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The Failure of the Tacoma Narrows Bridge

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Derive a mechanical model of the Bridge ... forcing term, f(t) to the second ... The nonlinear model ( for torsion) is: ?' = (-.01)?' (6k/m)cos ? sin ? f(t) ... – PowerPoint PPT presentation

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Title: The Failure of the Tacoma Narrows Bridge


1
The Failure of the Tacoma Narrows Bridge
By Thomas Fein Advisor Dr. Jyoti Champanerkar
2
Brief History of the Bridge
  • Built in Washington State
  • Construction started in November of 1938.
  • Completed on July 1, 1940.
  • Third Largest Suspension Bridge of its time.
  • Total Length1524 m (5000ft)
  • Length of Center Span854 m (2800 ft)
  • Width 11.9 m (39 ft, two lanes)
  • Cost 6.4M (90M in Todays Dollars)

3
Behavior of the Bridge on the Day of Collapse
  • Amplitude of Oscillations were about 5 feet
  • Frequency of oscillations were 38 cycles per
    minute
  • Vertical Oscillation of approximately 12 to14
    cycles per second
  • Instantaneous change from vertical to torsional
    motion

4
Classic Explanation for CollapseResonance
  • What is Resonace-The increase in amplitude of
    oscillation of an electric or mechanical system
    exposed to a periodic force whose frequency is
    equal or very close to the natural undamped
    frequency of the system.
  • Resonance fails to account to change from
    vertical to torsional motion
  • Also fails to account for vertical oscillation of
    bridge due to a single gust of wind

5
Outline
  • Derive a mechanical model of the Bridge
  • Choose Physical Constants for the differential
    equations based on Historical Data
  • Compare the linear and nonlinear models
  • Explain the instantaneous change from vertical to
    torsional oscillation by analyzing nonlinear model

6
Modeling the Bridge
7
Energy Equations from Bridge Model
  • Potential Energy-energy associated with position
  • Kinetic Energy-energy of motion
  • Potential Energy ky2/2, where kspring constant
    and y is the extended distance of the spring
  • Kinetic Energy(1/6)mL2(? ) 2, where a rod of
    mass m and length 2L rotates about its center of
    gravity (cg) with angular velocity ?

8
Energy Equations Continued
  • Let ?angle of the rod from horizontal and
    ydownward distance of the cg of the from the
    unloaded state (dashed lines)
  • Let y be the positive part of y
  • Gravitational Potential Energy is mgy
  • Extension of One Spring is (y-l sin ?) and the
    other is (y l sin ?)
  • Total PE(k/2) (y-l sin ?) (y l sin
    ?)2-mgy
  • Total KEm y2 /2 (1/6)mL2(? ) 2

9
Model Continued
(1/3)ml2? (kl) cos ? (y- L sin ?) - (y L
sin ?) y -k (y- L sin ?) - (y L sin ?)
mg Solving for ? and y adding a small
viscous damping term d? to the first equation
and dy to the second. Also, we are adding a
forcing term, f(t) to the second equation. ? -
d? (3k/m L) cos ? (y- L sin ?) - (y L sin
?) f(t) y - dy -(k/m) (y- L sin ?)
(y L sin ?) g Assuming cables never lose
tension ? - d? (6k/m)cos ? sin ? f(t) y
- dy - (2k/m)y g
10
Numerical Experiment Comparison of the Linear
and Non-linear Models
  • The nonlinear model ( for torsion) is
  • ? (-.01)? (6k/m)cos ? sin ? f(t)
  • Choosing Physical Constants
  • Letting K1000 and m2500, f(t) ?sin(µt) and
    .01 for the viscous damping terms. µ will be
    taken from 1.2 to 1.6.
  • ? (-.01)? (2.4)cos ? sin ? ? sin(µt)
  • For small ?, we let sin ? ? and cos ?1. Thus,
    the linearized version of the equation is
  • ? (- .01)? (2.4)? ? sin(µt)

11
  • Large torsional Push- ICs ? (0) 1.2 and ? (0)
    0 with µ1.2 and ?0.06
  • The linear model (Top) has settled down to an
    oscillation of 3 degrees.
  • The nonlinear model (Bottom) has settled into
    large amplitude periodic oscillation of about one
    radian.

