Expo

Numerical analysis of a locking-free mixedfem for a bending moment formulation of

Reissner-Mindlin plates

LOURENCO BEIRAO DA VEIGA1 , DAVID MORA2 , RODOLFO RODRIGUEZ3

1 Dipartimento di Matematica, Universit a degli Studi di Milano.2 Departamento de Matem atica, Universidad del Bıo Bıo.

3 Departamento de Ingenierıa Matem atica, Universidad de Concepci on.

Sesi on Especial de An alisis Num erico

XX Congreso de Matem atica Capricornio COMCA 2010

Universidad de Tarapac a, Arica, Chile

4–6 de Agosto de 2010.

Contents

• The model problem

• The continuous formulation

• Galerkin formulation

• Numerical tests

Mixed FEM for Reissner-Mindlin plates. – 2 – BEIRAO DA VEIGA, MORA, RODRIGUEZ

The model problem.

Given g ∈ L2(Ω), find β, γ and w

−div(C(ε(β))) − γ = 0 in Ω,

−div γ = g in Ω,

γ =κ

t2(∇w − β) in Ω,

w = 0, β = 0 on ∂Ω,

• w is the transverse displacement ,

• β = (β1, β2) are the rotations ,

• γ is the shear stress ,

• t is the thickness,

• κ := Ek/2(1 + ν) is the shear modulus (k is a correction factor),

• Cτ := E12(1−ν2) ((1 − ν)τ + νtr(τ )I) ,

• we restrict our analysis to convex plates.


The model problem. (cont.)

We introduce as a new unknown the bending moment σ = (σij)1≤i,j≤2, defined by

σ := C(ε(β)).

• C−1τ :=

12(1 − ν2)

E

(1

(1 − ν)τ −

ν

(1 − ν2)tr(τ )I

).

We rewrite the equation above as:

C−1σ = ∇β −

1

2(∇β − (∇β)t) = ∇β − rJ,

• r := − 12 rotβ,

• J :=

0 1

−1 0

.


The continuous formulation.

Continuous mixed problem

Find ((σ, γ), (β, r, w)) ∈ H × Q such that

∫

Ω

C−1σ : τ +

t2

κ

∫

Ω

γ · ξ +

∫

Ω

β · (div τ + ξ) +

∫

Ω

r(τ12 − τ21) +

∫

Ω

wdiv ξ = 0

∫

Ω

η · (div σ + γ) +

∫

Ω

s(σ12 − σ21) +

∫

Ω

vdiv γ = −

∫

Ω

gv,

for all ((τ , ξ), (η, s, v)) ∈ H × Q, where

H := H(div; Ω) × H(div ; Ω),

Q := [L2(Ω)]2 × L2(Ω) × L2(Ω),

with

H(div; Ω) := τ ∈ [L2(Ω)]2×2 : div τ ∈ [L2(Ω)]2,

and

H(div ; Ω) := ξ ∈ [L2(Ω)]2 : div ξ ∈ L2(Ω).


The continuous formulation. (cont.)

Continuous mixed problem

a((σ, γ), (τ , ξ)) + b((τ , ξ), (β, r, w)) = 0 ∀(τ , ξ) ∈ H,

b((σ, γ), (η, s, v)) = −

∫

Ω

gv ∀(η, s, v) ∈ Q,

where

a((σ, γ), (τ , ξ)) :=

∫

Ω

C−1σ : τ +

t2

κ

∫

Ω

γ · ξ,

b((τ , ξ), (η, s, v)) :=

∫

Ω

η · (div τ + ξ) +

∫

Ω

s(τ12 − τ21) +

∫

Ω

vdiv ξ.

In the analysis we will utilize the following t-dependent norm for the space H

‖(τ , ξ)‖H := ‖τ‖0,Ω + ‖div τ + ξ‖0,Ω + t‖ξ‖0,Ω + ‖div ξ‖0,Ω,

while for the space Q, we will use

‖(η, s, v)‖Q := ‖η‖0,Ω + ‖s‖0,Ω + ‖v‖0,Ω.



Additional regularity

Proposition 1 Suppose that Ω is a convex polygon and g ∈ L2(Ω). Then, there

exists a constant C , independent of t and g, such that a

‖w‖2,Ω+‖β‖2,Ω+‖γ‖H(div ;Ω)+t‖γ‖1,Ω+‖σ‖1,Ω+t‖div σ‖1,Ω+‖r‖1,Ω ≤ C‖g‖0,Ω.

aD. N. ARNOLD AND R. S. FALK, A uniformly accurate finite element method for the Reissner-Mindlin plate,

SIAM J. Numer. Anal., 26 (1989) 1276–1290.



Ellipticity in the kernel

V := (τ , ξ) ∈ H : ξ + div τ = 0, τ = τt and div ξ = 0 in Ω.

Lemma 1 There exists C > 0, independent of t, such that

a((τ , ξ), (τ , ξ)) ≥ C‖(τ , ξ)‖2H ∀(τ , ξ) ∈ V.

