IntroductionPE and KE energy of the rodEnergy to oscillation equationsTorsional oscillationsTrigonometric vs. LinearApplied to Tacoma NarrowsIn Conclusion1/24 Tacoma Narrows BridgeMatt Bates and Sean DonohoeSpring Semester 20032/24 IntroductionBefore the collapse of the Tacoma Narrows Bridge suspension bridgeswere commonly constructed using rule of thumb. This method proveddangerous, with many of these bridges exhibiting odd behavior consist-ing of large vertical and torsional oscillations. In the case of TacomaNarrows, it is believed that not the vertical oscillations but the torsionaloscillations ultimately led to its destruction. We can model these tor-sional oscillations with two different equations. These 2 equations willderived from the kinetic and potential energy equations for the bridge.3/24 PE and KE energy of the rodAdding the physics equations for vertical kinetic energy and rotationalenergy we get the total kinetic energy.T =12mv2+16mL2˙θ2(1)Modelling a cross section of the bridge (with length 2L) will help usvisualize how we get our potential energy.θθyLLy1y1There is a vertical deflection on both sides due to torsional oscillations.4/24 On the left side the vertical deflection is:yt= (y + y1)+, (2)and on the right the vertical deflection is:yt= (y − y1)+. (3)The + exponet refers to the fact that the cables only act like springswhen pulled in the downward direction.Looking at the picture depicting total displacement of a cross section ofthe bridge we see that y1is:L sin θ, (4)Subbing this into the ytequations above we get:yt= (y − L sin θ)+yt= (y + L sin θ)+.(5)5/24 From physics the total potential energy is:P E =12ky2− mgy. (6)Plugging ytin for y we get:P E = V =k2[((y − L sin θ)2)++ ((y + L sin θ)2)+] − mgy (7)6/24 Energy to oscillation equationsUsing the Lagrangian L = T − V we get:L =m ˙y22+16mL2˙θ2−12k[((y − L sin θ)2)++ ((y + L sin θ)2)+] + mgy(8)Applying the Euler Lagrange equationddtδLδ˙θ=δLδθ(9)and taking the partial derivatives we get:δLδθ=12k[2(y − L sin θ)+(−L cos θ) + 2(y + L sin θ)+(L cos θ)]= kL cos θ[(y − L sin θ)+− (y + L sin θ)+](10)7/24 andddtδLδ˙θ=13mL2¨θ.(11)With some intense simplifications we come to:¨θ =3kmLcos θ[(y − L sin θ)+− (y + L sin θ)+] (12)8/24 Applying the Euler Lagrange equationddtδLδ˙y=δLδy, (13)and taking the partial derivatives we get:δLδy= −12k[2(y − L sin θ)++ 2(y + L sin θ)+] + mg= −k[(y − L sin θ)++ (y + L sin θ)+] + mg(14)andddtδLδ ˙y=ddt(m ˙y) = m¨y. (15)With some intense simplifications we get:¨y = −km[(y − Lsinθ)++ (y + L sin θ)+] + g. (16)9/24 We then add a damping term, δ˙θ and δ ˙y, and we also add a externalforcing term f(t) to the torsional equation. Doing this we get:¨θ = −δ˙θ +3kmLcos θ[(y − L sin θ)+− (y + L sin θ)+] + f(t) (17)¨y = −δ ˙y −km[(y − Lsinθ)++ (y + L sin θ)+] + g. (18)Simplifying the torsional equation¨θ:¨θ = −δ˙θ +3kmLcos θ[(y − L sin θ) − (y + L sin θ)] + f (t)¨θ = −δ˙θ +3kmLcos θ[y − L sin θ − y − L sin θ] + f(t)¨θ = −δ˙θ +3kmLcos θ(−2L sin θ) + f (t)¨θ = −δ˙θ −6kmcos θ(sin θ) + f (t). (19)10/24 Torsional oscillationsThe one we are really interested in is the¨θ equation because this iswhat lead to the collapse of the bridge. Assuming that the cables neverlose tension and simplifying this equation we get:¨θ = −δ˙θ −6kmcos θ sin θ + f(t) (20)This is the non-linear equation. We will then assume small oscillationsand say sinθ is approximately θ and cosθ is approximately 1 to linearizeit to get:¨θ = −δ˙θ −6kmθ + f(t) (21)We chose the forcing term f(t) to be λsinµt, a term that accuratelymodels the frequency and amplitude of the oscillations observed beforethe collapse.11/24 Equation 21 is the linear model for the bridge and equation 20 is thenon-linear model.The builders of the Tacoma Narrows Bridge should have used the non-linear equation, but due to their inability to solve it they were forced touse the linear.This turned out to be a costly mistake. The linear equation gave thembad data as the model showed eventual torsional stability no matter whatdriving force and initial condition it was given.Since the bridge actually acted according to the non-linear equationnot the linear, the bridge did not settle down like the engineers of thebridge expected and the resulting torsional strain is believed to be amajor contributor to the failure of the bridge.12/24 13/24 Non-linear vs. LinearFirst we will explore the long-term behavior of the linear model, withdifferent initial conditions and driving terms of different periods (µ con-trols the period).0 200 400 600 800 1000−1.5−1−0.500.511.5Figure 1a: Linear model with initial conditions θ = 1.2,˙θ = 0, andλ = .05 (large push).Note: vertical axis is theta in radians and horizontal is time for all graphs.14/24 0 200 400 600 800 1000−2−1012Figure 1b Linear model with initial conditions θ = 2,˙θ = 0 and λ = .05(very large push).15/24 0 200 400 600 800 1000−2−1012Figure 1c Linear mo del, θ = 2,˙θ = 0 and λ = .08 (very large push andlarge driving term).We can easily see that the linear model always settles down whengiven enough time.16/24 Now let us explore the long-term behavior of the non-linear model.Here we see an interesting behavior, even if the period of the drivingterm is held constant the bridge can go crazy, all it needs is a largeinitial push. With sufficient push the torsional oscillations will never diedown.0 200 400 600 800 1000−1.5−1−0.500.511.5tθλ=0.05, µ=1.26Figure 2a: non-linear model with a large initial push (λ = .05 andµ = 1.26).17/24 0 200 400 600 800 1000−1−0.500.51tθλ=0.05, µ=1.26Figure 2b: non-linear model with a small initial push (λ = .05 andµ = 1.26).18/24 0 200 400 600 800 1000−1.5−1−0.500.511.5tθFigure 2c: non-linear model with a large initial push and a differentfrequency than in Figure 2a or 2b (λ = .05 and µ = 1.35).19/24 0 200 400 600 800 1000−0.2−0.100.10.2tθFigure 2d: non-linear model with a small initial push and