New Mathematical Models for Suspension Bridges - Gazzola (Alghero, 2013)

8/16/2019 New Mathematical Models for Suspension Bridges - Gazzola (Alghero, 2013)

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New mathematical models for suspension bridges

Filippo Gazzola

Dipartimento di Matematica - Politecnico di Milano

Alghero - June 2013

F. Gazzola - DipMat - PoliMi Maths & Bridges

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FIRST SUSPENSION BRIDGES

The first pictures of suspension bridges appear in Machinae Novae

(1615) by Fausto Veranzio from the Republic of Venice.


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However, these bridges were never built.


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However, these bridges were never built.

The first suspension bridges were built only about two centurieslater. Many bridges had serious problems under the action of thewind or of traffic loads.


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The most impressive failure of history is certainly the TacomaNarrows Bridge collapse in November 1940.




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Collapsed and current Tacoma Narrows Bridge:

The Official Report• O.H. Ammann, T. von Kármán, G.B. Woodruff, The failure of

the Tacoma Narrows Bridge , Federal Works Agency (1941)considers ...the crucial event in the collapse to be the suddenchange from a vertical to a torsional mode of oscillation.


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Collapsed and current Tacoma Narrows Bridge:

The Official Report• O.H. Ammann, T. von Kármán, G.B. Woodruff, The failure of

the Tacoma Narrows Bridge , Federal Works Agency (1941)considers ...the crucial event in the collapse to be the suddenchange from a vertical to a torsional mode of oscillation.

This leads to the following main question: why did torsional

oscillations appear suddenly?F. Gazzola - DipMat - PoliMi Maths & Bridges

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There have been many attempts to answer:


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- resonance due to the frequency of the wind


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- resonance due to the frequency of the wind- mistake in the project, structural failure


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- flutter theory




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- flutter theory

- vortex shedding, von Kármán vortices


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- flutter theory


- angle of attack of the wind




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- flutter theory


- angle of attack of the wind- ...

None of these explanations gives a full answer to the mainquestion:

why did the longitudinal oscillation transform suddenly into atorsional oscillation?


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Since then, many bridges have stiffening trusses which are veryrigid, heavy and... expensive!


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The problem has been solved, but... is it understood?


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There is a lack of reliable mathematical models which should

also display torsional oscillations.


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There is a lack of reliable mathematical models which should

also display torsional oscillations.

In particular, the bridge should not be modeled as a beam.


The cross section of the bridge may be modeled by

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• Y. Rocard, Dynamic instability: automobiles, aircraft, suspensionbridges , 1957



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• Y. Rocard, Dynamic instability: automobiles, aircraft, suspensionbridges , 1957

The equations describing this double oscillator are

m2

3 θ̈ = cos θ

f (y − sin θ) − f (y + sin θ)

m ÿ = −

f (y − sin θ) + f (y + sin θ)

.

• J. McKenna, American Mathematical Monthly, 1999F. Gazzola - DipMat - PoliMi Maths & Bridges



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The force f should take into account both the restoring elasticforce due to the hanger and gravity: McKenna takes f piecewiselinear, that is, linear unless the hanger slackens. He then adds

periodic forcing terms and damping.


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periodic forcing terms and damping.He is so able to numerically replicate the sudden transition fromvertical oscillations to torsional oscillations, but he writes that:...the range of parameters over which the transition from vertical

to torsional motion was observed was physically unreasonable ...

the restoring force due to the cables was oversimplified ... it was

necessary to impose small torsional forcing...


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We may conclude that

the McKenna model is quite promising, for the first time the

sudden transition between oscillations is visible.


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We may conclude that

the McKenna model is quite promising, for the first time the

sudden transition between oscillations is visible.

But it may be improved... some witnesses claim that the hangerswere never slacken at the Tacoma Bridge.




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Together with Gianni Arioli (PoliMi) we have stripped the modelby dropping both internal frictions (damping) and external sources(forcing).

The system is then isolated and conserves energy.


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Together with Gianni Arioli (PoliMi) we have stripped the modelby dropping both internal frictions (damping) and external sources(forcing).

The system is then isolated and conserves energy.

We assume that the moving rod is linked to two fixed rods throughan elastic membrane with elastic constant K > 0:


We considered nonlinear forces also for small elongations of the

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hangers: f (s ) = s + εφ(s ) with φ(s ) = o (s ) as s → 0. The valueof ε > 0 and the form of φ play no qualitative role.





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Basic requirements: f (0) = 0, f (s ) > 0.



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Basic requirements: f (0) = 0, f (s ) > 0. Let us take

f (s ) = s + s 2 + s 3

and m = = K = 1. Set F (s ) = s 0 f (σ)d σ and

U (θ, y ) = F (y + sin θ) + F (y − sinθ) + y 2 + θ2 .


