Structural instability of nonlinear plates modelling suspension bridges

Transcription

1 Structural instability of nonlinear plates modelling suspension bridges Dipartimento di Scienze e Innovazione Tecnologica, Università del Piemonte Orientale Amedeo Avogadro Modelli matematici per ponti sospesi, Settembre 2015, Torino

2 G. Arioli, F. Gazzola, A new mathematical explanation of what triggered the catastrophic torsional mode of the Tacoma Narrows Bridge collapse, Appl. Math. Modelling 39, 2015 G. Arioli, F. Gazzola, On a nonlinear nonlocal hyperbolic system modeling suspension bridges, to appear in Milan J. Math E. Berchio, A. F., F. Gazzola, Structural instability of nonlinear plates modelling suspension bridges: mathematical answers to some long-standing questions, to appear in Nonlinear Anal. Real World Appl. E. Berchio, F. Gazzola, A qualitative explanation of the origin of torsional instability in suspension bridges, to appear in Nonlinear Analysis TMA E. Berchio, F. Gazzola, C. Zanini, Which residual mode captures the energy of the dominating mode in second order Hamiltonian systems?, preprint A. F., F. Gazzola, A partially hinged rectangular plate as a model for suspension bridges, to appear in Disc. Cont. Dyn. Syst. A

3 A general Hamiltonian system Let U = U(y 1,..., y N ) be a potential energy. ÿ 1 (t) + U (y 1 (t),..., y N (t)) = 0, y 1. y N (t) + U (y 1 (t),..., y N (t)) = 0. y N The Hamiltonian is given by the total energy E(y 1,..., y N, y 1,..., y N ) := N i=1 ẏ i U(y 1,..., y N ). We may think to the function y i = y i (t) as the amplitude (fourier coefficient) of each of the a so-called oscillation modes.

4 A general Hamiltonian system (transfer of energy) Our main purpose is to study the transfer of energy from one mode to one or more of the other ones. For example suppose that for some k {1,..., N} at least one among y k (0) and y k (0) is different from zero and that y i (0) = ẏ i (0) = 0 for any i {1,..., N}, i k. Figure: The first longitudinal mode

5 A general Hamiltonian system (transfer of energy) A transfer of energy to other oscillation modes may occur. Figure: The second longitudinal mode

6 A general Hamiltonian system (transfer of energy) Figure: The third longitudinal mode

7 A general Hamiltonian system (transfer of energy) The transfer of energy from one oscillation mode to the other ones is simply due to the fact that the system does not admit any stationary wave solution corresponding to that mode. Suppose now that the system admits a solution Y 0 (t) = (y 0,1 (t),..., y 0,N (t)) satisfying y 0,N (t) = 0 for any t 0. Figure: The first torsional mode

8 A general Hamiltonian system (transfer of energy) Let Y 0 = Y 0 (t) be the solution considered before. Consider now the solution Y (t) = (y 1 (t),..., y N (t)) of the system satisfying y 1 (0) = y 0,1 (0),..., y N 1 (0) = y 0,N 1 (0), y N (0) = A, y 1 (0) = ẏ 0,1 (0),..., ẏ N 1 (0) = ẏ 0,N 1 (0), y N (0) = B. The main question is to understand if the solution Y 0 is stable or not with respect to perturbations of y N. We can say that Y 0 is a stable solution if for any ε > 0 there exists δ > 0 such that for any A + B < δ Y (t) Y 0 (t) + Y (t) Y 0(t) < ε for any t R.

