Elastic s in Advisor: Matthias Heinkenschloss Computational and Applied Mathematics Rice University 13 April 2007
Outline Elasticity in Differential Geometry of Shell Geometry and Equations The Plate Model and MITC Elements
by Alkis Rappas Figure: Scan by capture3d.com includes 975,410 points.
The Way of the Future? Figure: Balsa violin by Douglas Martin, 2005.
Acoustic Transfer Functions Radiativity (Pa/N) 10 2 10 1 10 0 10 1 Transfer Function (averaged over 5 microphones) Martin Balsa Barbieri Spruce 10 2 0 1000 2000 3000 4000 5000 6000 7000 Frequency (Hz) Figure: Acoustic transfer functions of Martin s Balsa violin #4 and a violin by Barbieri, as measured by George Bissinger.
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).
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.
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).
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.
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.
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
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.
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.
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 +.
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.
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.
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.
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 ).
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).
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.
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.
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).
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}.
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.
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 ).
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.
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.
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.
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
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.
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.
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 )}.
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
10 8 6 4 2 0 2 4 6 8 10 8 6 4 2 0 2 4 6 8 1.5 1 0.5 0 0.5 1 1.5 10 8 6 4 2 0 2 4 6 8 12 10 8 6 4 2 0 1.5 1 0.5 0 0.5 1 1.5 MITC7 Plate Elements 10 10 8 6 4 2 0 2 4 6 8 10 Displacement x 10 3 x 10 3 x 10 3 10 10 8 6 4 2 0 2 4 6 8 10 10 10 8 6 4 2 0 2 4 6 8 10 Rotation 1 Rotation 2 Figure: Clamped boundary, uniform pressure loading.
Convergence of MITC7 Elements Model Degrees of Freedom Thickness 219 499 907 1571.2 0.9736 0.9968 0.9994 1.0000.02 0.9723 0.9969 0.9996 1.0000.002 0.9723 0.9969 0.9996 1.0000 Table: MITC7: w(0)/w ref (0) Model Degrees of Freedom Thickness 219 499 907 1571.2 0.1778 0.5461 0.7951 0.8766.02 0.0023 0.2015 0.0946 0.4221.002 0.0000 0.1948 0.0492 0.2349 Table: Standard Galerkin (MITC7 without reduced integration): w(0)/w ref (0)
Implement MITC shell elements. Investigate convergence of eigenvalues. Optimize eigenvalues (plate tuning). Include air in the model.