Strain Energy 1

[desertfox]

Hi

Help an old fart here, I have an impact situation where a mass M travelling at X m/s collides with a spring mounted block, I went for the usual energy transfer ie:-

0.5mV^2 = 0.5 Kx^2

Then I started to consider what effect spring preload would have if any?

My final thoughts was that it would not have any effect because we have a certain amount of kinetic energy and that energy needs to be absorbed, yet if the mass hits the spring loaded stop and the spring has a higher preload than before surely this must have some effect.
All the googling I have tried seems only to deal with springs not preloaded or springs releasing energy.

slightly confused

desertfox

 

[LittleInch]

I think you're right about the energy part of it. I assume you mean that the spring is compressed, but held in a compressed state due to the reaction with some restraint. I think it would mean that the block that is spring mounted wouldn't move as far as one without the pre-load as it would take more distance to achieve the same force assuming identical springs, but might need a diagram to be sure.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[rb1957]

the obvious assumption is that the springs are along the line of flight of the impact.

i think spring preload (due to the weight of the block ?) doesn't affect the spring loads.

Quando Omni Flunkus Moritati

 

[desertfox]

Hi

Thanks for the responses

Assume the springs are in line with the colliding mass, what I am trying to workout is will altering the spring preload have an effect on a moving mass travelling towards it and finally hitting it.
If I consider 0.5 x k x x^2 this takes no account of the spring preload, so if I work this energy transfer out it seems to me that the spring is in a free state prior to the impact which is not the case in practice.

 

[rb1957]

won't spring deflection, x, have an initial value (from preload) ...
x_final = x_init+sqrt(m/k)*X

Quando Omni Flunkus Moritati

 

[desertfox]

Hi rb1957

That was my thought at one point also so in that case:-

0.5 k x^2 (intial) + 0.5 k x^2 (new) = 0.5 m V^2 (moving mass)

do we agree ?

Regards

desertfox

 

[rb1957]

no, its more final spring KE-initial spring KE = impact KE

Quando Omni Flunkus Moritati

 

[desertfox]

Hi rb1957

So the initial spring load won't help slow the impact down, at least not the way you've written it.
I was thinking if I have higher potential energy in the spring prior to impact that it would help reduce the impact energy but looking at how you've written it, it adds to the impact.

 

[rb1957]

i'm thinking (now ?!) that it might ... think about the triangle graph of F(x), if you're starting at x_initial going to x_final (absorbing the energy of the mass) then the difference (delta_x) gets smaller as x_initial increases (as the trapesium gets deeper) yes??

but that's not what i posted above !!?? 'cause i think i just did 1/2*k*(x_final^2-x_init^2) = 1/2*m*X^2
and this says x_final = x_init+dx (a fixed amount, =sqrt(m/k)*X)

but i think it's wrong ...
x_final^2-x_init^2 = m/k*X^2
x_final=x_init+delta_x
2*x_init*delta_x + delta_x^2 = m/k*X^2
so that the change in spring compression depends on the initial compression

Quando Omni Flunkus Moritati

 

[desertfox]

Hi

My thinking was preload the spring so it has potential energy, when the mass hits its ressisted by whatever potential energy is in the spring, so I know the kinetic energy of the moving mass and the initial potential energy of the spring, now if I subtract the potential energy of the spring from the kinetic energy of the moving mass the difference would be how much more the spring needs to compress.

I'm not sure what your last post is trying to say, so I put what I think in words in the above paragraph.

regards

desertfox

 

[rb1957]

words fail, see math ...

1/2*k*(xfinal-xinit)^2 = 1/2m*X^2

Quando Omni Flunkus Moritati

 

[desertfox]

Hi

So back to the preload does not help reduce impact. if I rearrange your formula then the potential energy in the spring adds to the impact energy.

 

[rb1957]

no, x_init affects delta_x

and it shouldn't've been ...
1/2*k*(xfinal-xinit)^2 = 1/2m*X^2 ... instead
1/2*k*(xfinal^2-xinit^2) = 1/2m*X^2

math fails, see sketch ...
the two hashed areas are the same size, the same energy, but one is narrower than the other ... the preload reduced the compression due to the impact.

Quando Omni Flunkus Moritati

 

[msquared48]

The impact force has to be greater than the pre-load spring tension before the spring further compresses to resist the impact force. If the impact force is less than the pre-load, the spring will not move.

Mike McCann
MMC Engineering

 

[rb1957]

i figured the preload was the weight of the block (being impacted)

Quando Omni Flunkus Moritati

 

[msquared48]

And my comments were in a frictionless universe, the impact load concentric to the major axis of the spring, and with the spring not compressed into the inelastic range, among other things....

Mike McCann
MMC Engineering

 

[rb1957]

"spherical chickens in a vaccuum" ?

Quando Omni Flunkus Moritati

 

[msquared48]

That works too...

Mike McCann
MMC Engineering

 

[desertfox]

Hi guys

Thanks for the responses!

Msquared48 yes I totally agree, that the moving mass needs enough force to overcome the spring preload otherwise it stops.
That's why used the potential energy of the preload spring and subtracted It from the moving mass kinetic energy, then whatever kinetic energy was left had to be absorbed by the spring.

Rb1957 looks like we agree now!

Regards

Desertfox