Finite Elements for Elastic Shell Models in
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1 Elastic s in Advisor: Matthias Heinkenschloss Computational and Applied Mathematics Rice University 13 April 2007
2 Outline Elasticity in Differential Geometry of Shell Geometry and Equations The Plate Model and MITC Elements
3 by Alkis Rappas Figure: Scan by capture3d.com includes 975,410 points.
4 The Way of the Future? Figure: Balsa violin by Douglas Martin, 2005.
5 Acoustic Transfer Functions Radiativity (Pa/N) Transfer Function (averaged over 5 microphones) Martin Balsa Barbieri Spruce Frequency (Hz) Figure: Acoustic transfer functions of Martin s Balsa violin #4 and a violin by Barbieri, as measured by George Bissinger.
6 Elasticity in Cartesian The elastic body occupies the region ˆΩ ˆR 3. Ĉ is the tensor in the constitutive relation. ˆɛ(û(ˆx)) = 1 ( 2 û(ˆx) + ( û) T (ˆx) ) is the symmetric strain tensor. ˆf H 1 (ˆΩ) is the body forcing. ĥ H 1/2 ( ˆΩ) is the boundary traction. The weak formulation is: (Ĉ : ˆɛ(û(ˆx))) : ˆɛ(ˆv(ˆx)) dˆx = ˆΩ ˆΩ ˆΩ ˆf (ˆx) ˆv(ˆx) dˆx+ ĥ(ˆx) ˆv(ˆx) dˆγ, ˆv H 1 (ˆΩ).
7 Three-dimensional Differential Geometry Figure: The chart function Θ maps Ω R 3 to ˆR 3. This transformation takes a point x to a vector ˆx. The vector g i is tangent to coordinate line i.
8 Notation The Einstein summation convention is used (Greek letters over {1, 2}, Latin letters over {1, 2, 3}). Physical coordinates (ˆx ˆR 3 = R 3 ) are distinguished from reference coordinates (x R 3 ) through the use of hats. are related via the chart function Θ : R 3 ˆR 3 by ˆx = Θ(x). Function domains are made explicit via use of hats, e.g. ˆv(ˆx) = v(x) = v i (x)g i (x) = v i (x)g i (x).
9 Three-dimensional Differential Geometry The reference domain Ω R 3 is mapped to the physical domain ˆΩ ˆR 3 via the chart function Θ : R 3 ˆR 3. The local covariant basis is given by g i = Θ x i. It is assumed that Θ is an immersion, i.e., that the g i are linearly independent for all x Ω. The contravariant basis vectors g i are uniquely defined by g i g j = δ i j.
10 Naghdi s Model Naghdi s model starts from linear elasticity on a thin shell, and makes several additional assumptions: Integration over thin dimension is replaced with the kinematic assumption. Stress is assumed to be planar (σ 33 = 0). Transverse shear strain terms are truncated to first order in the thickness variable.
11 Elasticity in Refered to these coordinates, the weak form now becomes C ijkl ɛ ij (u(x))ɛ kl (v(x)) g dx = Ω f i (x)v i (x) g dx+ Ω h i (x)v i (x) g n i g ij n j dγ, v H 1 (Ω). Ω
12 Differential Geometry of The reference domain ω R 2 is mapped to the physical domain ˆΩ ˆR 3 via the chart function θ : R 2 ˆR 3. The local covariant basis is given by a α = Θ x α. It is assumed that θ is an immersion, i.e., that the a α are linearly independent for all x ω. The surface normal is given by a 3 = a 1 a 2 a 1 a 2. The contravariant basis vectors a i are uniquely defined by a i a j = δ i j.
13 Figure: Two-dimensional differential geometry: the chart function θ maps a point x ω R 2 to a vector ˆx ˆR 3. The vector a α is tangent to coordinate direction α. a 3 is the unit normal to the surface.
14 Description of Shell Geometry In a general shell model, the surface chart function θ : ω ˆR 3 defines the middle surface. The entire shell is then the image of the set Ω {x = (x 1, x 2, x 3 ) R 3 : (x 1, x 2 ) ω, x 3 < e(x 1, x 2 )/2}, under the mapping Θ(x 1, x 2, x 3 ) θ(x 1, x 2 ) + x 3 a 3, where the thickness of the shell is given by the bounded function e : ω R +.
15 Figure: A shell model combines the two approaches: the chart function Θ is given in terms of the chart function θ and a thickness function e : ω R +. The vectors g i and a i coincide on the middle surface (Θ(ω {0})), but generally (if the middle surface is curved) in following the x 3 coordinate direction (normal to the middle surface), g α differs from a α.
