The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems: Thermomechanics

Transcription

1 The Finite Element Method for the Analysis of Non-Linear and Dynamic Systems: Thermomechanics Prof. Dr. Eleni Chatzi Dr. Giuseppe Abbiati, Dr. Konstantinos Agathos Lecture December, / 30

2 Forewords Rail buckling driven by thermal expansion. 2 / 30

3 Forewords Thermal expansion joint. Nuclear reactor containment vessel. Chemical plant in fire. 2 / 30

4 Learning Goals Constitutive equations of linear thermoelasticity Linear and nonlinear thermomechanical finite element derivation 3 / 30

5 Constitutive equations The Hooke and Fourier laws are constitutive models and together define describe the behaviour of a linear thermoelastic material: ( ) σ ij + αθδ ij ɛ ij = 1 2G ɛ = ɛ mec + ɛ the : strain tensor σ : stress tensor G = E 2(1+ν) ν : Poisson ratio E : Young modulus Notation : shear modulus ν 1 + ν σ kkδ ij φ i = kθ,i θ : relative temperature φ : heat flux α : thermal expansion coef. k : thermal conductivity δ ij : Kronecker delta ( ),i is a shorthand for the partial derivative operator ( ) i. Summation over repeated index i.e. σ kk = tr (σ). 4 / 30

6 Balance equations Force and heat flux balance equations are physical laws instead: σ + P = ρü φ + R T 0 ( 1 + θ T 0 ) β ɛ ii = c ɛ ρ θ σ : stress tensor P : body force ü : acceleration ρ : material density Notation φ : heat flux vector R : heat source T 0 : room temperature β = αe 1 2ν c ɛ : heat capacity : stress temp. coef. ( ),i is a shorthand for the partial derivative operator ( ) i. Summation over repeated index i.e. ɛ ii = tr (ɛ). 5 / 30

7 Linear thermoelastic truss element FE discretization of displacement and temperature fields for a linear bar element: E 2 u 2 + P = ρü k 2 θ 2 + R = ρ c ɛ θ A and L are cross-section area and length of the bar. 6 / 30

8 Finite element discretization The same functional space i.e., set of shape functions, is used to approximate displacement and temperature fields: n u u (x) ũ (x) = ũ i h i (x) i=1 m u t m m nu ũ t m = i=1 d m ũ i dt m h i (x) Nodal values carry time dependency. θ (x) θ (x) = n θ i=1 m θ t m m nθ θ t m = i=1 θ i h i (x) d m θ i dt m h i (x) 7 / 30

9 Method of weighted residuals (strong form) Approximated solutions ũ (x) and θ (x) minimize the weighted averages of residuals r u and r θ : E 2 u 2 + P ρü = 0 E 2 ũ 2 + P ρ ü = r u 0 ( ) E 2 ũ 2 + P ρ ü δũd k 2 θ 2 + R ρc ɛ θ = 0 k 2 θ 2 + R ρc θ ɛ = rθ 0 ( ) k 2 θ 2 + R ρc θ ɛ δ θd However it is more convenient to approximate the solutions of the differential equations in weak form. 8 / 30

10 Method of weighted residuals (weak form) The weak formulation generates symmetric stiffness matrices. Only the upper/lower triangulars need to be computed and stored. For the unidimensional case, the weak form can be obtained via integration by parts. E ũ δ ( ) ũ d+ ρ üδũd = [ Pδũd + E ũ ] δũ Integration by parts [f (x) g (x)] Γ = Γ ( ) k θ δ θ d+ ρc ɛ θδ θd = [ ] Rδ θd + k θ δ θ f (x) g d + g (x) f d Γ 9 / 30

11 Method of weighted residuals (weak form) The weak formulation generates symmetric stiffness matrices. Only the upper/lower triangulars need to be computed and stored. For the unidimensional case, the weak form can be obtained via integration by parts. E ũ δ ( ) ũ d+ ρ üδũd = [ Pδũd + E ũ ] δũ Restoring force and heat flux Γ ( ) k θ δ θ d+ ρc ɛ θδ θd = [ ] Rδ θd + k θ δ θ The Hooke s law describes the mechanical restoring force while the Fourier s law dictates the heat flux response. Both are constitutive models. Γ 9 / 30

12 Method of weighted residuals (weak form) The weak formulation generates symmetric stiffness matrices. Only the upper/lower triangulars need to be computed and stored. For the unidimensional case, the weak form can be obtained via integration by parts. E ũ δ ( ) ũ d+ ρ üδũd = [ Pδũd + E ũ ] δũ Γ ( ) k θ δ θ d+ ρc ɛ θδ θd = [ Rδ θd + k θ δ θ Mass and thermal inertia Mass is responsible of inertia force while heat capacity governs thermal inertia. ] Γ 9 / 30

13 Method of weighted residuals (weak form) The weak formulation generates symmetric stiffness matrices. Only the upper/lower triangulars need to be computed and stored. For the unidimensional case, the weak form can be obtained via integration by parts. E ũ δ ( ) ũ d+ ρ üδũd = [ Pδũd + E ũ ] δũ Element load Γ ( ) k θ δ θ d+ ρc ɛ θδ θd = [ ] Rδ θd + k θ δ θ Body loads, such as gravity (mechanical) or heat flux generate by electric current flow (thermal), are distributed within the domain. Γ 9 / 30

