Can we change the volume of a wire by applying force on it in order to stretch it, if yes is there any way to calculate it ??


The volume will change while tension force is still being applied. You can calculate the change using Poisson's ratio for the material.

But when you release the tension the wire will return to its original volume, even if you applied enough force to permanently increase its length.

  • 2
    $\begingroup$ You are probably right, but isn't it possible that plastic deformation could change density? I don't know, but it might be possible. Also doesn't Poisson's ratio assume linear elastic behavior? $\endgroup$ – Bob D Dec 15 '19 at 15:25
  • $\begingroup$ @BobD From what I briefly read, plastic deformation doesn't really induce a change in volume. At that point, the bonds slide past each other/rearrange so the shape deforms, but it's the same volume as they rearrange. The linear elastic behaviour is actually where you expect volume to increase/decrease. The bonds are actually becoming slightly further apart AFAIK; but then as soon as you release the elastic stress the volume begins to go back to normal. $\endgroup$ – JMac Dec 15 '19 at 15:32
  • $\begingroup$ @JMac like I said to Alphazero I wasn't sure. So you are saying plastic deformation does not affect material density? $\endgroup$ – Bob D Dec 15 '19 at 15:36
  • $\begingroup$ @BobD From what I understand, no. And sorry, I wasn't trying to correct you, just share what I had heard about it. It seems kind of intuitive. The elastic deformation is basically stretching the bonds further apart, which is why it goes back when you remove the load. The plastic deformation is a rearranging of bonds; but once the load is let off, the inter molecular distances should go back to near normal, just in some different patterns. $\endgroup$ – JMac Dec 15 '19 at 15:51
  • $\begingroup$ Is this assuming the wire is made out of metal? Surely there are substances whose volumes would permanently change. For instance, if you have a foam where tension causes bubbles to pop. $\endgroup$ – Acccumulation Dec 16 '19 at 6:38

While the diameter may be reduced when increasing the length of the wire, the volume should remain unchanged. The only way the volume could change that I am aware of is if somehow stretching the wire changed the density of the wire material.

As far as determining whether or not there is an actual change in volume, I suppose you could make direct measurements before and after stretching. Length shouldn't be problem, but the change in diameter may not be uniform, particularly at or near the areas where the wire is grasped during stretching (e.g., the jaws of a tensile tester). Perhaps you could use some sort of displacement method, such as measuring the volume of water displaced with the wire submerged before and after stretching.

Hope this helps.

  • 2
    $\begingroup$ But stretching (within the elastic limits) does change the density! $\endgroup$ – Peter - Reinstate Monica Dec 16 '19 at 7:42
  • $\begingroup$ does not .. copper remains copper - and even the insulation material will keep its density ... as long as you stay in elastic limits - over exceeding the elastic limits may change density permanently (of the insulation ... ) $\endgroup$ – eagle275 Dec 16 '19 at 9:28
  • $\begingroup$ @eagle275 "copper remains copper". Sure. So does Helium remains Helium, but it density is not constant. $\endgroup$ – Bob D Dec 16 '19 at 12:35
  • $\begingroup$ hm a metal doesn't change density as easily as a gas... the crystal grid is pretty rigid $\endgroup$ – eagle275 Dec 16 '19 at 12:39
  • $\begingroup$ @eagle275 Sure you can consider copper incompressible (tension may be another matter). But you're missing my point, which is to say "copper remains copper" is not a valid argument that a materials density can not change. $\endgroup$ – Bob D Dec 16 '19 at 12:53

No, assuming you are talking theoretically, then stretching the wire will not change the volume. The same amount of material exists except now the wire is $dh$ longer, but $dA$ will proportionally decrease with it resulting in the same volume.

  • $\begingroup$ This is incorrect for elastic deformation. $\endgroup$ – Peter - Reinstate Monica Dec 16 '19 at 7:43
  • $\begingroup$ This is wrong -- consider Poisson's ratio etc. $\endgroup$ – Toffomat Dec 16 '19 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.