16 Assumptions The displacement field u satisfies the kinematic assumption, i.e., u(x 1, x 2, x 3 ) = U 1,i (x 1, x 2 )a i + x 3 U 2,α (x 1, x 2 )a α. The material is assumed isotropic, with Young modulus E, and Poisson ratio ν. Using the plane-stress assumption (σ 33 = 0), the strain energy 1 C ijkl ɛ ij (u)ɛ kl (u) g dx 2 Ω can be written as 1 {C αβλµ ɛ αβ (u)ɛ λµ (u) + (C 3α3β + C α3β3 + 2 Ω C 3αβ3 + C α33β ) ɛ α3 (u)ɛ β3 (u)} g dx.
17 Covariant Differentiation The functions v i j are called the the covariant derivatives of the vector field v. They are used to express derivatives in the local basis: j (v i g i ) = v i j g i, and to write derivatives in ˆR 3 in terms of derivatives in R 3 : ˆ j ˆv i (ˆx) = v k l [g k ] i [g l ] j, where They are given by [g k ] i g k ê i, [g k ] i g k ê i. v i j = j v i Γ p ij v p, with the Christoffel symbols Γ q lk defined via Γ q lk = lg q g k.
18 Truncation of Shear Strain Expansions Due to the special form of the chart function Θ, the strain terms take the form ɛ αβ (U 1, U 2 ) = γ αβ (U 1 ) + x 3 χ αβ (U 1, U 2 ) (x 3 ) 2 κ αβ (U 2 ), ɛ α3 (U 1, U 2 ) = ϕ α (U 1, U 2 ), where γ αβ (U 1 ) = 1 2 (U 1,α β + U 1,β α ) b αβ U 1,3, χ αβ (U 1, U 2 ) = 1 2 (U 2,α β + U 2,β α b λ β U 1,λ α b λ αu 1,λ β ) + c αβ U 1,3, κ αβ (U 2 ) = 1 2 (bλ β U 2,λ α + b λ αu 2,λ β ), ϕ α (U 1, U 2 ) = 1 2 (U 2,α + U 1,3 α + b λ αu λ ).
19 The Naghdi Bilinear Form (Finally) Truncation to first order in x 3 and integration yields the Naghdi bilinear form A(U, V ) = 1 { C αβλµ {eγ αβ (U 1 )γ λµ (V 1 )+ 2 ω e 3 12 χ αβ(u 1, U 2 )χ λµ (V 1, V 2 )}+ e D λµ ϕ λ (U 1, U 2 )ϕ µ (V 1, V 2 )} a dx, where C αβλµ = D λµ = E 2(1 + ν) (aαλ a βµ + a αµ a βλ + 2ν 1 2ν aαβ a λµ ), Eaλµ 2(1 + ν).
20 The Scaled Naghdi Define e min = min x ω e(x), and let the number ɛ be given by ɛ = e min /L, with L some characteristic length of the shell. Then we can define the scaled thickness t by t = e e min L. We can thus make clear the dependence of the Naghdi problem on the thickness.
21 The Scaled Naghdi Let V = (H 1 (Ω)) 3 (H 1 (Ω)) 2 BC (BC prohibits rigid body motions), F V, e : ω R +. Problem (The Naghdi Problem) Find U = (U 1, U 2 ) in V such that for all V in V, A(U, V ) = ɛa m (U, V ) + ɛ 3 A b (U, V ) = F (V ), where A m (U, V ) = A b (U, V ) = ω ω t{ C αβλµ γ αβ (U 1 )γ λµ (V 1 )+ D λµ ϕ λ (U 1, U 2 )ϕ µ (V 1, V 2 )} a dx, t 3 12 { C αβλµ χ αβ (U 1, U 2 )χ λµ (V 1, V 2 )} a dx.
22 The Scaled Naghdi Theorem (Solutions to the Naghdi Problem) The Naghdi bilinenar form A is bounded and coercive on V. Assume that The boundary conditions prohibit rigid body motion. The forcing f L 2 (ω), with F (V ) = ω ef i v i dx. Then the Naghdi problem has a unique solution U V, and U V C f L 2 (ω).
23 Scaling of the Loading For the Naghdi problem to be well-posed as ɛ 0, we must scale the loading in the Naghdi problem as ɛ 3 A b (U, V ) + ɛa m (U, V ) = F ɛ (V ) = ɛ ρ G(V ), for G independent of ɛ, and consider the contents of the closed pure-bending subspace V 0 = {V V : A m (V, V ) = 0}.
24 The Membrane-Dominated Case Problem (Membrane Limit Problem) When V 0 = {0}, we can define the membrane energy norm via V m = A m (V, V ), and study the problem: find U m V m such that A m (U m, V ) = G(V ), V V m, where V m is the completion of V with respect to m. This corresponds to the scaling ρ = 1, provided that G V m.
