Some Remarks on the Reissner Mindlin Plate Model

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1 Some Remarks on the Reissner Mindlin Plate Model Alexandre L. Madureira LNCC Brazil Talk at the Workshop of Numerical Analysis of PDEs LNCC February 10,

2 Outline The 3D Problem and its Modeling Full Discretization Playing Around with the Timoshenko Beam Conclusions 2

3 The 3D Problem and its Modeling 3D linearly elastic plates A Hellinger Reissner Principle Reissner Mindlin Model Derivation Convergence Estimate Full Discretization Playing Around with the Timoshenko Beam Conclusions 3

4 The Domain Ω P ε x = (x, x 3 ) 2ε Middle surface: Ω R 2, 3D Domain: P ε = Ω ( ε, ε), Lateral Boundary: P ε L = Ω ( ε, ε), Top and bottom boundaries: P ε ± = Ω { ε, ε}. 4

5 Bending of Linearly Elastic Plate Find u ε : P ε R 3 and σ ε : P ε R 3 3 sym such that div σ ε = 0, σ ε = Ce(u ε ), in P ε, σ ε n = g ε on P ε ±, u ε = 0 on P ε L, where g ε = (0, 0, g 3 ) : P ε ± R 3 is the traction load. Also, e(u ε ) = 1 2 ( uε + T u ε ) and Cτ = 2ντ + λ tr(τ)δ, where µ and λ are the Lamé coeff., and δ is the 3 3 ident. matrix. Assume g 3 (x, ε) = g 3 (x, ε). 5

6 Dimension Reduction The goal of dimension reduction is to pose a system of equations in the middle surface Ω that is easy to solve and which solution approximates the original 3D solution. The most popular models are variants of Reissner Mindlin Kirchhoff Love (also known as biharmonic model) formal higher order models 6

7 Notation: u = u u 3 R2 R, σ = σ σ σ t σ 33 R2 2 R 2 R 2t R, 7

8 Hellinger Reissner Principle Let S g (P ε ) = { τ H(div, P ε ) : τn = g on P± ε }. Hellinger-Reissner Principle: (σ ε, u ε ) is critical point of L(τ, v) = 1 C τ : τ dx + div τ v dx 2 P ε P ε on S g (P ε ) L 2 (P ε ). So σ ε S g (P ε ) and u ε L 2 (P ε ) satisfy C σ ε : τ dx + P ε u ε div τ dx = 0 P ε for all τ S 0 (P ε ), div σ ε v dx = 0 P ε for all v L 2 (P ε ). 8

9 We now derive a model of Reissner Mindlin type by looking for critical points of L within subspaces of S g (P ε ) L 2 (P ε ). Consider subspaces composed of functions which are polynomial in the transverse direction, with the following degrees: 1, Then our solution will be of the form θ u R = (x )x 3, σ R = ω(x ) + ω 2 (x )r(x 3 ) σ(x)x 3 σ(x )q(x 3 ), g 3 x 3 /ε + σ 33 s(x 3 ) where r(z) = 3(z 2 /ε 2 1/5)/2, q(z) = 3(1 z 2 /ε 2 )/2, and s(z) = 5(z/ε z 3 /ε 3 )/4. 9

10 Consider the eqtn P ε div σ v dx = 0, where v is a test function. Choose v = (ψ (x )x 3, 0), with ψ L 2 (Ω) arbitrary, to obtain ε2 3 Ω div σ ψ dx + 2ε Ω σ ψ dx = 0 for all ψ L 2 (Ω). Choosing v = (0, 0, w(x )), with w L 2 (Ω) arbitrary, it follows that ε Ω div σ w dx = Ω g 3 w dx, for all w L 2 (Ω) Finally, with v = (0, 0, w 2 (x )r(x 3 )), and w 2 L 2 (Ω) arbitrary, Ω σ 33 w 2 dx = 2 5 Ω g 3 w 2 dx, for all w 2 L 2 (Ω) 10

11 Hence ε2 3 div σ ψ dx + 2ε σ ψ dx = 0 for all ψ L2 (Ω) Ω Ω ε div σ w dx = g 3 w dx, for all w L 2 (Ω) Ω Ω σ 33 w 2 dx = 2 g 3 w 2 dx Ω 5, for all w 2 L 2 (Ω) Ω 11

