Let's consider a spring. I am a strong man(well, lets assume) and I am pulling the spring. the work I do is being stored in the spring in the form of its elastic potential energy. Then suddenly, elastic limit is crossed and the spring reaches to the yielding region. then, at that very moment, i become exhausted and release the spring. then I realise that the spring doesn't budge. the energy i stored is lost. its gone. Nowhere to be found. Mind helping me find it?
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ It's been used to deform the spring permanently. $\endgroup$– LagerbaerApr 10, 2013 at 15:49
-
$\begingroup$ in what form it will store then? $\endgroup$– neweraApr 10, 2013 at 15:51
-
1$\begingroup$ Each arrangement of atoms in a material has a certain amount of energy stored in it (call it the structural energy) which basically comes down to the electrostatic energy of the atoms and their electrons. When deforming a material, you change the arrangement and thus change the electrostatic energy. By pulling our your spring too hard, you bring the atoms in a new "metastable" state where they won't revert to the old position on their own, but still have a higher energy than before. $\endgroup$– LagerbaerApr 10, 2013 at 15:56
-
$\begingroup$ ah, nice answer man $\endgroup$– neweraApr 10, 2013 at 15:57
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
When a metal spring is stretched beyond it's elastic limit, the metal begins to undergo some plastic deformation. This is a permanent deformation of metal crystals caused by the creation and motion of crystal lattice dislocations. These processes are partially irreversible and some of the work performed to deform the spring is lost as heat.
$\begingroup$
$\endgroup$
The way you have stated the question is unrealistic. If you let go at the very instant that you hit the yield stress of the material, it will still recover all of the elastic deformation. Only the deformation beyond yield will be permanent.
That said, if you continue deforming the material beyond the yield point of the material, any extra work ($\dot{W} = \sigma:\dot\epsilon$ assuming a perfectly plastic material) will be dissipated as heat. The energy is spent rearranging the microstructure of the material, and it cannot be recovered.
To address the title of your question, Hooke's law only applies in the linear elastic region of the material behavior. Beyond that point, it becomes invalid as it does not capture the plasticity.
$\begingroup$
$\endgroup$
Each arrangement of atoms in a material has a certain amount of energy stored in it (call it the structural energy) which basically comes down to the electrostatic energy of the atoms and their electrons. When deforming a material, you change the arrangement and thus change the electrostatic energy. By pulling our your spring too hard, you bring the atoms in a new "metastable" state where they won't revert to the old position on their own, but still have a higher energy than before.
answered Apr 10, 2013 at 15:58
$\begingroup$
$\endgroup$
when a material is deformed the energy required to overcome the structural energy(i.e binding energy) is provided by the part of the stored potential energy in the spring and the most of the remaining part of the stored energy(as already said by mark rovelta) gets dissipated in the form heat from the material.