12
  • µ1.2 and ?.06
  • Small initial conditions ,?(0)1.2 and ?(0)0.
    Similar to linear model from previous graph (Top)
  • Large torsional push combined with the same small
    torsional forcing term results in a large
    amplitude oscillation. (Bottom)
  • ConclusionNonlinear Model can have different
    responses to same forcing term depending on
    whether or not there is a single push

13
  • Increase µ 1.3 but decrease ? to 0.02 produces
    an oscillation of 1 degrees in the linear model.
    Here the linear model settles down to near
    equilibrium (Bottom)
  • Nonlinear model,-with its large initial push-
    produces a large torsional oscillation.
  • The period is 4.83 which is somewhat above that
    observed when the bridge started its torsional
    oscillation.

14
  • µ is increased to 1.4-same results.
  • Period is reduced to 4.5 which is much closer to
    the reported 4.3.
  • Similar result if µ is increased to 1.5.
  • Results vanish until integral multiples of µ are
    reached.

15
  • Historical Observation- vertical motions that
    came before the torsional ones had a frequency of
    about 40 per minute( µ near 4).
  • When the initial conditions are large or small
    combined with a forcing term of ? sin (4t) the
    results in either a small linear response or a
    large nonlinear response respectively.

16
  • Compare the transient behavior for the nonlinear
    (correct) model and the linear model.
  • Different initial conditions are used and the
    forcing term, when used, takes the form ?
    sin(1.2t).
  • Figure Below- Large Initial Push and No Forcing
    Term. There is very little difference as both
    models settle back to equilibrium.

17
  • Now, a forcing term is introduced, ?.05 but with
    initial conditions at equilibrium.
  • As seen in Figure below, the results for both
    models are again similar.

18
  • In Figure Below both effects are combined.
  • As predicted by the Principle of Superposition,
    the linear system (top) dies out.
  • The nonlinear model goes into large torsional
    oscillation, eventually settling down to a large
    periodic motion.

19
  • The sensitivity of the nonlinear system to the
    amplitude of the forcing term is exhibited in the
    Figure below.
  • If ? is reduced from 0.05 (top) to 0.04 (bottom)
    the motion is reduced to near-equilibrium.

20
The Transition to Torsional Motion
  • Investigate initial value problem for nonlinear
    system if cables lose tension briefly
  • Start with large vertical push y(0)26 and tiny
    torsional forcing
  • As seen in figure below, the cables lose tension
    for about 2 periods but torsional motions are
    around zero.

21
Vertical and Torsional Responses to large push
(y(0)26) in vertical direction
22
The Transition to Torsional Motion
  • Repeat Experiment but change y(0)31.
  • At first almost no torsional oscillations then it
    instantaneously approaches 1 radian
  • With no torsional forcing term, the motion is
    damped and settles down to near equilibrium

23
Vertical and Torsional Responses to large push
(y(0)31) in vertical direction
24
  • Repeat Experiment with tiny displacement ? (0)
    .001 and no torsional forcing term.
  • Results Below- Again, no torsional oscillation
    for a while, then and instantaneous jump to large
    torsional oscillation up to around 1 radian
  • With no torsional forcing term, oscillation dies
    down due to damping.

25
  • Now, we combine a small periodic forcing term and
    a large vertical push.
  • Results- A quick transition to large torsional
    oscillation as the vertical oscillation is damped
    away.

26
Conclusion
  • Linearizing for small oscillations was a good
    approximation but for large oscillations it
    should be categorized as an error.
  • The error of treating cables, in both their
    loaded and unloaded state, as springs that obey
    Hookes law was corrected.
  • Most important conclusion-Within frequencies
    corresponding to those of the bridge, the final
    oscillations produced by the nonlinear model can
    be both very large or very small.

27
  • Thank You For Your Time.
  • Questions?
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