Proof . Given (τ , ξ) ∈ V , using tr(τ )2 ≤ 2(τ : τ ) ∀τ ∈ [L2(Ω)]2×2, we obtain

a((τ , ξ), (τ , ξ)) ≥12(1 − ν)

E‖τ‖2

0,Ω +t2

κ‖ξ‖2

0,Ω.

Since ‖div τ + ξ‖0,Ω = 0 and ‖div ξ‖0,Ω = 0, we get

a((τ , ξ), (τ , ξ)) ≥ C(‖τ‖2

0,Ω + ‖div τ + ξ‖20,Ω + t2‖ξ‖2

0,Ω + ‖div ξ‖20,Ω

),

⇒ a((τ , ξ), (τ , ξ)) ≥ C‖(τ , ξ)‖2H,



Inf-Sup condition

Lemma 2 There exists C > 0, independent of t, such that

sup(τ ,ξ)∈H(τ ,ξ) 6=0

|b((τ , ξ), (η, s, v))|

‖(τ , ξ)‖H≥ C‖(η, s, v)‖Q ∀(η, s, v) ∈ Q.

Well-posedness

Theorem 2 There exists a unique ((σ, γ), (β, r, w)) ∈ H × Q solution of the

continuous mixed problem and

‖((σ, γ), (β, r, w))‖H×Q ≤ C‖g‖0,Ω,

where C is independent of t.


Galerkin formulation.

• Thh>0: regular family of triangulations of the polygonal region Ω.

• hT : diameter of the triangle T ∈ Th.

• h := maxhT : T ∈ Th.

• Ω =⋃T : T ∈ Th.

Finite elements subspaces

Hγh := ξh ∈ H(div ; Ω) : ξh|T ∈ RT0(T ), ∀T ∈ Th,

Qwh := vh ∈ L2(Ω) : vh|T ∈ P0(T ), ∀T ∈ Th,


Galerkin formulation. (cont.)

We consider the unique polynomial bT ∈ P3(T ) that vanishes on ∂T and is

normalized by∫

TbT = 1.

B(Th) := τ ∈ H(div ; Ω) : (τi1, τi2)|T ∈ spancurl(bT ), i = 1, 2,∀T ∈ Th ,

where curl v := (∂2v,−∂1v).

Hσ

h := τh ∈ H(div ; Ω) : τh|T ∈ [RT0(T )t]2, ∀T ∈ Th ⊕ B(Th),

Qβh := ηh ∈ [L2(Ω)]2 : ηh|T ∈ [P0(T )]2, ∀T ∈ Th,

Qrh :=

sh ∈ H1(Ω) : sh|T ∈ P1(T ), ∀T ∈ Th

,

Note that Hσ

h × Qβh × Qr

h correspond to the PEERSa finite elements.aD. N. ARNOLD, F. BREZZI AND J. DOUGLAS, PEERS: A new mixed finite element for the plane elasticity,

Japan J. Appl. Math., 1 (1984) 347–367.



Discrete mixed problem:

Find ((σh, γh), (βh, rh, wh)) ∈ Hh × Qh such that

a((σh, γh), (τh, ξh)) + b((τh, ξh), (βh, rh, wh)) = 0 ∀(τh, ξh) ∈ Hh,

b((σh, γh), (ηh, sh, vh)) = −

∫

Ω

gvh ∀(ηh, sh, vh) ∈ Qh.

Hh := Hσ

h × Hγh ,

Qh := Qβh × Qr

h × Qwh .



Discrete kernel

Vh :=

(τh, ξh) ∈ Hh :

∫

Ω

ηh · (div τh + ξh) +

∫

Ω

sh(τ12h − τ21h)

+

∫

Ω

vhdiv ξh = 0 ∀(ηh, sh, vh) ∈ Qh

.

Let (τh, ξh) ∈ Vh. Taking (0, 0, vh) ∈ Qh and using that (div ξh)|T is a constant,

since vh|T is also a constant, we conclude that div ξh = 0 in Ω.

Now, taking (ηh, 0, 0) ∈ Qh, since div τh = 0 in Ω ∀τh ∈ B(Th), we have that

(div τh)|T is a constant vector. Since div ξh = 0, we have that ξh|T is also a

constant vector. Therefore, since ηh|T is also a constant vector, we conclude that

(div τh + ξh) = 0 in Ω. Thus, we obtain

Vh =

(τh, ξh) ∈ Hh : ξh + div τh = 0,

∫

Ω

sh(τ12h − τ21h) = 0 ∀sh ∈ Qrh

and div ξh = 0 in Ω

.



Ellipticity in the discrete kernel

Lemma 3 There exists C > 0 such that

a((τh, ξh), (τh, ξh)) ≥ C‖(τh, ξh)‖2H ∀(τh, ξh) ∈ Vh,

where the constant C is independent of h and t.

