Where Does Energy Go in Plastic Deformation of Materials?

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When bending a steel bar, the energy expended primarily goes into plastic deformation and generates heat due to hysteresis losses. The process involves breaking atomic bonds and creating shear stresses, which increases the entropy of the system. While some energy is converted into heat, a portion is also used to deform the material, meaning not all energy is lost without purpose. The conservation of energy principle holds, as the energy input results in both heat and structural changes in the material. Thus, spending 100 kJ on bending the bar is analogous to spending the same amount on heating it, both increasing the system's entropy.
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I have a simple question that I cannot find a compelling answer for.

If I expend (let's say) 100 kJ bending a steel bar into a right angle, where did the energy go?

Some will obviously be expended as noise and heat. Will the rest be expended in the plastic deformation of the bar? No useful mechanical energy can be obtained from the bar after the 100 kJ is invested in it. Did almost all the energy go into elongating and compressing the steel grains and migrating dislocations? If so, that almost seems like 100 kJ disappeared from the universe without any being stored for later use or any heat being created. That doesn’t satisfy E = Q + W, right?
 
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Due to loading and unloading you would get hysteresis losses which manifest as heat.
 
Well, you moved molecules of the steel around (F x d). I'd say that was work. Heat was expended in the process. I think E = Q + W is satisfied.
 
rock.freak667 said:
Due to loading and unloading you would get hysteresis losses which manifest as heat.

So in other words, 100 kJ of heat is created when 100 kJ of energy is spent bending a bar?
E = 100 kJ, Q = 100 kJ, and W = 0 kJ?

I'll have to read more on hysteresis as well. Thanks for the reply.
 
SethGrace said:
So in other words, 100 kJ of heat is created when 100 kJ of energy is spent bending a bar?
E = 100 kJ, Q = 100 kJ, and W = 0 kJ?

I'll have to read more on hysteresis as well. Thanks for the reply.

No, you are doing work on the bar, some of which goes into deforming the bar and some is converted to heat.
 
Thanks again for the reply.

I suppose I'm basically stumped on how (on the atomic scale) deforming material consumes energy.

I understand how heating an object results in increasing the RMS velocity of its constituent atoms as well as how compressing a gas increases its pressure and temperature and gives it the capacity to do useful work. I don't understand how plastically deforming a piece of metal results in the equivalent creation of heat and stored energy however. I also cannot visualize that conservation of energy on the atomic scale.

Again, thanks for all the replies.
 
At the atomic level, metals are really like crystal lattices. They have atomic bonds and arrange themselves in planes. In plastic deformation, you are creating shear stresses between atomic planes and breaking bonds. That's where the energy is going. The atoms during bending try to configure themselves into the lowest energy state possible.
 
I think I'm starting to see it.

When the atoms scramble to find the lowest energy state after plastic deformation, heat is created which increases the entropy of the system. Additionally, by breaking bonds in the lattice structure, the entropy of the system has also been significantly increased, right? That way energy consumed by deforming the material is not 'protected' from entropy.

If I were to spend 100kJ heating up the steel bar I would increase the entropy of the system as much as I would if I spent 100 kJ bending the bar, correct?
 
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