We considered nonlinear forces also for small elongations of the( ) ( ) ( ) ( )

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f (s ) = s + s 2 + s 3



Then the equations becomeθ̈

3 + U θ(θ, y ) = 0 , ÿ + U y (θ, y ) = 0

which have a (conserved) energy given by

E =θ̇2

6 +

ẏ 2

2 + U (θ, y ) .


We considered nonlinear forces also for small elongations of theh f ( ) φ( ) i h φ( ) ( ) 0 Th l

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f (s ) = s + s 2 + s 3



Then the equations becomeθ̈

3 + U θ(θ, y ) = 0 , ÿ + U y (θ, y ) = 0

which have a (conserved) energy given by

E =θ̇2

6 +

ẏ 2

2 + U (θ, y ) .

Finally, we consider initial data such as

(y (0), ẏ (0)) = (y 0, y 1) , 0 < |θ(0)| + |θ̇(0)|


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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92


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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92

(y (0), ẏ (0)) = (0, 3) ⇒ E = 4.5


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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92

(y (0), ẏ (0)) = (0, 3) ⇒ E = 4.5

(y (0), ẏ (0)) = (0, 3.2) ⇒ E = 5.12


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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92

(y (0), ẏ (0)) = (0, 3) ⇒ E = 4.5

(y (0), ẏ (0)) = (0, 3.2) ⇒ E = 5.12There is a sudden transfer of oscillations which becomes morerelevant for increasing energy; no transfer is visible if E 3.56.


ENERGY TRANSFER (only for large energies).

E E ( ) E ( ) E ( )

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We decompose E T = E y (t ) + E θ(t ) + E y θ(t ).



W d E E ( ) E ( ) E ( )

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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92



W d E E ( ) + E ( ) + E ( )

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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92

(y (0), ẏ (0)) = (0, 3) ⇒ E = 4.5



W d E E (t) + E (t) + E (t)

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(y (0), ẏ (0)) = (0, 2.8) ⇒ E = 3.92

(y (0), ẏ (0)) = (0, 3) ⇒ E = 4.5

(y (0), ẏ (0)) = (0, 3.2) ⇒ E = 5.12F. Gazzola - DipMat - PoliMi Maths & Bridges

Can we explain this fact?

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Can we explain this fact? The phase space is 4D but, due to theconservation of energy, the dynamics takes place in a 3D manifold.

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conservation of energy, the dynamics takes place in a 3D manifold.We take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 at

energies E = 3.4, 3.5, 3.6, 3.8.



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gy, y pWe take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 at

energies E = 3.4, 3.5, 3.6, 3.8.





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gy, y pWe take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 at

energies E = 3.4, 3.5, 3.6, 3.8.



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gy y pWe take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 atenergies

E = 3.4, 3.5, 3.6, 3.8.



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We take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 atenergies E = 3.4, 3.5, 3.6, 3.8.



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We take a Poincaré section: we represent the (θ̇, θ)-intersectionsof the solutions of the system with the plane y = 0 when ẏ > 0 atenergies E = 3.4, 3.5, 3.6, 3.8.

The origin is a stable fixed point for the Poincaré map in the toppictures, whereas it is unstable in the bottom pictures.


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Since the system is conservative, det(JP E (0, 0)) = 1, so eitherboth its eigenvalues have modulus 1, or they are both real.


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• if |λ1| = |λ2| = 1 and λ2 = λ1 the system is torsionally stable;• if λ1, λ2 ∈ R and |λ1|


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• if |λ1| = |λ2| = 1 and λ2 = λ1 the system is torsionally stable;• if λ1, λ2 ∈ R and |λ1| 0 such that the system is torsionally stable

whenever 0 < E


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whenever 0 < E


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whenever 0 < E


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1) Why did we first exclude resonance and now we speak againabout resonance?


NATURAL QUESTIONS:

1) Wh did fi l d d k i

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1) Why did we first exclude resonance and now we speak againabout resonance?One must distinguish between external resonance and

internal resonance.


NATURAL QUESTIONS:

1) Wh did fi l d d k i

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internal resonance.

2) What happens if we change the parameters in the model?


NATURAL QUESTIONS:

1) Wh did fi t l d d k i

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internal resonance.

2) What happens if we change the parameters in the model?Quantitative variations of the critical energy threshold, the

kind of bifurcation may be different. The model is robust.


NATURAL QUESTIONS:

1) Wh did fi t l d d k i

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internal resonance.



3) What happens if we add damping and forcing terms?


NATURAL QUESTIONS:

1) Why did we first exclude resonance and now we speak again

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internal resonance.