25 The Bending-Dominated Case Problem (Bending Limit Problem) When V 0 {0}, we consider the problem: find U 0 V 0 such that A b (U 0, V ) = G(V ), V V 0. Since A b is coercive on V 0, this problem has a unique solution, and the scaling is ρ = 3 (except in the special case where G(V ) = 0 V V 0 ).
26 Convergence of Displacement-Based Finite Elements This theorem guarantees convergence of finite-element methods for Naghdi s model. Theorem (Céa s Lemma) Let the finite element space V h V, and let U be the unique solution to the Naghdi problem. Then the Naghdi problem posed over V h has a unique solution U h, which satisfies U U h V C ɛ inf U V V, V V h for some C ɛ > 0. Unfortunately, the constant C ɛ depends on the thickness.
27 Finite Elements in the Limiting Cases In the membrane-dominated case, there is (under fairly benign assumptions) an ɛ-uniform convergence result, and displacement-based finite elements work. The bending-dominated case causes trouble: here, the solution to the limit problem lies in V 0. The trouble is that in general, V 0 V h = {0}! Must formulate a mixed method for the bending-dominated problem.
28 The Bending Limit Problem Recall the scaling for the bending problem: ɛ 3 A b (U, V ) + ɛa m (U, V ) = ɛ 3 G(V ). Represent the membrane stress as an element of T + = L 2 (ω), and find (U, Σ) V T + such that A b (U, V ) + B(V, Σ) = G(V ), V V B(U, Ξ) ɛ 2 D(Σ, Ξ) = 0, Ξ T +. The bilinear forms B and D are closely related to A m. This looks a lot like a simpler constrained problem.
29 Theorem (Mixed Finite Element Convergence) Let T be a Hilbert space with T T +, and assume that A b is coercive on V. The continuous inf-sup condition holds: δ > 0 such that B(V, Ξ) inf sup δ Ξ T,Ξ 0 V V Ξ T V V,V 0 The discrete inf-sup condition holds: δ > 0 such that inf sup Ξ T h,ξ 0 V V h,v 0 B(V, Ξ) V V Ξ T δ Then the finite element solution (U h, Σ h ) satisfies U ɛ U ɛ h V + Σ ɛ Σ ɛ h T + ɛ Σ ɛ Σ ɛ h T + inf { U ɛ V V + Σ ɛ Ξ T + ɛ Σ ɛ Ξ T +}. V V h,ξ T h
30 The Plate Model This model is simply Naghdi s for a flat shell, and neglecting the in-plane displacements of the middle surface. Problem () Find (U 1, U 2 ) in V RM = H 1 (ω) (H 1 (ω)) 2 BC such that for all V V RM, ɛ 3 A RM b (U, V ) + ɛa RM m (U, V ) = F (V ). Unfortunately, A RM b is not coercive on V, and the inf-sup condition does not hold for T RM = L 2 (ω), so the previous theorem does not apply.
31 MITC Properties The Mixed Interpolation of Tensorial Components (MITC) elements can overcome these difficulties by Suitable choice of spaces for V h, T h so that the inf-sup condition holds. The use of a carefully chosen reduction operator acting on the membrane strains ensures coercivity while allowing for consistency.
32 MITC7 Plate Elements For the MITC7 triangular element, the finite element spaces chosen are V h =W h Θ h, T h ={δ : δ T TR 1 (T ) T, δ τ where continuous at element boundaries, δ τ = 0 on ω}, W h = {ζ H 1 : ζ T P 2, T }, Θ h = {η (H 1 ) 2 : η T (S 7 (T )) 2, T } TR 1 (T ) = {δ :δ 1 = a 1 + b 1 x 1 + c 1 x 2 + x 2 (dx 1 + ex 2 ) δ 2 = a 2 + b 2 x 1 + c 2 x 2 x 1 (dx 1 + ex 2 )}.
33 MITC7 Plate Elements The reduction operator R : (H 1 (T )) 2 TR 1 (T ) is defined via (η Rη) τp 1 (s) ds = 0 e edge of T, p 1 P 1 (e), e (η Rη)dx = 0. T
34 MITC7 Plate Elements Displacement x 10 3 x 10 3 x Rotation 1 Rotation 2 Figure: Clamped boundary, uniform pressure loading.
35 Convergence of MITC7 Elements Model Degrees of Freedom Thickness Table: MITC7: w(0)/w ref (0) Model Degrees of Freedom Thickness Table: Standard Galerkin (MITC7 without reduced integration): w(0)/w ref (0)
36 Implement MITC shell elements. Investigate convergence of eigenvalues. Optimize eigenvalues (plate tuning). Include air in the model.
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