the limit problem is too strong.

8 Indeed, if 8 w h = θ h then, w h piecewise constant and continuous But, w h constant θ h H 1 0 (Ω)2 w h = θ h = 0 CONSEQUENCE: For t small θ h, w h 0. This is called LOCKING Remark 2. : Indeed, now the continuity constant of the bilinear form is too large. It seems that we have a problem similar to the original 3D problem! AND SO, WHAT IS THE ADVANTAGE OF USING THE 2D MODEL?

9 9 SOLUTION: Mixed Interpolation or Reduced Integration. IDEA: Relax the restriction of the limit problem. w θ = 0 replaced by Π( w θ) = 0 Π is some interpolation or projection onto some space Γ h. So, in the discrete problem, the restriction is verified only at some points or in some average sense. FINITE ELEMENT APPROXIMATION: θ h H h H0 1 (Ω)2, w h W h H0 1 (Ω) are such that a(θ h, η)+t 2 (Π( w h θ h ), Π( v η)) = (f, v) η H h, v W h

10 In the usual methods 10 So, W h Γ h a(θ h, η)+t 2 ( w h Πθ h, v Πη) = (f, v) η H h, v W h MIXED FORM: Introducing the shear stress γ = t 2 ( w θ) { a(θh, η) + (γ h, v Πη) = (f, v) γ h = t 2 ( w h Πθ h ) η H h, v W h

11 11 MAIN PROBLEM: How to choose the spaces H h, W h, Γ h and the operator Π? EXAMPLE: The Bathe-Dvorkin MITC4 rectangular elements (Mixed Interpolation Tensorial Components). H h and W h are the standard Q 1 elements and Γ h is locally defined by (a + by, c + dx) (is a rotated Raviart-Thomas space). The operator Π is defined by Πη t l = η t l l for all side l of an element, where t l is the unit tangent vector on l. l

12 GENERAL ERROR ANALYSIS Continuous problem: 12 { a(θ, η) + (γ, v Πη) = (f, v) + (γ, η Πη) γ = t 2 ( w θ) η H 1 0 (Ω)2, v H 1 0 (Ω) Discrete Problem: { a(θh, η) + (γ h, v Πη) = (f, v) γ h = t 2 ( w h Πθ h ) η H h, v W h Error equation: a(θ θ h, η) + (γ γ h, v Πη) = (γ, η Πη) η H h, v W h

13 13 Lemma 1. Let θ I H h, w I W h and γ I = t 2 ( w I Πθ I ) Γ h. Suppose and γ Πγ 0 Ch γ 1 (γ Πγ, η) = 0 η P 2 k 2 Let P be the L 2 projection into P 2 k 2. Then, θ θ h 1 + t γ γ h 0 C( θ I θ 1 +t γ I γ 0 +h γ Pγ 0 Proof. a(θ I θ h, η) + (γ I γ h, v Πη) = a(θ I θ, η)+(γ I γ, v Πη)+(γ, η Πη) η H h, v W h Take η = θ I θ h and v = w I w h. So, γ I γ h = t 2 ( v Πη) Using the coercivity of a we obtain θ I θ h t2 γ I γ h 2 0

14 = a(θ I θ, θ I θ h ) + t 2 (γ I γ, γ I γ h ) +(γ P γ, θ I θ h Π(θ I θ h )) and the lemma follows. To apply the Lemma we need to find approximations θ I and w I such that the associated γ I be also a good approximation. This will follow from the existence of approximations satisfying the following property 14 and w I Πθ I = Π( w θ) θ θ I 1 Ch k θ k+1 γ Πγ 0 Ch k γ k This property can be seen as a generalization of the known Fortin property basic in the analysis of mixed methods. In fact, introducing the operators I(θ, w) = (θ I, w I ) B(θ, w) = w θ

15 and B h (θ h, w h ) = w h Πθ the property can be summarized by the following commutative diagram: 15 (H0 1)2 H0 1 I B (L 2 ) 2 Π H h W h B h Γ h When Π is an L 2 projection, this is exactly the Fortin property, which is known to be equivalent to the inf-sup condition. This will be the case in one of our examples (the Arnold-Falk elements). In that case error estimates for the shear stress γ can also be obtained. Remark 3. Indeed it is enough to have the commutative diagram with Π replaced by some operator with good approximation properties.

