# Volume of metal in a spring changes or not?

1. Jul 2, 2009

### physical1

Springs deform and basically change shape through atom rearrangement, but I wonder if the actual metal volume changes? Some of the atoms are pulled apart on the outer edges of the spring, while some of the atoms are rammed closer together on the inner portion. The total volume is conserved, or changes?

If a massive spring was compressed in bath water I wonder if it would change the water level, for example - and if so, significantly or not?

2. Jul 2, 2009

### Danger

The volume of the metal won't change, although the distribution of that volume will. As for the water scenario, that depends upon how much of the spring is in the water (as opposed to partial submersion) before and after compression.

3. Jul 2, 2009

### maverick_starstrider

That sounds like something that is very testable. I have a hunch that there actually might be a small decrease in volume.

4. Jul 2, 2009

### Mapes

It's interesting that you chose a spring as an example. Since the torsional load on a compression spring is essentially all shear stress, relatively little volume change is expected.

$$V^\prime=V(1+\epsilon_1)(1+\epsilon_2)(1+\epsilon_3)$$

where $V^\prime$ and $V$ are the new and old volumes, respectively, and $\epsilon_i$ ($i=1,2,3$) are the normal strains. For the small strains we see in a metal, the normalized change in volume can be simplified to

$$\frac{V^\prime-V}{V}=\epsilon_1+\epsilon_2+\epsilon_3$$

Continuing, the normal strain at any plane of an isotropic, homogeneous solid is

$$\epsilon_1=\frac{1}{E}\sigma_1-\frac{\nu}{E}\sigma_2-\frac{\nu}{E}\sigma_3$$

etc., $\nu$ and $E$ are the Poisson's ratio and Young's elastic modulus, respectively, and $\sigma_1$, $\sigma_2$, and $\sigma_3$ are the normal stresses at any plane. Again, shear stresses aren't included (as modeled in mechanics of materials, shear causes a change in shape, not a change in size).

So the normalized change in volume is

$$\frac{\Delta V}{V}=\frac{1-2\nu}{E}(\sigma_1+\sigma_2+\sigma_3)$$

(note the correspondence to the bulk modulus $K=E/3(1-2\nu)$, which is the triaxial pressure needed to achieve unit volumetric compression). Thus, normal stresses will generally cause a normalized volume change in metals of order $\sigma/E$, as the Poisson's ratio $\nu$ is often near 0.3.

For plastic loading, the volume is usually assumed to be constant, but the problem becomes more complicated here. It sounds like you're interested specifically in elastic loading.

Last edited: Jul 2, 2009
5. Jul 2, 2009

### Naty1

Well the shaope doesn't change because of "atoms rearrangement"....maybe if you permanently damage the spring, bend a portion of it for example, then some molecular rearrangement of a lattice type structure has taken place but the atoms are still atoms of metal...

The spring will change volume only very slightly as a result thermal heating do to frictional forces within the spring...not enough to be seen in a tub of water....but immersing a "hot" spring in water via thermal expansion would expand the volume of water while the spring actually cools to some slightly different volume than when it was initially immersed....all this is normally infinitesimal....

6. Jul 6, 2009

### physical1

Was thinking of something that doesn't exist today in our technology.. liquid filled hollow "vein" springs. Wonder what happens when a spring is a vein and filled with different fluids (gases, liquids, gels, etc) to get different effects than a regular solid core spring. Could be useful in the automotive industry or could just be a silly thought experiment of mine that popped up due to lack of enough rest.

Last edited: Jul 6, 2009
7. Jul 6, 2009

### Phrak

There's a corollary to the Original Post question. Would materials be elastic if there were no volumetric changes, though they may be small?

In some retrospect, on the molecular level, I'd think volumetric changes could be larger or small, and only circumstancial. Stressing bonds to a higher energy state could either increase or decrease volume.

Last edited: Jul 7, 2009
8. Jul 7, 2009

Staff Emeritus
I don't think so. Think of Play-Doh.

9. Jul 7, 2009

### Mapes

I'm not quite sure what you mean here, but cork (and cellular solids in general) is elastic, but has a Poisson's ratio of 0.5, so there's essentially no volume change during deformation (plus $\nu=0.5$ into the equations above).

10. Jul 11, 2009

### physical1

I thought a cork squeezed into a wine bottle would contract? It just changes in length and no shrinkage? Or some air bubbles inside the cork get compressed?

11. Jul 16, 2009

### Mapes

I goofed above; cork and cellular solids have a Poisson's ratio close to 0, not 0.5. (I had my extreme examples mixed up.)

12. Jul 17, 2009

### Ranger Mike

nice post
can't address the outer limits stuff on molecular changes of springs...but i can tell you that the push isfor more MPG on cars...(automobiles)...so ifin you want to up the MPG..reduce the vehicle weight...unsprung weight has always been the devil for racers so any reduction of unspeung weight will help make you rich and famous..
as of this date..no real substitute for spring steel...no real benifit of tricking out the springs as the additional requirment of a pump and associated acssesories to plumb cooling fluid to a spring would have minimla benefit to the effort

13. Jul 17, 2009

### BobG

Liquids and gases will change their volume a lot easier than solids. Why include the springs at all when the liquid and the gas could be your spring (similar to shock absorbers, eh?). In fact, the ease in which they change volume is why cars also have to use coil springs or leaf springs to absorb the bulk of the shocks to the car.

Magnets actually make better springs than solids, liquids, or gases if you're just looking for better springs (like poles repulse each other).

14. Jul 17, 2009

### physical1

I was thinking not of cooling the springs but filling the springs with different pressures, to change the effects of the spring in real time - therefore allowing the spring to become softer or harder without the springs having to be changed. However I do not quite understand what all would happen if different pressures were applied inside the spring with fluid still yet. The pressure would not have to be changed via only a pump pressurizing the springs, it could be aided by gravity storage tube/tank like how our water pressure is adjusted in a tall tank.

For MPG improvements on cars - I wish it were true! I would buy a new car if they actually considered MPG today. But my old cars get close to the same mileage - so no car sale for them! They seem to be stuck in 1984 from my perspective. Even worse than 1984 actually. Back in 1982-1984 there were cars that got 40mpg-50mpg and a Volkswagen diesel truck that got 50-60mpg, and high mileage special civic that got 50-60mpg in the late 1980's. If cars got 80-100mpg today I would buy one. No sale! They blame it on accessories and the weight of the car. If people were worried about MPG then why do the auto makers make the cars so heavy with useless heavy accessories that MPG owners don't care about - I guess because not enough MPG buyers are out there. You can find a proper size small car in 1984 that runs gasoline which gets pretty much the same mileage or better than the tiny miniature Smart Car of today! Can't blame it on weight I guess, since the smart car weighs (or should weigh) nothing. http://www.mpgomatic.com/2007/10/09/1982-a-banner-year-for-high-mpg-cars/

HOWEVER I hope this changes in say 2010, 2011 and they actually consider the recession and etc.

Last edited: Jul 17, 2009
15. Jul 17, 2009

### physical1

Because springs have a solid shape and are easier to install/swap with existing technologies in place.. for example springs in the head under camshaft, springs at the back of the car for suspension. Shocks that have air in them tend to leak and get worse over time. Well the idea of filling springs with pressure fluid is purely a mind experiment and may come to nothing at all. But was useful to think about for academic twiddling anyway.