Continuum mechanics VI. Typical problems of elasto-statics AleˇsJanka office Math 0.107 [email protected] http://perso.unifr.ch/ales.janka/mechanics Apr 13, 2011, Universit´ede Fribourg AleˇsJanka VI. Problems of elasto-statics 1. Elasto-statics in small deformations, linear material Let us formulate the simplest problem of elasticity in 3D: given a computational domain Ω IR3, find a displacement field ⊂ u :Ω IR3 such that: → Force equilibria: symmetric Euler stress-tensor τ, τ ij = F i ie. div τ = F in Ω −∇j − ij ijk` Constitutive law: Hooke’s law, τ = E ek`, e.g. 2µ τ = 2µe + λ tr(e) Id where λ = K − 3 Kinematic equation: Cauchy strain tensor e: 1 1 e = [ u + u ] ie. e = u + ( u)T in Ω k` 2 ∇k ` ∇` k 2 ∇ ∇ Boundary conditions h i u = u¯ on Γ ∂Ω (Dirichlet-type), D ⊂ τ n = g on Γ = ∂Ω Γ (Neumann-type). · N \ D AleˇsJanka VI. Problems of elasto-statics 1.1. Weak formulation of elasto-statics: howto a) Non-homogenous Dirichlet condition homogeneous: → Instead of finding u [H1(Ω)]3 with u = u¯ on Γ , let us offset u ∈ D by the known u¯ (extended somehow onto the whole Ω). u = u¯ + δu in Ω, with δu = 0 on ΓD Define the appropriate functional subspace of [H1(Ω)]3 by V = w [H1(Ω)]3 : w = 0 on Γ . 0 { ∈ D } Instead of finding u, find δu V , u = u¯ + δu and ∈ 0 b) Multiply force equilibria by any test function v V to get ∈ 0 force equilibria in weak form div τ v dx = F v dx v V . − · · ∀ ∈ 0 ZΩ ZΩ ie. (component-wise) τ ij v dx = F i v dx v V . −∇j i i ∀ ∈ 0 ZΩ ZΩ AleˇsJanka VI. Problems of elasto-statics 1.1. Weak formulation of elasto-statics: howto c) Apply Green’s Theorem (integration by parts) τ ij v dx τ ij n v dΓ = F i v dx v V · ∇j i − j i i ∀ ∈ 0 ZΩ Z∂Ω ZΩ d) Split to and and apply boundary conditions: ∂Ω ΓD ΓN R R Rv = 0 on ΓD , ij i τ nj = g on ΓN . ij i i to get τ j vi dx = F vi dx + g vi dΓ. Ω · ∇ Ω ΓN Z ij Z1 ij ji Z e) Use symmetry of τ = 2 τ + τ to get ij 1 ij 1 ji τ (u) j vi dx = τ (u) j vi dx + τ (u) j vi dx Ω · ∇ 2 Ω · ∇ 2 Ω · ∇ Z Z Z τ ij (u) v ·∇i j 1 = τ ij (u) ( v + v ) dx| {z } · 2 ∇j i ∇i j ZΩ eij (v) AleˇsJanka |VI. Problems{z of elasto-statics} 1.1. Weak formulation of elasto-statics: howto f) Use constitutive law to express τ: find δu V , so that u = u¯ + δu and ∈ 0 E ijk` e (u) e (v) dx = F i v dx + g i v dΓ, v V k` ij i i ∀ ∈ 0 ZΩ ZΩ ZΓN g) Use linearity of e(u) to separate u¯ from the unknown δu: find δu V so that ∈ 0 E ijk`e (δu) e (v) dx = F i v dx+ g i v dΓ E ijk` e (u¯) e (v) dx v V k` ij i i − k` ij ∀ ∈ 0 ΩZ ΩZ ΓZN ΩZ AleˇsJanka VI. Problems of elasto-statics 2. Nonlinear elasto-statics: classification Origins of nonlinearity of elasticity problems: Geometrical nonlinearity: also called kinematic nonlinearity for large displacements and/or large deformations Material nonlineraity: also called physical nonlinearity due to constitutive laws (nonlinear materials). Classification: Material nonlinearity + geometrical linearity: use “small deformations” theory, iterate on material nonlinearity Large displacements + small deformations: big shifts and rotations as a rigid body, but small deformations. Use nonlinear kinematics (Green or Almansi strain). Linear constitutive law (e.g. Hooke’s law) can be used. Large displacements + large deformations: need nonlinear kinematics (Green or Almansi strain) and nonlinear constitutive law (non-linear material). AleˇsJanka VI. Problems of elasto-statics 2. Nonlinear elasto-statics (Lagrange formulation) Force equilibria: T j` uk + δk = F k in Ω −∇` ∇j j 0 0 h i Kinematic equation: Green strain 1 ε (u) = u + u + uk u ij 2 ∇j i ∇i j ∇i · ∇j k Nonlinear constitutive law: nonlinear material ∂W T ij = = T ij (ε) ∂εij Boundary conditions u = u¯ on Γ ∂Ω (Dirichlet-type), D ⊂ 0 Id + u T n = σ n = g on Γ = ∂Ω Γ (Neumann-type). ∇ · 0 · 0 N 0 \ D AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) Same steps as in Section 1.1.: a)-b) find δu V s.t. u = u¯ + δu and ∈ 0 j` k k k ` T j u + δj vk dx = F0 vk dx , v V0. Ω −∇ ∇ Ω ∀ ∈ Z 0 h i Z 0 c) Apply Green’s Theorem (integration by parts): T j` uk +δk v dx T j` uk +δk n0 v dΓ = F k v dx ∇j j ∇` k − ∇j j ` k 0 k ΩZ0 ∂ZΩ0 ΩZ0 d) Split to and , apply boundary conditions v V0 ∂Ω0 ΓD ΓN ∈ on Γ and σ n = g on Γ to get D R · 0 R NR T j` uk +δk v dx = F k v dx+ g k v dΓ v V ∇j j ∇` k 0 k k ∀ ∈ 0 ΩZ0 ΩZ0 ΓZN AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) j` 1 j` `j e) Use symmetry of T = 2 T + T T j` uk +δk v dx = T j` uk v dx + T k` v dx ∇j j ∇` k ∇j ∇` k ∇` k ΩZ0 ΩZ0 ΩZ0 uk v + v uk v + v = T j` ∇j ∇` k ∇j k ∇` dx + T k` ∇` k ∇k ` dx 2 2 ΩZ0 ΩZ0 v + v + uk v + uk v = T ij ∇i j ∇j i ∇i ∇j k ∇j ∇i k dx 2 ΩZ0 Dεij (u,v) Here, we have introduced | {z } 1 Dε (u, v) = v + v + uk v + uk v ij 2 ∇i j ∇j i ∇i ∇j k ∇j ∇i k How does it relate to εij (u)? AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) f) Use constitutive law to express T ij : T ij = ∂Ψ(u) ∂εij find δu V so that u = u¯ + δu and ∈ 0 ∂Ψ(u) Dε (u, v) dx = F k v dx + g k v dΓ ∂ε ij 0 k k ZΩ0 ij ZΩ0 ZΓN f’) Dεij (u, v) to εij (u): directional (Gˆateaux)derivative: What is the variation of εij (u+α v), seen as a function of α IR? Get the slope at u, ie. at α = 0: ∈ d [εij (u+α v)]α 0 = dα → d 1 = (u +αv ) + (u +αv ) + (u +αv ) (uk +αv k ) dα2∇i j j ∇j i i ∇i k k · ∇j u uk +α( uk v + uk v )+α2 v k v ∇i k ·∇j ∇j ·∇i k ∇i ·∇j k ∇j ·∇i k α 0 1 → = v + v + uk v + uk | v = D{zε (u, v) } 2 ∇i j ∇j i ∇i · ∇j k ∇j · ∇i k ij AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) f”) Directional derivative DΨ(u, v) of Ψ(u) at u along the direction v: use chain rule to differentiate Ψ(ε(u)): d ∂Ψ(u) [Ψ(u+αv)]α 0 = Dεij (u, v) dα → ∂εij T ij f?) Introduce the total potential energy:| {z } Π(u) = Ψ(u) dx F k u dx + g k u dΓ − 0 k k ZΩ0 ZΩ0 ZΓN strain energy work by external forces | {z } | {z } AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) Weak formulation and its equivalent minimization problem: The weak formulation: find u u¯ + V such that ∈ 0 ∂Ψ(u) Dε (u, v) dx = F k v dx + g k v dΓ v V ∂ε ij 0 k k ∀ ∈ 0 ZΩ0 ij ZΩ0 ZΓN simplifies to finding u u¯ + V such that ∈ 0 DΠ(u, v) = 0 v V ∀ ∈ 0 This is in fact the optimality condition for an equivalent minimization problem: find u u¯ + V such that ∈ 0 Π(u) min . → AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) g) Linearize around the known state u,¯ express the problem in the unknown δu V : find u u¯ + V such that ∈ 0 ∈ 0 ∂Ψ(u) Dε (u, v) dx = F k v dx + g k v dΓ v V ∂ε ij 0 k k ∀ ∈ 0 ZΩ0 ij ZΩ0 ZΓN with limited Taylor expansion around u¯: ∂Ψ(u) ∂Ψ(u¯) ∂2Ψ(u¯) = + Dεk`(u¯, δu) + o(δu). ∂εij ∂εij ∂εij ∂εk` · T ij (u¯) E ijk`(u¯) and | {z } | {z } 2 Dεij (u, v) = Dεij (u¯, v) + D εij (u¯, v, δu) + o(δu) where 2 d D εij (u¯, v, δu) = [Dεij (u¯ + αδu, v)]α 0 dα → 1 = δuk v + δuk v 2 ∇i · ∇j k ∇j · ∇i k AleˇsJanka VI. Problems of elasto-statics 2.1 Weak formulation of elasto-statics (Lagrange formul.) g) Linearize around the known state u,¯ express the problem in the unknown δu V : find δu V such that ∈ 0 ∈ 0 T ij (u¯)+E ijk`(u¯) Dε (u¯, δu) Dε (u¯, v)+D2ε (u¯, v, δu) dx k` · ij ij ΩZ0 = F k v dx + g k v dΓ + o(δu). v V 0 k k ∀ ∈ 0 ΩZ0 ΓZN Collect known terms onto the right-hand side, neglect o(δu) terms: Linearized problem for δu (one iteration of Newton’s method): Given initial guess u¯, find a correction δu V for which ∈ 0 ijk` ij 2 E (u¯) Dεk`(u¯, δu) Dεij (u¯, v)+T (u¯) D εij (u¯, v, δu) dx Z Ω0 = F k v dx + g k v dΓ T ij (u¯)Dε (u¯, v) v V .