What's the Effect of a Suddenly Applied Load on a Uniaxial Spring?

Are you talking about a mass placed on top of the spring? If so, you've got to thing about the inertia of the moving mass.

I'm not sue what sort of spring you mean and what sort of loading.

A 'suddenly applied load' could well have mementum which will cause the difference.

If the load is a weight being put on a spring...if the weight is placed on gently it will give a deflection (say x) If the same weight is allowed to fall a short distance onto the spring (ie. Applied suddenly) the deflection will be greater than x but it will then settle to x.

Del

consider a ball being placed on a spring.... now whn i place the ball on the spring slowly ie i allow the load on the spring to increase gradually by holding the ball and allowing it to settle down i get a deflection x.

if i were to jus place the ball on it from a vertical distance of even 1 mm without holding on to it i get a deflection 2x.

im interested in knowing why this happens.. the momentum factor does not answer the question entirely as the deflection should be propotional to the momentum and thus the distance it traverses before 'hitting' the spring. ...

Hello Guest,

That 2x deflection is caused by the kinetic energy of the falling ball, caused by gravitational attraction, as already explained above.

Remember the ball and the Earth (Planet) mutually attract each other, the distances each move is proportional to the inverse ratio of the respective masses. (A basic Law of Gravitational Fields)

Thus the small ball moves much further towards the Planet Earth, than the Planet Earth moves towards the small ball.

The ball yields up that energy, and in the process, deflects the spring further.

The spring (in the nature of springs), returns the energy, less some lost as heat, and the spring with the ball sitting on it, rises back to the deflection of a slowly lowered load = the 1x position.....

as the deflection should be propotional to the momentum and thus the distance it traverses before 'hitting' the spring. ...

Yes..but ...it still has momentum..during the travel of distance x, and onward to 2x, it is admittedly slowing down, but it is still moving...You would need to do some integration to work it all out? Or maybe you can just think about the logic of it...after all the maths just quantifies the effect..you have to understand before you can apply the math!

If you have a car with springs, take the shock absorbers off and drive up a rough street. Then look at harmonic motion.

Look at Hooke's law:-

http://en.wikipedia.org/wiki/Hooke's_law

and simple harmonic motion as TVP45 said in post #6

http://en.wikipedia.org/wiki/Simple_harmonic_motion

As an aside, I knew these guys were protective of their "intellectual property", but, what about that: first publishing the law as an anagram in 1696 "ceiiinossssttuu," before the Latin 2 years later: "ut tensio, sic vis"; "as the deflection, so is the force."; and no real explanation until after his death (1705).

To obtain this effect the load MUST be a mass in an homogenous force fiel.

What happens is very easy to explain with an energy balance:

In the gravitational field the mass gets an energy = G x H where

G- the weight of the mass and H= the height between the starting point, which has to be the beginning of the spring, and the position where the velocity will be =0

The energy accumulated in a spring is E= F x H / 2 where

F is the maximal force in the spring and H the spring deflection

The factor 1/2 comes from the fact that during compression (or stretching) the force in the spring is proportional to the deflection.

If you set that both energies are equal you get:

G x H = F x H/2 thus G=F/2 or F=2 x G (qed).

It has nothing to do with inertia only with energy balance.

Since F=2xG after it came to a stop the mass will be accelerated with a force =G but in the oposite direction and it starts to swing.

"It has nothing to do with inertia ..."

Are you sure about that?

I think you didn't ask your question clearly enough. Do you mean the case where you have an extension spring hanging vertically and you fasten a weight onto the end and then release the weight? And then do you mean to compare this to the case where you slowly lower the weight until the spring fully supports the weight?

Despite the poorly worded Q, I feel it has been answered pretty well already, as the same comments apply to any spring/load system...

We are being rewarded for our consideration by a deafening silence (as is so often the case )

Grumpy Del who wants his 'good answer'

Hello harish.s14

Here is your megaphone so we can hear.....

Harish, where'd you go?

When you do what I think you're doing, you're making a harmonic oscillator with no damping. If you start at 0 and oscillate about x, then it will go to 2x. Without damping, the amplitude is always symmetrical about the rest point.

sorry bout da silence guys... haven been ol lately..

well i was talkin bout an extension spring placed on a flat surface and a load applied on it axially. anway the qn has been answered .. thnx!

n @ del u got ur good ans vote! ;)