3) What happens if we add damping and forcing terms?Only quantitative variations of the width of the oscillations

and of the critical energy threshold.


NATURAL QUESTIONS:


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internal resonance.





50 100 150 200 250 300 350

0.6

0.4

0.2

0.2

0.4


NATURAL QUESTIONS:


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internal resonance.





50 100 150 200 250 300 350

0.6

0.4

0.2

0.2

0.4

50 100 150 200 250 300 350

0.4

0.2

0.2


FULL BRIDGE MODEL We link together the n cross sectionswhich are subject to the restoring force of the hangers.

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We take m = = 1, we put y 0 = y n+1 = θ0 = θn+1 = 0, and we

obtain a system of 2n second order equations:





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We take m = = 1, we put y 0 = y n+1 = θ0 = θn+1 = 0, and we

obtain a system of 2n second order equations:13 θ̈i + U θi (Θ, Y ) = 0

ÿ i + U y i (Θ, Y ) = 0(i = 1, ..., n) ,

where (Θ, Y ) = (θ1 . . . , θn, y 1, . . . , y n) ∈ R2n and

U (Θ, Y ) =n

i =1

F (y i + sin θi ) + F (y i − sin θi )

+12

ni =0

K y (y i − y i +1)2 + K θ(θi − θi +1)

2

.


We fix n = 16 and K y = K θ = 320. We choose initial data close toa longitudinal mode: the k -th mode may be approximated by



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g y pp y

˙Y (0) = 0 , y i (0) = y sin

ik π17

(i = 1, ..., 16)

and Θ̇(0) = Θ(0) = 0. The value of y measures the initialpotential energy (and hence the energy E of the system).These initial conditions give rise to a solution Y which is not

periodic; we seek a periodic solution “close” to Y and we denoteby T = T (k , E ) its period.


We fix n = 16 and K y = K θ = 320. We choose initial data close toa longitudinal mode: the k -th mode may be approximated by



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g y pp y

˙Y (0) = 0 , y i (0) = y sin

ik π17

(i = 1, ..., 16)

and Θ̇(0) = Θ(0) = 0. The value of y measures the initialpotential energy (and hence the energy E of the system).These initial conditions give rise to a solution Y which is not

periodic; we seek a periodic solution “close” to Y and we denoteby T = T (k , E ) its period.

Then we perturb the torsion by taking initial data θ̇i (0) and θi (0)in the interval [−10−4, 10−4]. We define the transfer map

Ψk E : R2n → R2n , Ψk E ( Θ̇(0), Θ(0)) = ( Θ̇(T ), Θ(T )) .


We compute the eigenvalues λi = λi (k , E ) of J Ψk E (0, 0), the

Jacobian of the generalized Poincaré map at the origin

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Jacobian of the generalized Poincare map at the origin.




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We display the plot of E → max |λi | for k = 1, 2, 3, 4.




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In the damped version one sees similar behaviors.




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In the damped version one sees similar behaviors.Is this a reliable method to determine the critical energy threshold?


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Define critical energy threshold E k of the k -th mode by

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E k = inf E > 0; max

i |λi (k , E )| > 1

.



| ( )|

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E k = inf E > 0; max

i |λi (k , E )| > 1

.

The threshold E k depends on k and the effective critical energythreshold E of the bridge satisfies

E ≤ min1≤k ≤n

E k .



E i f

E 0 |λ (k E)| 1

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E k = inf

E > 0; max

i |λi (k , E )| > 1

.

The threshold E k depends on k and the effective critical energythreshold E of the bridge satisfies

E ≤ min1≤k ≤n

E k .

Our explanation of the TNB collapse. The TNB has collapsedbecause on November 7, 1940, the wind inserted enough energy to

overcome the critical energy threshold. That amount of energycreated an internal resonance which was the starting spark for theappearance of destructive torsional oscillations.


But this is a conference on variational methods!



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Together with Alberto Ferrero (UniPMN) we set up a continuousmodel for suspension bridges

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model for suspension bridges.


Together with Alberto Ferrero (UniPMN) we set up a continuousmodel for suspension bridges

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Prior works by McKenna, Walter, Lazer, Micheletti, Pistoia,Saccon...


Together with Alberto Ferrero (UniPMN) we set up a continuousmodel for suspension bridges.

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Prior works by McKenna, Walter, Lazer, Micheletti, Pistoia,Saccon...

We view the bridge as a plate

Ω = (0, π) × (−, ) ( π)

which is hinged on the small edges and free on the large edges:


According to the Kirchhoff-Love theory of elasticity the bendingenergy of the plate subject to an external load f ∈ L2(Ω) is

1 2 2

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E(u ) = Ω1

2(∆u )2 + (1 − σ)(u 2xy − u xx u yy ) − fu dxdy

where σ ∈ (0, 1/2) is the Poisson ratio. Due to the free part of theboundary, we neglect here the stretching energy.