9 A general Hamiltonian system (transfer of energy) Figure: The second torsional mode

10 Oscillation modes u tt + Lu = f (u) in Ω (0, ), +Boundary Conditions on Ω. Here by oscillation mode we mean linear oscillation modes. We may choose the eigenfunctions of the following eigenvalue problem Lu = λu in Ω, +Boundary Conditions on Ω. (EP) Denote by w 1,..., w n,... a complete orthonormal system of eigenfunctions of (EP) and by λ 1 λ n... the corresponding eigenvalues. u(, t) = + n=1 y n (t)w n ( )

11 A model with interacting oscillators ml 2 3 θ = lcosθ [f (y l sin θ) f (y + l sin θ)] my = [f (y l sin θ) + f (y + l sin θ)]

12 A model with interacting oscillators θ i (t) + 3 U (Θ(t), Y (t)) = 0 θ i y i (t) + U (Θ(t), Y (t)) = 0. y i where (Θ, Y ) = (θ 1,..., θ n, y 1,..., y n ) R 2n and n U(Θ, Y )= [F (y i +sin θ i )+F(y i sin θ i )]+ 1 n [K y (y i y i+1 ) 2 +K θ (θ i θ i+1 ) 2 ] 2 i=1 with the notation y 0 = y n+1 = θ 0 = θ n+1 = 0. i=0

13 Some references for the previous model A possible choice for f is f (s) = k 1 s + k 2 s 2 + k 3 s 3. G. Arioli, F. Gazzola, A new mathematical explanation of what triggered the catastrophic torsional mode of the Tacoma Narrows Bridge, Appl. Math Model. 38, 2015 G. Bartoli, P. Spinelli, The stochastic differential calculus for the determination of structural response under wind, J. Wind Eng. Ind. Aerodyn. 48, 1993 A. C. Lazer, P. J. McKenna, Large-amplitude periodic oscillations in suspension bridges: some new connections with nonlinear analysis, SIAM rev. 32, 1990

14 Some models of suspension bridges y: deflection in the vertical direction; θ: angle of torsion. My tt + EIy xxxx + f (y + l sin θ) + f (y l sin θ) = 0 Ml 2 3 θ tt µl 2 θ xx + l cos θ(f (y + l sin θ) + f (y l sin θ)) = 0. x (0, L), t > 0. E. Berchio, F. Gazzola, A qualitative explanation of the origin of torsional instability in the suspension bridges, Nonlin. Anal. T.M.A., 2015

15 The Kirchhoff-Love model for the plate Ω := (0, L) ( l, l), 2l = L 150. E d 3 ( ) 1 E B (u) = 12(1 σ 2 ) Ω 2 ( u)2 + (σ 1) det(d 2 u) dxdy u: deflection in the vertical direction; E > 0: Young modulus; 0 < σ < 1 2 : Poisson ratio; d: thickness; E T (u) = E B (u) fu dxdy. Ω {0} ( l, l) and {L} (l, l): hinged edges. (0, L) { l} and (0, L) {l}: free edges. G. R. Kirchhoff, Uber das gleichgewicht und die bewegung einer elastischen scheibe, J. Reine Angew. Math. 40, 1850 A. E. H. Love, A treatise on the mathematical theory of elasticity (Fourth edition), Cambridge Univ. Structural Press instability (1927) of nonlinear plates modelling suspension bridges

16 Some comments on the plate model - In the model of the plate only vertical displacements are allowed and in our case, when torsional oscillations appear, horizontal displacements are not negligible. - The deck of the bridge is not actually a plate. - Rocard writes: The plate as a model is perfectly correct and corresponds mechanically to a vibrating suspension bridge. - Due to the fact that the width of the deck is quite small with respect to its length, the deformation of any cross section is negligible. Y. Rocard, Dyamic Instability: Automobiles, Aircraft, Suspension Bridges, Crosby Lockwood, 1957

17 The plate model E d 3 12(1 σ 2 ) 2 u = f in Ω u(0, y) = u xx (0, y) = u(l, y) = u xx (L, y) = 0 for y ( l, l) u yy (x, ±l) + σu xx (x, ±l) = 0 for x (0, L) u yyy (x, ±l) + (2 σ)u xxy (x, ±l) = 0 for x (0, L)

18 The plate model Figure: Different positions of the bridge. - u 0 equilibrium position if, the plate had no weight, there were no loads acting on the plate. - u w equilibrium position of the plate if it was only subject to its own weight w. - u h equilibrium position if the plate had no weight and if was subject to the restoring force of the cables-hangers. - If both the weight and the action of the cables-hangers are considered, the two effects cancel and the equilibrium position u 0 0 is recovered. E. Berchio, A. F., F. Gazzola, Structural instability of nonlinear plates modelling suspension bridges: mathematical answers to some long-standing questions, to appear in Nonlinear Anal. Real World Appl.