Discrete inf-Sup condition

Lemma 4 There exists C > 0, independent of h and t, such that

sup(τh,ξh)∈Hh

(τh,ξh) 6=0

|b((τh, ξh), (ηh, sh, vh))|

‖(τh, ξh)‖H≥ C‖(ηh, sh, vh)‖Q ∀(ηh, sh, vh) ∈ Qh.



Theorem 3 There exists a unique ((σh, γh), (βh, rh, wh)) ∈ Hh × Qh solution of

the discrete mixed problem. Moreover, there exist C, C > 0, independent of h and t,

such that

‖((σh, γh), (βh, rh, wh))‖H×Q ≤ C‖g‖0,Ω,

and

‖((σ, γ), (β, r, w)) − ((σh, γh), (βh,rh, wh))‖H×Q

≤ C inf(τh,ξh)∈Hh

(ηh,sh,vh)∈Qh

‖((σ, γ),(β, r, w)) − ((τh, ξh), (ηh, sh, vh))‖H×Q,

where ((σ, γ), (β, r, w)) ∈ H × Q is the unique solution of the continuous mixed

problem.



Rate of convergence

Theorem 4 Let ((σ, γ), (β, r, w)) ∈ H × Q and

((σh, γh), (βh, rh, wh)) ∈ Hh ×Qh be the unique solutions of the continuous and

discrete mixed problem, respectively. If g ∈ H1(Ω), then,

‖((σ, γ), (β, r, w)) − ((σh, γh), (βh, rh, wh))‖H×Q ≤ Ch‖g‖1,Ω.


Numerical tests.

• Isotropic and homogeneous plate.

• Ω := (0, 1) × (0, 1).

• t = 0.001.

• E = 1, ν = 0.30 and k = 5/6.

Figure 1: Square plate: uniform meshes.


Numerical tests. (cont.)

Choosing the load g as:

g(x, y) =E

12(1 − ν2)

[12y(y − 1)(5x2 − 5x + 1)

(2y2(y − 1)2

+x(x − 1)(5y2 − 5y + 1))

+ 12x(x − 1)(5y2 − 5y + 1)(2x2(x − 1)2

+y(y − 1)(5x2 − 5x + 1))]

,

so that

w(x, y) =1

3x3(x − 1)3y3(y − 1)3

−2t2

5(1 − ν)

[y3(y − 1)3x(x − 1)(5x2 − 5x + 1)

+ x3(x − 1)3y(y − 1)(5y2 − 5y + 1)],

β1(x, y) =y3(y − 1)3x2(x − 1)2(2x − 1),

β2(x, y) =x3(x − 1)3y2(y − 1)2(2y − 1).



e(σ) := ‖σ − σh‖0,Ω, e(γ) := ‖γ − γh‖t,H(div ;Ω),

e(β) := ‖β − βh‖0,Ω, e(r) := ‖r − rh‖0,Ω, e(w) := ‖w − wh‖0,Ω,

rc(·) := −2log(e(·)/e′(·))

log(N/N ′),

where N and N ′ denote the degrees of freedom of two consecutive triangulations with

errors e and e′.



Table 1: Errors and experimental rates of convergence for variables σ and γ, computed

on uniform meshes.

N e(σ) rc(σ) e(γ) rc(γ)

1345 0.40270e-04 – 0.31715e-02 –

5249 0.19649e-04 1.054 0.15876e-02 1.016

20737 0.09760e-04 1.019 0.07942e-02 1.008

82433 0.04868e-04 1.008 0.03971e-02 1.004

328705 0.02431e-04 1.004 0.01986e-02 1.002



Table 2: Errors and experimental rates of convergence for variables β, r and w, com-

puted on uniform meshes.

N e(β) rc(β) e(r) rc(r) e(w) rc(w)

1345 0.39713e-04 – 0.87462e-04 – 0.66226e-05 –

5249 0.18189e-04 1.147 0.39217e-04 1.178 0.27707e-05 1.280

20737 0.08884e-04 1.043 0.15009e-04 1.398 0.13136e-05 1.086

82433 0.04416e-04 1.013 0.05491e-04 1.457 0.06478e-05 1.025

328705 0.02205e-04 1.004 0.01991e-04 1.466 0.03228e-05 1.007



1e-08

1e-07

1e-06

1e-05

0.0001

0.001

0.01

0.1

1

10

1000 10000 100000 1e+06

e

degrees of freedom N

σ

♦♦

♦♦

♦

♦γ

++

++

+

+β

r

××

××

×

×w

h

⋆⋆

⋆⋆

⋆ ⋆



Figure 2: Approximate transverse displacement (left) and first component of the rotation

vector (right).



Figure 3: Approximate shear vector: first component (left) and second component

(right).



Figure 4: Approximate bending moment: σ11h(left) and σ12h

(right).



Figure 5: Approximate bending moment: σ21h(left) and σ22h

(right).


Many thanks for your attention.


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