16 16 If approximations satisfying the properties above (i.e, commutative diagram + approximation properties) exist, then it follows from the Lemma: ERROR ESTIMATES θ θ h 1 + t γ γ h 0 + w w h 1 Ch k ( θ k+1 + t γ k + γ k 1 ) With constant C independent of h and t. EXAMPLES Example 1: MITC3 or DL element: H h : P 2 1 {λ 2λ 3 t 1, λ 3 λ 1 t 2, λ 1 λ 2 t 3 }, C 0 Degrees of freedom: θ(v i ), l i θ t i W h : P 1 C 0 Γ h : (a by, c + bx) C 0 t. c. Degrees of freedom: l i γ t i Π defined by

17 17 l Πγ t = l γ t Example 2: Arnold-Falk element: H h : P1 2 {λ 1λ 2 λ 3 }, C 0 Degrees of freedom: θ(v i ), T θ W h : P 1 Non Conforming Degrees of freedom: l w Γ h : P 2 0 Π defined by L 2 -projection! T Πγ = T γ Error estimates for Examples 1 and 2: θ θ h 1 + w w h 1 Ch Example 3: Bathe-Brezzi second order rectangular elements:

18 H h : Q 2, C 0 W h : Q r 2 : {1, x, y, xy, x2, y 2, x 2 y, xy 2 }, C 0 Γ h : {1, x, y, xy, y 2 } {1, x, y, xy, x 2 } Π defined by { l Πγ tp 1 = l γ tp 1 R Πγ = R γ Error estimates: 18 θ θ h 1 + w w h 1 Ch 2 Remark 4. Here the constant C is NOT independent of t because the right hand side involves higher order norms of the solution which depend on t.

19 L 2 - ERROR ESTIMATES 19 θ θ h 0 + w w h 0 Ch 2 Recall the error equation a(θ θ h, η)+(γ γ h, v Rη) = (γ, η Rη) (η, v) H h W h, Duality argument: (ϕ, u) H 1 0 (Ω)2 H 1 0 (Ω) solution of a(η, ϕ) + ( v η, δ) = (v, w w h ) + (η, θ θ h ) δ = t 2 ( u ϕ). (η, v) H 1 0 (Ω)2 H 1 0 (Ω), A priori estimate (we assume Ω is convex) ϕ 2 + u 2 + δ 0 + t δ 1 C( θ θ h 0 + w w h 0 ),

20 20 Take v = w w h and η = θ θ h in the dual problem and use the error equation with (η, v) = (ϕ I, u I ): w w h θ θ h 2 0 = a(θ θ h, ϕ ϕ I ) + t 2 (γ γ h, δ Πδ) + (θ h Rθ h, δ) + (γ, ϕ I Πϕ I ), where we have used the commutative diagram property Πδ = t 2 ( u I Πϕ I ). PROBLEM: The last two terms. For the MITC3 elements we have: Lemma 2. If γ H(div, Ω), ψ H 1 0 (Ω) and ψ A is a piecewise-linear average interpolant: (γ, ψ A Πψ A ) Ch 2 div γ 0 ψ 1

21 ISOPARAMETRIC ELEMENTS 21 F K : ˆK K ˆK = (0, 1) (0, 1): reference element K: shape regular quadrilateral F K : Q 2 1 transformation W h : Standard isoparametric Q 1 elements: v F K = (a + bˆx + cŷ + dˆxŷ) Γ h : Rotated Raviart-Thomas elements: η F K = DF T K (a + bŷ, c + dˆx) The operator Π is defined by Πη t l = η t l l for all side l of an element, where t l is the unit tangent vector on l. For H h we consider two cases: l

22 MITC4: Standard Q 2 1 isoparametric elements DL4: Q 2 1 p 1, p 2, p 3, p 4 p i = (ˆp i F 1 K )t i ˆp i is a cubic polynomial vanishing on ˆl j, j i. ERROR ESTIMATES 22 θ θ h 1 + t γ γ h 0 + w w h 1 Ch( θ 2 + t γ 1 + γ 0 ) ASSUMPTIONS: For DL4 general shape regular meshes. For MITC4 the mesh T h is a refinement of a coarser T 2h obtained by joining the edge midpoints and T 2h is obtained in the same way from T 4h. L 2 - ERROR ESTIMATES θ θ h 0 + w w h 0 Ch 2 ASSUMPTION: Asymptotically parallelogram meshes.