19 The plate model Figure: The plate Ω and its subset ω (dark grey) where the hangers act. - ω := (0, L) [( l, l + ε) (l ε, l)]; - Υ = Υ(y): characteristic function of the set ω; - h(y, u) = Υ(y)(k 1 u + k 2 u 3 ): restoring force due to the cables-hangers system. G. Bartoli, P. Spinelli, The stochastic differential calculus for the determination of structural response under wind, J. Wind Eng. Ind. Aerodyn. 48, 1993

20 The plate model u ( k1 H(y, u) := Υ(y) h(y, t) dt = Υ(y) 0 2 u2 + k ) 2 4 u4 E T (u) = E B (u) + (H(y, u) fu) dxdy. Ω. E d 3 12(1 σ 2 ) 2 u + h(y, u) = f in Ω u(0, y) = u xx (0, y) = u(l, y) = u xx (L, y) = 0 for y ( l, l) u yy (x, ±l) + σu xx (x, ±l) = 0 for x (0, L) u yyy (x, ±l) + (2 σ)u xxy (x, ±l) = 0 for x (0, L). (PS) Problem (PS) admits a unique solution which is also the unique minimizer for the total energy E T.

21 The plate model A(u) := E u (t) := m ut 2 dxdy + E T (u). 2 Ω Ω T ( ) m T ut 2 dxdy dt E T (u) dt. 2 Ω Ω 0 0 m Ω u tt + E d 3 12(1 σ 2 ) 2 u+h(y, u)=f in Ω (0, T ) u(0, y, t)=u xx (0, y, t)=u(l, y, t)=u xx (L, y, t)=0 for (y, t) ( l, l) (0, T ) u yy (x, ±l, t)+σu xx (x, ±l, t)= 0 for (x, t) (0, L) (0, T ) u yyy (x, ±l, t)+(2 σ)u xxy (x, ±l, t)=0 for (x, t) (0, L) (0, T ) u(x, y, 0)=u 0 (x, y), u t (x, y, 0)=u 1 (x, y) for (x, y) Ω (PE) Problem (PE) is well posed and every solution u of (PE) is globally defined in time.

22 The evolution problem with a damping term m Ω u tt + δu t + E d 3 12(1 σ 2 ) 2 u+h(y, u)=f in Ω (0, T ) u(0, y, t)=u xx (0, y, t)=u(l, y, t)=u xx (L, y, t)=0 for (y, t) ( l, l) (0, T ) u yy (x, ±l, t)+σu xx (x, ±l, t)= 0 for (x, t) (0, L) (0, T ) u yyy (x, ±l, t)+(2 σ)u xxy (x, ±l, t)=0 for (x, t) (0, L) (0, T ) u(x, y, 0)=u 0 (x, y), u t (x, y, 0)=u 1 (x, y) for (x, y) Ω In presence of a damping term the solution converges as t + to the unique solution of the stationary problem. A. F., F. Gazzola, A partially hinged rectangular plate as a model for suspension bridges, to appear in Disc. Cont. Dyn. Syst. A

23 Three fundamental questions (Q1) why do longitudinal oscillations suddenly transform into torsional oscillations? From Smith-Vincent we read: the only torsional mode which developed under wind action on the bridge or on the model is that with a single node at the center of the main span. This raises the following question: (Q2) why do torsional oscillations appear with a node at midspan? F. C. Smith, G. S. Vincent, Aerodynamic stability of suspension bridges: with special reference to the Tacoma Narrows Bridge, Part II: Mathematical analysis, Structural Research Laboratory, University of Washington Press, 1950