According to the Kirchhoff-Love theory of elasticity the bendingenergy of the plate subject to an external load f ∈ L2(Ω) is

1 2 2

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E(u ) = Ω1

2(∆u )2 + (1 − σ)(u 2xy − u xx u yy ) − fu dxdy

where σ ∈ (0, 1/2) is the Poisson ratio. Due to the free part of theboundary, we neglect here the stretching energy.

For σ ∈ (−1, 1) the quadratic part of the functional E(u ) is

positive. We define the space

H 2∗ (Ω) :=

w ∈ H 2(Ω); w = 0 on {0, π} × (−, )

which is a Hilbert space when endowed with the scalar product

(u , v )H 2∗(Ω) :=

Ω

[∆u ∆v + (1 − σ)(2u xy v xy − u xx v yy − u yy v xx )] dxdy .


By the Lax-Milgram Theorem, for any f ∈ L2(Ω) there exists aunique u ∈ H 2∗ (Ω) such that

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Ω

[∆u ∆v + (1 − σ)(2u xy v xy − u xx v yy − u yy v xx )] = Ω

fv

for all v ∈ H 2∗ (Ω).


By the Lax-Milgram Theorem, for any f ∈ L2(Ω) there exists aunique u ∈ H 2∗ (Ω) such that

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Ω

[∆u ∆v + (1 − σ)(2u xy v xy − u xx v yy − u yy v xx )] = Ω

fv

for all v ∈ H 2∗ (Ω).

The strong form of this problem reads

∆2u = f

u (0, y ) = u xx (0, y ) = u (π, y ) = u xx (π, y ) = 0

u yy (x , ±) + σu xx (x , ±) = u yyy (x , ±) + (2 − σ)u xxy (x , ±) = 0.

for, respectively, (x , y ) ∈ Ω, y ∈ (−, ), x ∈ (0, π).


What can be done for this problem?

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F Gazzola - DipMat - PoliMi Maths & Bridges

What can be done for this problem?• If f = f (x ) then one can separate variables and find explicitsolutions through Fourier series (Navier 1823, Lévy 1899, Zanaboni1940) This is possible only because two opposite edges are hinged

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1940). This is possible only because two opposite edges are hinged.The beam equation is somehow the limit of the plate equation as → 0.


What can be done for this problem?• If f = f (x ) then one can separate variables and find explicitsolutions through Fourier series (Navier 1823, Lévy 1899, Zanaboni1940). This is possible only because two opposite edges are hinged.

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1940). This is possible only because two opposite edges are hinged.The beam equation is somehow the limit of the plate equation as → 0.

• Nonlinear restoring forces due to the hangers and the failure of the Hooke law, suggest to study the nonlinear problem

∆2

u + Υ(x )g (u ) = f (x ) in Ω;

here one should think of Υ to be the characteristic function of some proper subset ω ⊂ Ω, the set where the hangers act. Theproblem is well-posed.


What can be done for this problem?• If f = f (x ) then one can separate variables and find explicitsolutions through Fourier series (Navier 1823, Lévy 1899, Zanaboni1940). This is possible only because two opposite edges are hinged.

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) p y pp g gThe beam equation is somehow the limit of the plate equation as → 0.

• Nonlinear restoring forces due to the hangers and the failure of the Hooke law, suggest to study the nonlinear problem

∆2

u + Υ(x )g (u ) = f (x ) in Ω;

here one should think of Υ to be the characteristic function of some proper subset ω ⊂ Ω, the set where the hangers act. Theproblem is well-posed.

• Introducing the kinetic energy, yields the “hyperbolic” problem

u tt + ∆2u + δ u t + Υ(x )g (u ) = f (x ) in Ω × (0, T );

existence, uniqueness, regularity, asymptotic behavior...


What about the eigenvalues and eigenfunctions of the linearoperator? ∆2u = λu and (bc).



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What about the eigenvalues and eigenfunctions of the linearoperator? ∆2u = λu and (bc).

The spectrum is made by a sequence (in fact many sequences) of



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The spectrum is made by a sequence (in fact, many sequences) of

eigenvalues

(1 − σ)2 < λ1 < λ2


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eigenvalues

(1 − σ)2 < λ1 < λ2


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eigenvalues

(1 − σ)2 < λ1 < λ2


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eigenvalues

(1 − σ)2 < λ1 < λ2


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THANK YOU FOR YOUR ATTENTION!

F Gazzola DipMat PoliMi Maths & Bridges
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New Mathematical Models for Suspension Bridges - Gazzola (Alghero, 2013)

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