14 Method of weighted residuals (weak form) The weak formulation generates symmetric stiffness matrices. Only the upper/lower triangulars need to be computed and stored. For the unidimensional case, the weak form can be obtained via integration by parts. E ũ δ Nodal load ( ) ũ d+ ρ üδũd = [ Pδũd + E ũ ] δũ Γ ( ) k θ δ θ d+ ρc ɛ θδ θd = [ ] Rδ θd + k θ δ θ Nodal load vectors represent Neumann (natural) boundary conditions. Γ 9 / 30

15 Stiffness and thermoelastic matrices K uu n u n u u + K uθ n u 1 L θ = n u n θ n θ 1 0 = L 0 L 0 ( ũ δ h 1. h nu h 1. h nu ) ( ũ EA EA ) α θ dx [ ] ũ 1 h 1 h... nu. ũ nu αea [ θ 1 ] h 1... h nθ. θ nθ dx dx 10 / 30

16 Thermal conductivity matrix L K θθ θ = n θ n θ n θ 1 0 = L 0 ( δ θ ) h 1. h nθ ka ka θ dx [ h 1... h nθ ] θ 1. θ nθ dx 11 / 30

17 Mass matrix L M uu ü = n u n u n u 1 0 = L 0 δũρ üdx h 1. ρa [ ü 1 ] h 1... h nu. h nu ü nu dx 12 / 30

18 Heat capacity matrix L C θθ θ = n θ n θ n θ 1 0 = L 0 δ θρc ɛ θdx h 1 θ. ρc ɛ A [ ] 1 h 1... h nθ. h nθ θ nθ dx 13 / 30

19 Mechanical and thermal body loads F u = n u 1 = L 0 L 0 δũpdx h 1. h nu Pdx F θ = n θ 1 = L 0 L 0 δ θrdx h 1. h nθ Rdx 14 / 30

20 Semi-discretized linear thermoelastic system This is the set of semi-discretized equations for a linear thermoelastic system. For the sake of simplicity ũ u and θ θ hereinafter. [ Muu ] [ü θ ] [ [ ] Cuu C θθ] u θ [ Kuu K uθ 0 K θθ ] [ ] u = θ [ Fu F θ ] M uu : mass matrix C uu : damping matrix K uu : stiffness matrix F u : mech. load vector C θθ : heat capacity matrix K θθ : thermal cond. matrix K uθ : thermoelasticity matrix F θ : thermal load vector 15 / 30

21 Modal analysis of thermoelastic systems Modal analysis is applied to both thermal and mechanical equations to provide a deep insight into the dynamics of the thermoelastic system: E = 210GPa, ρ = 7800kg/m 3,α = K 1 k = 43W / (mk),c ɛ = 445J/ (kgk) 16 / 30

22 Thermal eigenvalue problem For a generic time modulating function ψ (t) the heat transfer equation reads: C θθ θ + K θθ θ = F θ ψ (t) The solution of the associated eigenvalue problem allows for describing the temperature response in modal coordinates: (K θθ + λ j C θθ ) Ψ θθ,j = 0 λ j = Ψ T θθ,j K θθψ θθ,j ˆΨ θθ,j = Ψ θθ,j Ψ T θθ,j C θθψ θθ,j π j = Ψ T θθ,j F θ ξ j + λ j ξ j = π j ψ (t) S T = {j 1,..., j s } π j > π j S T 17 / 30

23 Thermal modal participation factor n θ = 19 m θ = / 30

24 Thermal mode shapes Thermal Mode #1: λ 1 = , τ 1 = λ 1 1 = 182s (Excited) 19 / 30

25 Thermal mode shapes Thermal Mode #2: λ 2 = , τ 2 = λ 1 2 = 179s (Not excited) 19 / 30

26 Mechanical eigenvalue problem The mechanical response of the system subjected to thermal loading reads: M uu ü + C uu u + K uu u = K uθ θ The solution of the following eigenvalue problem allows for describing the displacement response induced by thermal loading in modal coordinates: ( Kuu + ω 2 i M uu ) Ψuu,i = 0 ω 2 i = Ψ T uu,ik uu Ψ uu,i ˆΨ uu,i = Ψ uu,i Ψ T uu,i M uuψ uu,i η ij = Ψ T uu,ik uθ Ψ θθ,j q i + 2ζ i ω i q i + ωi 2 q i = η ij ξ j (t) Γ i = η ij j S T j S T 20 / 30