24 Three fundamental questions From the report by O. H. Ammann, T. von Kármán, G. B. Woodruff we read: the motions, which a moment before had involved a number of waves (nine or ten) had shifted almost instantly to two. This raises a third natural question (Q3) are there longitudinal oscillations which are more prone to generate torsional oscillations? O. H. Ammann, T. von Kármán, G. B. Woodruff, The failure of the Tacoma Narrows Bridge, Federal Works Agency, 1941

25 The eigenvalue problem 2 w = λw in Ω w(0, y) = w xx (0, y) = w(π, y) = w xx (π, y) = 0 for y ( l, l) w yy (x, ±l) + σw xx (x, ±l) = 0 for x (0, π) w yyy (x, ±l) + (2 σ)w xxy (x, ±l) = 0 for x (0, π). The set of eigenvalues may be ordered in an increasing sequence {λ k } of strictly positive numbers diverging to + and any eigenfunction belongs to C (Ω). A. F., F. Gazzola, A partially hinged rectangular plate as a model for suspension bridges, to appear in Disc. Cont. Dyn. Syst. A

26 The eigenvalue problem (i) For any m 1 there exists a sequence of eigenvalues λ k,m + such that λ k,m > m 4 for all k 1; the corresponding eigenfunctions are of the kind [ ( ) ( ) a cosh y λ 1/2 k,m + m2 + b sinh y λ 1/2 k,m + m2 ( ) ( ) ] +c cos y λ 1/2 k,m m2 + d sin y λ 1/2 k,m m2 sin(mx) for suitable constants a, b, c, d, depending on m and k; (ii) if the unique positive solution m of tanh( ( ) 2 σ 2ml) = 2 σ 2ml is an integer m N, then λ = m 4 is an eigenvalue with corresponding eigenfunction [ σl sinh( 2m y) + (2 σ) sinh( ] 2m l) y sin(m x) ;

27 The eigenvalue problem (iii) for any m 1, there exists an eigenvalue λ m ((1 σ) 2 m 4, m 4 ) with corresponding eigenfunction [ ( cosh( λm (1 σ)m 2) + ( ( λm + (1 σ)m 2) cosh y y m 2 + ) λ m ( cosh l m 2 + ) λ m m 2 λ m ) ( cosh l m 2 ) λ m ] sin(mx) ; (iv) for any m 1, satisfying lm 2 coth(lm 2) > ( ) 2 σ 2, σ there exists an eigenvalue λ m (λ m, m 4 ) with corresponding eigenfunction ( ( λm (1 σ)m 2) sinh y m 2 + ) λ m ( + ( ( λm + (1 σ)m 2) sinh y sinh l m 2 + ) λ m m 2 λ m ) ( sinh l m 2 ) λ m ] sin(mx) ; (v) There are no eigenvalues other than the ones characterized in (i) (iv).

28 The eigenvalue problem

29 The eigenvalue problem

30 The eigenvalue problem

31 The eigenvalue problem

32 The eigenvalue problem

33 Finite dimensional reduction In our analysis we consider the first 14 longitudinal eigenvalues /eigenfunctions and the first 2 torsional eigenvalue/eigenfunctions. In increasing order the first 10 eigenvalues are longitudinal; then we have the first torsional eigenvalue, then other 4 longitudinal eigenvalues and finally the second torsional eigenvalue. eigenvalue λ 1 λ 2 λ 3 λ 4 λ 5 λ 6 kind µ 1,1 µ 2,1 µ 3,1 µ 4,1 µ 5,1 µ 6,1 eigenvalue eigenvalue λ 7 λ 8 λ 9 λ 10 λ 11 kind µ 7,1 µ 8,1 µ 9,1 µ 10,1 ν 1,2 eigenvalue eigenvalue λ 12 λ 13 λ 14 λ 15 λ 16 kind µ 11,1 µ 12,1 µ 13,1 µ 14,1 ν 2,2 eigenvalue Table: Approximate value of the least 16 eigenvalues of for σ = 0.2 and l = π 150.

