INTERVAL ARITHMETIC APPLIED TO STRUCTURAL DESIGN OF UNCERTAIN MECHANICAL SYSTEMS

Μέγεθος: px

Εμφάνιση ξεκινά από τη σελίδα:

Download "INTERVAL ARITHMETIC APPLIED TO STRUCTURAL DESIGN OF UNCERTAIN MECHANICAL SYSTEMS"

Ευδοκία Τοκατλίδης
5 χρόνια πριν
Προβολές:

1 INTERVAL ARITHMETIC APPLIED TO STRUCTURAL DESIGN OF UNCERTAIN MECHANICAL SYSTEMS O. Dessombz, F.Thouverez J-P. Laîné, L.Jézéquel Laboratoire de Tribologie et Dynamique des Systèmes UMR CNRS 5513 École Centrale de Lyon Ecully France

2 Uncertainties in mechanical systems 2 Modeling: Finite Element Method Hypothesis Deterministic Structure Uncertain Data Variable Variations of the parameters, the geometry Variations of the results - constraints - displacements Computations made for the mean value of the parameters - eigenfrequencies Different types of uncertainties - stochastic parameters: Young s modulus, density, thickness... - unknown parameters : boundary conditions, assembly... - evolutive parameters : aging, variation in time - model uncertainties: behavior law, choice of the mesh... Modeling - stochastic variables generic laws (Gaussian, uniform) particular laws (bounded domain) - Intervals uncertain and bounded variables

3 Finite Element Problems 3 Γ Formulation : PDE u M Ω L(u) on Ω C(u) on Γ Stationarity of an integral functional J(u) R(u)dΩ δj(u) u E u E u space of the admissible fields Discretization of E u sub domains (FE) and interpolation û nodes U i N i (M) node

4 Finite Element Problems 4 Elementary formulation Assembly global formulation U1 U2 U3 U4 - Static problems KU F K EI L L 12 6L 6L 4L 2 6L 2L L 12 6L 6L 2L 2 6L 4L 2 - Dynamic problems MÜ + R U + KU F harmonic load F F e iωt ( K +iωr ω 2 M ) U F Frequency Response Function U - Properties of M and K Symmetric, (semi) definite, positive

5 One DOF example 5 x η m F k F 1 m 1 k 99, 11] η.2 Static problem kx F x F 1 k 11, 1 ] 9 Dynamic problem (Frequency response function) ( k(1 + iη) ω 2 m ) x F ] ] k ω 2 m ηk x r real problem: ηk k ω 2 m x i F ] k appears four times in the matrix! Direct interval resolution (Intlab) Analytic solution: x r x i ] (k ω 2 m)f (η 2 +1)k 2 2kω 2 m+ω 4 m 2 ηkf (η 2 +1)k 2 2kω 2 m+ω 4 m 2 x r x i x r x i ] ] ].3425,.3425].678,.396] overestimation ].24,.197].51,.397]

6 Formulation adapted to Finite Element Method 6 Young s modulus in E ] varying Stiffness matrix E k 11 k 12 k 21 k 22 ] ] Ek 11 Ek 12 E 1 k 11 E 2 k 12 Ek 21 Ek 22 E 3 k 21 E 4 k 22 E i E E i independent Symmetric General form: A] A ]+ N n1 ɛ na n ] {b {b + P p1 β p{b p K] K ]+ N n1 ɛ nk n ] FEM Definite (or semi-definite) Positive for ɛ n ɛ n K] corresponds to a stiffness matrix

7 Symmetry, Vertex A] ] 3, 4] 1, 2] 1, 2] 3, 4] x 2 2 P 4 P 1 ] 3 1 {x 2 4 { P 3 1 P x 1 { ] ] e 1 {x { x 1 x 2 { 1 2, 1 9 5, 13 3 ] 3 ] { 44 2 e 1 1 e 1 e 1 1 x x x x x x sup bound sup bound inf bound inf bound

8 The algorithm of Rump 8 Rump s inclusion Fix point theorem Initialization ɛ (, 1) Y : g, g] A]{x {b {x {x + {x {x G]{x + {g X : 1 ɛ, 1+ɛ] Y Y : G X g with G] I] R]A] {g R]({b A]{x ) No Y X In practice R] A 1 ] {x R]{b Yes x Y

9 Algorithm adapted to the mechanical formulation 9 Taking into a account the factorization example with one (centered) parameter α A] A ]+αa 1 ] R] mid(a]) 1 A ] 1 {x R]{b A ] 1 {b Convergence ρ( G ) < 1 partition of e: e i e i ρ( G(e i ) ) < 1 e i x i x i {x 1 G]{x + {g G] I] R]A] αa ] 1 A1] {g R]({b A]{x ) αa ] 1 A 1 ]{x G] ɛ 1 A ] 1 A 1 ] ρ( G] ) ρ(η, ω) k 1 k (1+η 2 ) ω 2 m +ηk 1 ω 2 m (1+η 2 )k 2 2k ω 2 m+ω 4 m {x n {x n 1 +( 1) n+1 α n+1 ɛ n B] n+1 {x B] A ] 1 A 1 ] ρ( G] ) η ω