12 Similarly for the stresses, choosing arbitrary test functions with polynomial profile, we gather that A σ : τ 2εθ div τ dx = ν (σ g 3 ) tr(τ E ) dx Ω Ω for all τ H1 (Ω), ( ) 1 6 Ω 2µ 5 σ + θ τ + ω div τ dx = 0 for all τ H 1 (Ω), ω 2 τ 33 dx = ε ( 1 Ω E 3 g σ 33 ν ) 6 tr(σ) τ 33 dx Ω for all τ 33 H 1 (Ω). Integrating by parts, we can write the above equations in terms of the displacement unknowns only. 12

13 Reissner Mindlin Model: Find θ and ω such that 2ε3 3α div e (θ ) + 4εµ5 6 (θ ω) = ε εµ 5 6 div(θ ω) = 2g 3 in Ω, θ = 0, ω = 0 on Ω. β α g 3 in Ω, After that we can perform the following computations. σ = 2εA e (θ) λ 2µ + λ g 3δ, σ 33 = 2 5 g 3, σ = 2µ5 6 ( θ + ω), ω 2 = ε ( 1 E 3 g σ 33 ν ) 6 tr(σ). 13

14 Convergence of Reissner Mindlin Theorem 1 Assume g 3 = ε α ĝ 3, where ĝ 3 is ε-independent and smooth, and α is a nonnegative number. Then u ε u R E(P ε ) u ε E(P ε ) + σε σ R L 2 (P ε ) σ ε L 2 (P ε ) Cε 1/2. The constant C depends on Ω, and ĝ 3 only. 14

15 The proof of this theorem relies on the two-energies principle. This principle says that if, given a stress that is statically admissible, and a displacement that is kinematically admissible, then the sum of the squared energy and complementary energy norms is equal to the complementary energy norm of the constitutive residual. The complementary energy norm is simply [ σ C ε = (C σ) : σ dx P ε ] 1/2 15

16 Theorem 2 (The two energies principle.) Suppose that σ H(div, P ε ), is statically admissible, i.e. divσ = 0 in P ε, σn = g ε on P ε ±, and suppose u H 1 (P ε ) is kinematically admissible, i.e. u = 0 on P ε L. Then u ε u 2 E ε + σε σ 2 C ε = σ Ce(u) 2 C ε. 16

17 Our derivation of the Reissner Mindlin system yields an statically admissible stress field. Also, we add a boundary layer to make the displacement field kinematically admissible. 17

18 The constitutive residual ρ = C σ R e(u R ) is given by ρ = 0, ρ = 5 8µ (x2 3 ε )[µ(θ w)], ρ 33 = x 3 2µ + λ (λ div θ + g 3). We shall need the following a priori estimates: Lemma 3 Let γ = µ(θ w). Then there exists a constant C only dependent on Ω such that θ H 1 + w H 1 + ε γ L 2 C g 3 L 2. 18

19 The 3D Problem and its Modeling Full Discretization Playing Around with the Timoshenko Beam Equations Full Discretization Results for Constant Traction Conclusions 19

20 Consider the beam (2D elasticity) problem of finding such that u ε = u 1 and σ ε = σ 1,1 σ 1,2 u 2 σ 1,2 σ 2,2 div σ ε = 0, σ ε = C e(u ε ), in (, 1) ( ε, ε), σ ε ε n = g on (, 1) { ε, ε}, u ε = 0 on {, 1} ( ε, ε), where g ε = (0, g 2 ) is the traction, and e (u ε ) = 1 2 ( u ε + T u ε ). 20

21 Hellinger-Reissner Principle: (σ ε, u ε ) is critical point of L(τ, v ) = 1 2 on S g (P ε ) L 2 (P ε ). P ε C τ : τ dx + div P ε τ v dx Full discretization by choosing σ = σ 11 σ 12 P 0 2 (, 1) x 2 P0 1 (, 1) q(x 2 ) σ 12 σ 22 P(, 0 1) s(x 2 ) u = u 1 P (, 1 1) x 2 P(, 0 1) {1, r(x 2 )} u 2 P k 0 (, 1): continuous piecewise polynomials of degree k in (, 1) P k (, 1): discontinuous polynomials of degree k in (, 1) 21

