Spring Force Graphs

[Pete]

Hi all,

I want to use a spring and a solenoid in opposition to create a locking mechanism. Nice and simple - the solenoid has to pull a peg vertically out of a hole when activated, and the spring will supply the returning force.

I know that my solenoids (at least the ones I have been looking at on the RS website) have a very steep fall off in force applied as the Stroke length increases. For example, the solenoid might apply a 50N force over 1mm, but by 10mm, the force could have dropped by 75%.

My question is, do springs behave in a similar way? If I have a 20N spring in conjunction with this solenoid, will the spring apply more or less force as it is compressed or is it constant?

Make sence? I'm sure I have the answer in a physics revision guide but that’s AAAALL the way at home!

Ta!

[Pete]

Ahh, I might be able to answer my own question...

Springs follow Hooke's Law, so a spring within its elastic limit would provide a linear increase in force...

http://www.diracdelta.co.uk/science/source/h/o/hookes%20law/hookes-law-001.gif

Ok, next question!

If a spring is rated at 20N, is that the maximum force?

Thanks again!

[mvalenti]

Rate? or Rated?

Spring Rate is the change in load per unit deflection in pounds per inch (lb. /in.) or Newtons per millimeter (N/mm).


Rated may be the allowable load at the working height of the spring.


-Mark

[Pete]

Thanks for answering...

Ever have one of those moments where the penny drops, the fog clears and it all becomes painfully obvious? I'm pretty sure I’ve worked it all out now... but wouldn't mind someone checking my logic!

A spring is applies a 50N at its work height. So, if a 50N force is applied to this spring, it will compress to its working height.

If you then apply a larger force to the spring, it will compress and apply and equal and opposite force.

And this reaction force is linear... (as in hookes law?)

(Mechanical engineering 101, I know, but please bear with me).

Back to my original problem: If i have a 50N solenoid, which when turned on, retracts a pin, and compresses a spring, which then returns the pin to the hole when the solenoid is disengaged. The maximum force applied by the spring in its fully compressed state, would need to be smaller than the 50N of the Solenoid. So if the spring has a SPRING RATE of 25 N/mm, for every mm compressed, the spring applies another 25N.

The phrase "mountain out of a molehill" springs to mind... But thanks for your help!

[mvalenti]

Not nowing your exact design, I would think that your spring only needs to apply as much force as it takes to move the mass of the rod. Keeping in mind friction and such... Make sense?

[Pete]

Yep! :D

I was thinking around the problem too much and just made it way to complicated in my head!

Thanks again!

[mvalenti]

Glad I could help. I often forget the acronym K.I.S.S.... Keep It Simple Stupid.... Its good to get fresh ideas from others especially when you become mired in the design. Good Luck!

-Mark