Chronic Logic

I found this out by first looking at a vertical bar fixed at one end, without damping applied; if you go to 0.5 g and then take it to 1g as the oscillation is at its lower extremity, it will come to rest immediately
My second and third paragraphs are do do with this effect. Applying the second increase at the lower extremity of the oscillation cancels out each other due to the waves of oscillation of equal amplitude being 90 degrees apart. The same can be said if you apply the additional force in 4 stages of 0.25 G at 45 degree increments. Extending this you can see that if an equal amount of force is applied each step, and that the steps are applied evenly across the whole oscillation, they will cancel each other out. Although we don't know the frequency of the oscillation of each bridge, if we assume that all steps apply the same force evenly and that the time to apply these steps is greater than the minimum frequency of any bridge, this effect will come into play.  

This is all without damping though. When you introduce damping, the remaining force from the initial step that is still oscillating in the system at a point in the future is proportional to the damping co-efficient (another way of stating the definition of the damping co-efficient). Since the amount of additional force is equal at each step, the net resulting force is equal to the amount of force damped over half an oscillation. The higher the damping effect, the higher the net resulting force.

If you managed to work out the damping co-efficient of a bridge, you could also proportionally reduce the force at each step by that amount, therefore the resulting force from half an oscillation ago would equal the new force being added, thus returning the equation back into equilibrium. However, from some quick tests the damping co-efficient seems to vary across bridge designs (proportional to the amount of regidity across the whole bridge ?).

So from all that, what's the conclusion? If you had the amount of force applied each step slowly decreasing, and the time taken to apply the whole gravity greater than one oscillation, you would get a pretty stable result. The smaller the steps the better, but means the time to apply all the gravity is more, approaching infinity (and no-one is going to wait that long ;) ). I think around 3-4 oscillations worth would be reasonable, so assuming an oscillation is one second (I've seen oscillations of .5 seconds and as long as 2 secons) and that 50% or the resulting force was damped over that time, you would have the bridge gradually sink due to the strain for 4 seconds, then have a remaining oscillation of ~4% the size of the oscillation experienced now.  

(Edited by VRBones at 4:26 pm on Nov. 7, 2001)

(Edited by VRBones at 4:27 pm on Nov. 7, 2001)

Although the frequency of the bounce is not known for all bridges, if an even amount of gravity was applied on each step over the duration of one cycle, the resulting waves would more or less cancel each other out with the remaining force being proportional to the damping co-efficient. That means that if there was no damping the bridge would be in equilibrium, the more damping the less equilibrium (as the remaining force from the initial step would be dissapated by the damping co-efficient over time), but that would also have a smaller amplitude due to the damping effect.

Another way to look at it would be to have the amount of energy added to the system (gravity) to be decremented each step by the damping effect. This would keep the above equilibrium intact, regardless of the natural frequency of the bridge. This would also make an asymptotic force (like the 50%, 75% scenario) which technically would never be fully applied, but I'm sure you could drop it to G when it's as close as practical.

Update: The last section is incorrect as the damping effect for each bridge is different, thus unknown.

(Edited by VRBones at 6:25 pm on Nov. 6, 2001)