22 Our (Timoshenko Beam Model) solution will be of the form σ T (x 1, x 2 ) = σ 11(x 1 )x 2 σ 12 (x 1 )q(x 2 ), g 3 x 3 /ε + σ 22 s(x 2 ) θ(x u T (x 1, x 2 ) = 1 )x 2, ω(x 1 ) + ω 2 (x 1 )r(x 2 ) where σ 11 P0 2 (, 1), σ 12 P0 1 (, 1), σ 22 P(, 0 1), θ P(, 1 1), ω P(, 0 1) 22

23 ε ε 1 dσ 11 dx 1 ψ dx 1 + 2ε dσ 12 dx 1 w dx 1 = σ 22 w 2 dx 1 = σ 12 ψ dx 1 = 0 ψ P 1 (, 1), g 2 w dx 1 w P 0 (, 1), g 2 w 2 dx 1 w 2 P 0 (, 1), ασ 11 τ 11 βσ 22 τ 11 2εθ dτ 11 dx 1 = 2β g 2 τ 11 dx 1, dx 1 1 ( ) 1 6 2µ 5 σ 12 + θ τ 12 + ω dτ 12 dx 1 = 0, dx 1 1 ( )[ 1 1 ω 2 τ 22 dx 1 = ε 2µ β 3 g ] 21 σ 22 τ 22 β ε 6 σ 11τ 22 dx 1, for all τ 11 P 2 0 (, 1), τ 12 P 1 0 (, 1), τ 22 P 0 (, 1). 23

24 Assume g 2 constant by parts. It is possible to solve for the stresses using the equations above: σ h 11 = 6 ε 2 P ε c 0x 1 + c 1, σ h 12 = P ε + c 0, σ h 22 = 2 5 g 2, where P (x) = x g 2(ξ) dξ, P 2 (x) = x P (ξ) dξ, and we compute c 0 and c 1 from (1+ 5µα 3ε 2 )c 0 = 1 2ε 1 P dx 1 5µα 2ε 3 1 P 2 x 1 dx 1 βµ 1 g 2 x 1 dx 1. 2αc 1 = β g 2 dx 1 6α ε 2 1 P 2 dx 1. 24

25 Hence, we have to solve for θ P 1 (, 1) and ω P 0 (, 1), where 1 1 2εθ dτ 11 dx 1 = dx 1 1 ω dτ 12 dx 1 dx 1 = 1 (βσ 22 ασ βg 2 ) τ 11 dx 1, ( 1 2µ for all τ 11 P 2 0 (, 1), τ 12 P 1 0 (, 1). ) 6 5 σ 12 + θ τ 12 dx 1, 25

26 For the plate problem, things are not that easy. Indeed, the planar stress is no longer a scalar, but a 2 2 symmetric matrix with rows in H(div, P ε ). It then becomes nontrivial to find stable elements for stresses and displacements. An option is to solve for the planar stress in terms of the displacement and shear stress. This is the same as start with Reissner Mindlin, and include shear stress, and it s traditionally considered to develop stable elements. 26

27 The 3D Problem and its Modeling Full Discretization Playing Around with the Timoshenko Beam Conclusions 27

28 Conclusions The derivation of the Reissner Mindlin Model using Hellinger Reissner Principle is mathematically consistent. The model obtained is actually the first in a hierarchical sequence. It also makes the derivation of the modeling error through the two energies principle easier, since it yields statically admissible stress field. Reissner Mindlin is better than the biharmonic: if there is shear, the biharmonic fails to deliver a meaningful result. 28

29 Example: General load Domain: (0, 1) ( ε, ε), with ε = 1/40 Traction: g ε = (1, 10 3 ) on the top and g ε = (, 10 3 ) on the bottom Clamped lateral boundary conditions 29

30 Planar elasticity, Reissner Mindlin and Kirchhoff Love solutions:

31 Reissner Mindlin is the same as the biharmonic: for purely transverse load, both models converge at the same rate. I now of no proof indicating that one is better than the other in this case. Both suffer from a quite low convergence rate: O(ε 1/2 ) in relative energy norm. Is there a way to improve this rate? 31

32 Thank You! 32

Questions (And some Answers) (And lots of Opinions) On Shell Theory

1 Questions (And some Answers) (And lots of Opinions) On Shell Theory 2 Shells Q: What is a shell? A: A three-dimensional elastic body occupying a thin neighborhood of a two-dimensional submanifold of