27 Thermomechanical modal participation factor n u = 51 m u = / 30

28 Mechanical mode shapes Mechanical Mode #1: f 1 = 8.45Hz, T 1 = f 1 1 = s (Not excited) 22 / 30

29 Mechanical mode shapes Mechanical Mode #4: f 4 = 58.85Hz, T 4 = f 1 4 = s (Excited) 22 / 30

30 Model reduction Excited mechanical and thermal modes can be used to reduce system matrices preserving the thermoelastic response: ˆK uu = ˆΨ T uuk uu ˆΨ uu, ˆM uu = ˆΨ T uum uu ˆΨ uu, ˆK uθ = ˆΨ T uuk uθ ˆΨ θθ ˆK θθ = ˆΨ T θθ K θθ ˆΨ θθ, Ĉθθ = ˆΨ T θθ C θθ ˆΨ θθ, ˆF θ = ˆΨ T θθ F θ n θ = 19 m θ = 10, n u = 51 m u = / 30

31 The example of structures subjected to fire Linear models are not able to capture phenomena such as thermo-visco-plastic deformation, which are encountered in real structures: Fire test of a steel-framed composite structures at the RAF Cardington facilities, England. Fire test of a steel column at UniLige. 24 / 30

32 Constitutive models In addition, material parameters can be temperature dependent: Steel thermal conductivity λ (EN ) Steel heat capacity c ɛ (EN ) 25 / 30

33 Constitutive models In addition, material parameters can be temperature dependent: Concrete thermal conductivity λ (EN ) Concrete heat capacity c ɛ (EN ) 26 / 30

34 Thermomechanical analysis: 1- and 2-way coupling Typically, both mechanical and heat transfer equations are nonlinear and coupling between both can be: 1-way: only the thermal affects the mechanical response 2-way: thermal and mechanical responses affect each other 1-way coupling allows for de-coupling the solution of thermal and mechanical equations. First, the transient thermal response is solved by integrating the heat transfer equation: R θ (θ, θ ) = F θ Then, the temperature response enters as parameter of the restoring force and the mechanical system response is computed by solving: R u (u, u, θ) = F u or M uu ü + R u (u, u, θ) = F u 27 / 30

35 Transient thermal analysis Trapezoidal rule algorithm with Newton-Raphson iterations: 1: for k = 1 to step do 2: θ k+1 0 3: θ k+1 θ k + θ k (1 γ) t + θ k+1 γ t 4: ε F θ (t k+1 ) R θ (θ k+1, θ k+1 ) 5: while ε >= tol do 6: θ k+1 (C θθ + K θθ γ t) 1 ε 7: θk+1 θ k + θk+1 8: θ k+1 θ k + θ k+1 γ t 9: ε F θ (t k+1 ) R θ (θ k+1, θ k+1 ) 10: end while 11: end for Assembly of heat flux vector and Jacobian matrix loops over elements: 1: for i = 1 to I do 2: R θ,i ) elemforce (Z θ,i θ k+1, Z θ,i θk+1 3: R θ R θ + Z T θ,ir θ,i 4: end for 1: for i = 1 to I do 2: [K θθ,i, C θθ,i ] elemjac (Z θ,i θ k+1 ) 3: K θθ K θθ + Z T θ,ik θθ,i Z θ,i 4: C θθ C θθ + Z T θ,ic θθ,i Z θ,i 5: end for 28 / 30

36 Nonlinear static analysis Newton-Raphson algorithm: 1: for k = 1 to step do 2: u k+1 u k 3: ε F u(u k+1 ) R u(t k+1 ) 4: while ε >= tol do 5: u k+1 K 1 uu ε 6: u k+1 u k + u k+1 7: ε F u(u k+1 ) R u(t k+1 ) 8: end while 9: end for Assembly of restoring force and stiffness matrix loops over elements: 1: for i = 1 to I do 2: R u,i elemforce (Z u,i u k+1, Z θ,i θ k+1 ) 3: R u R u + Z T u,ir u,i 4: end for 1: for i = 1 to I do 2: K uu,i elemjac (Z u,i u k+1, Z θ,i θ k+1 ) 3: K uu K uu + Z T u,ik uu,i Z u,i 4: end for 29 / 30

37 Nonlinear dynamic analysis Newmark algorithm with Newton-Raphson iteration: 1: for k = 1 to step do 2: ü k+1 0 3: u k+1 u k + u k t + ü k ( 1 2 β) t 2 + ü k+1 β t 2 4: u k+1 u k + ü k (1 γ) t + ü k+1 γ t 5: ε F u (t k+1 ) R u (u k+1 ) C uu u k+1 M uuü k+1 6: while ε >= tol do 7: ü k+1 ( M uu + C uuγ t + K uuβ t 2) 1 ε 8: ü k+1 ü k+1 + ü k+1 9: u k+1 u k+1 + ü k+1 γ t 10: u k+1 u k+1 + ü k+1 β t 2 11: ε F u (t k+1 ) R u (u k+1 ) C uu u k+1 M uuü k+1 12: end while 13: end for Assembly of restoring force and stiffness matrix loops over elements: 1: for i = 1 to I do 2: R u,i elemforce (Z u,i u k+1, Z θ,i θ k+1 ) 3: R u R u + Z T u,ir u,i 4: end for 1: for i = 1 to I do 2: K uu,i elemjac (Z u,i u k+1, Z θ,i θ k+1 ) 3: K uu K uu + Z T u,ik uu,i Z u,i 4: end for 30 / 30