34 Finite dimensional reduction We write u(x, y, t) 16 k=1 y k(t)w k (x, y) where u solves ( u tt + γ 2 u + Υ(y)(u + u 3 ) = 0 in (0, π) π 150, π ) [0, + ). 150 We also write ϕ k (t) = y k (t) for k = 1,..., 10, τ 1 (t) = y 11 (t), ϕ k (t) = y k+1 (t) for k = 11,..., 14, τ 2 (t) = y 16 (t).

35 Finite dimensional reduction ϕ k (t) + γ µ k,1ϕ k (t) + Φ k ( ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, k 1,..., 14, τ k (t) + γ ν ( k,2τ k (t) + Γ k ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, k = 1, 2. We study analytically the interaction between each longitudinal mode ϕ k, k = 1,..., 14, and one of the torsional modes τ l, l = 1, 2. We give a suitable definition of (linearized) stability for the k-th longitudinal mode k = 1,..., 14 with respect to the l-th torsional mode. We prove that for small values of the total energy, the k-th mode, k = 1,..., 14, is stable with respect to the l-th torsional mode, l = 1, 2.

36 Numerical results for the complete system ϕ τ j (t) + γ µ j,1 ϕ j (t) + Φ j ( ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, j {1,..., 14}, ( l (t) + γ ν l,2 τ l (t) + Γ l ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, l = 1, 2, ϕ k (0) = A, ϕ k (0) = 0, ϕ j (0) = ϕ j (0) = 0, j {1,..., 14}, j k, τ l (0) = τ l (0) = 0, l = 1, 2. (P 0,k )

37 Numerical results for the complete system ϕ τ j (t) + γ µ j,1 ϕ j (t) + Φ j ( ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, j {1,..., 14}, ( m(t) + γ ν m,2 τ m (t) + Γ m ϕ1 (t),..., ϕ 14 (t), τ 1 (t), τ 2 (t) ) = 0, m = 1, 2, ϕ k (0) = A, ϕ k (0) = 0, ϕ j (0) = ϕ j (0) = 0, j {1,..., 14}, j k, τ m (0) = τ m(0) = δ if m = l, τ m (0) = τ m(0) = 0 otherwise. (P δ,k,l )

38 Numerical results for the complete system k A 1 (k) A 2 (k) Table: Numerical values of A 1 (k) and A 2 (k) when γ = k A 1 (k) > 10 > 10 > A 2 (k) > 10 > 10 > 10 > 10 Table: Numerical values of A 1 (k) and A 2 (k) when γ =

39 Instability of longitudinal modes ϕ1 ϕ2 1 ϕ3 ϕ ϕ5 ϕ6 1 ϕ7 ϕ ϕ9 ϕ ϕ11 ϕ ϕ13 ϕ τ1 τ Figure: Problem P δ,k,l with k = 10, l = 2, A = 2 and δ =

40 Instability of longitudinal modes 4 x ϕ1 ϕ2 1.5 x ϕ3 ϕ4 6 x ϕ5 ϕ6 3 x ϕ7 ϕ ϕ9 ϕ x ϕ11 ϕ12 1 x ϕ13 ϕ x τ1 τ Figure: Problem P δ,k,l with k = 10, l = 1, A = 2 and δ =

41 Answers to the three questions (Q1) why do longitudinal oscillations suddenly transform into torsional oscillations? Answer: presence of stability thresholds. (Q2) why do torsional oscillations appear with a node at midspan? Answer: for k = 8, 9, 10 we have A 1 (k) > A 2 (k). (Q3) are there longitudinal oscillations which are more prone to generate torsional oscillations? Answer: the critical amplitude of oscillation of a longitudinal mode depends on the ratio between the torsional and longitudinal frequencies. This ratio reaches its minimum for the 10th longitudinal mode.