10 The proposed algorithm 7 G e i B i Initialization ɛ (, 1) Y : g, g] ρ( G ) < 1 ρ( e i B i ) < 1 Yes X : 1 ɛ, 1+ɛ] Y Y : G X g No Y X No el e, ] A l A +.5eA 1 er, e] A r A +.5eA 1 Yes x Y x inf(x i ) x sup(x i )

11 Example 11 EI L 3 ]{ 12 6L d 6L 4L 2 θ { F M EI , ]e4 (1.898e4 ± 2%) F 12, 98] M 44.1, 45.9] L 1 θ F d M ]{ { 2.178, 2.36] 1.18, 1.8] d 12, 98] 1e5 1.18, 1.8].72,.787] θ 44.1, 45.9] ]{ { 12 6 d 12, 98] EI 6 4 θ 44.1, 45.9] Rotation θ.5 1 x 1 3 verifylss Σ (A], {b) { d θ { d θ { 1.163,.44] e ,.24] {.659,.494] e 3.381,.157].5 1 Hull of the mechanical set Mechanical exact solution set Σ (A], {b) Displacement d x 1 4

12 Algorithm adapted to the mechanical formulation 12 x 1 4 Rotation θ hull for the modified Rump s algorithm Solution sets for a, ] and,a] 1 x verifylss 3 Rotation θ.5 Σ (A], {b) Hull containing the mechanical solution 5 Exact mechanical solution exact hull Exact mechanical solution Displacement d x Σ (A], {b) Displacement d x 1 4

13 Clamped free Bar. Comparison 13 x.5 x.75 x 1 E 1 E 2 x E 2 21 ± 1% GP a F Original Rump s algorithm Adapted algorithm 1.8 x x D 1 D x x

14 Example 1 14 f 1 f 2 k 1 k 2 k 3 m 1 m 2 Static Case ( ] ] k1 + k2 k2 k1 1 k2 k2 + + e 1 + k 3 ] ]) { k2 1 k2 1 e 2 + e k2 1 k2 1 3 k3 1 x 1 x 2 { Monte Carlo{ ,.11333] x MC ,.19962] Monte Carlo{ ,.1136] x MC ,.19925] Proposed algorithm {.1331,.11353] x lss.18394,.19968] x MC2 x MC1 real bound x lss f 1 f 2 x 1 x 2 Real part(m/n) Imaginary part (m/n) ω (rad/s) Dynamic Case ( ] (1 + iη 1 )k1 +(1+iη 2 )k2 (1 + iη 2 )k2 (1 + iη 2 )k2 (1 + iη 2 )k2 +(1+iη 3 )k3 ] ] (1 + iη 1 )k1 1 (1 + iη 2 )k2 1 (1 + iη 2 )k2 1 +e 1 +e 2 (1 + iη 2 )k2 1 (1 + iη 2 )k2 1 ] ]) { + e 3 ω 2 m 1 (1+iη 3 )k3 1 m 2 H 1 H 2 { f 1 f 2 ω (rad/s)

15 Example Real part (m/n) ω (rad/s) Imaginary part (m/n) ω (rad/s)

16 Example 2 : Truss structure L1 L F 12 L2 L2 L2 L2 1 3 L1 L2 1m F 1 3 Nm I π 1 8 /4 m 4 S π 1 4 m 2 H(3, 3) (m/n) E 21 ± 1% GP a ω (rad/s)

17 Modulus of the FRF H(3, 3) of the truss. The stiffness is uncertain (E 21 ± 1% GPa). Min and max values, FRF for several values of E H(3, 3) (m/n) ω (rad/s)

18 Example 3: Multiple eigenvalues system L 1m S π1 4 m 2 6 FRF H(1, 3) ρ 78 kg/m 3 E 21 GP a 1 1 η 2% 1 11 For the first blade E E ± 1% frequency ω (rad/s) x 1 5

19 Modulus of the transfer function H(1, 3) E E ± 1% for the first blade Crisp cases, envelope calculated with the modified algorithm FRFH(1, 3) frequency ω (rad/s)

20 Gyroscopic System 2 ]{ (J 1 ]{ Ψ + J B ) + J 1 Θ 1 ]{ Ψ aω + K z L 1 Θ 2 µ Ψ 1 Θ { r J B L Ω2 cos Ωt sin Ωt Ω m K y 1 x K z unbalanced rotating disk Ωandµ K y K z intervals 2 x ( ] Ω M + K Ω G ηk Ω r M + e 1 Ω G + ηk Ω M + K Ω r G ] ]) { { Ω r G K 1 ηk 1 Ψ + e 2 Ω Ω r M ηk 1 K 1 Θ F r F i + e 1 Ω r { F r F i

21 Conclusion 21 Use of Interval Arithmetic overestimations resolution of linear systems Mechanical Formulation A] A ]+αa 1 ] Modification of the algorithm of Rump smallest Hull convergence efficiency robust envelopes Static cases Dynamic cases (FRF)