# How do springs work?

1. Sep 23, 2012

### joeh1971

If you take a piece of steel wire and pull it with your fingers it is very difficult to stretch. However, if you coil that steel wire around a pencil it now becomes very easy to stretch. Why is a coiled wire (i.e.- a spring) really easy to stretch or squeeze but the same piece of metal, in the form of a wire, was so reluctant to change shape in the first place?

I know that the interatomic forces in the straight wire are very large making it very difficult to stretch the wire by a large distance but how does changing the wire shape decrease those forces? Any insight into the physics mechanism involved will be very helpful. An explanation in terms of the forces involved is what I am looking for.

2. Sep 24, 2012

### Reogl

it is a mechanics of material that is involve .. a long round bar stretched along its ends..lengthens & thins ,when compressed.. it shortens & fattens, (the change in shape in not noticeable) . Releasing.. it return to original structure (wear/tear builds up). If twisted or bended it also deforms the same way..If you shape it to spiral with a certain radius and push or pull it along its ends..the spring will be shorten or lengthens (depend on the direction of force)
But with noticeable change in spring length..force usually directly proportion to change of length (hooks law i think).
The spring incorporate all this type (stretch,compress,twist or bend -the weakest resistant) of stress into one action like pulling it (or pushing) with ease and observable change in shape.

Furthermore if you make it into a "zigzag" shape..same thing happen but its like you are just bending it.

Last edited: Sep 24, 2012
3. Sep 24, 2012

### sophiecentaur

The reason that a spring is 'easier' to stretch is to do with the geometry. When a coil spring is stretched, the actual wire is twisted, rather than stretched. You can demonstrate this to yourself by using your two index fingers as models of one coil of a spring and see how they would need to twist as you pull your hands apart.
So, what you have with a coil spring is effectively a very long wire that is twisted rather than stretched. Instead of all the material over the whole cross section being deformed by the same amount, only the outside layers are stretched fully (in a spiral sense) and the very core is hardly deformed at all. What you have is a much thinner piece of material that is also longer (the spiral distance). The formula which gives the force for a stretching a straight wire is
F = A 'strength' factor X Cross sectional area X fractional increase in length

For a helical spring, (a twisted wire) less of the CSA is used and the amount that the material stretches (actual fractional increase in length) in much less - so less force.

Note: you can get 'torsion bar' springs on car suspension which use a bar in the 'twisting' mode. This can give a nice, soft ride - as in the Citroen 2CV and Renault 4 of old.

4. Sep 24, 2012

### Staff: Mentor

I had been under the impression that the shear strain within the helical wire is uniform over the cross section, rather than higher at the outside than in the center. If this were the case, the shear stress would be uniform over the cross section. Is this not correct?

5. Sep 24, 2012

### sophiecentaur

The stress would be proportional to displacement / = strain (no?). If the displacement varies over the cross section (and it does with torsional movement) then I would expect the stress not to be uniform. The same thing even applies to bending a bar where the centre doesn't alter its length at all. That's why an 'I' girder or a box section is as strong as a rectangular bar.
There is very little actual shear displacement, I should have thought, in a spring, compared with the rotational strain. I was trying to visualise what would cause shear distortion and I couldn't. I could be missing something but won't a helical spring support a much greater load than the equivalent piece of wire, straightened out and used as a cantilever (in shear)?

6. Sep 24, 2012

### Staff: Mentor

Opps. My mistake. Of course you're right. The strain is rdθ/dl, where l is measured along the length of the wire, and where dθ/dl is constant. Thanks for setting me straight.