# Is strain energy the same as potential energy, or something differen?

Sorry for this stupid, basic question. But I got hung up on it.

We know from Clapeyron's theorem that the strain(internal) energy of an elastically deformed body is equal to half the work

$$U=\frac{W}{2}$$

What I'm effectively confused about is, where does the rest of the energy from the work go? From conservation of energy the other half of the work has to be somewhere. Can you say that the other half is potential energy? Or is strain energy and potential energy effectively the same? Or am I thinking about this all wrong?

This has been asked and answered numerous times before: https://www.quora.com/Why-is-strain-energy-equal-to-1-2*force*displacement-What-about-the-remaining-half basically, it's because you're calculating the area under a triangle, because the strain energy increases linearly as the displacement increases. The Work term here refers to the final position, but the energy stored is from a relaxed state, and all of the intermediate positions in between.

• Indeed I to understand why the strain energy ends up being half the work both mathematically and graphically, as you illustrate. But I don't quite understand from a physical standpoint how only half of the energy applied to the system is retained. Where is the other half? What would a complete energy balance look like? – dan479 Oct 22 '18 at 21:48
• @dan479: There is no other half. All the input energy is stored in the elastic material. The total work done is not $W = Px$ but $W = 1/2 Px$ and therefore U = W. – Biswajit Banerjee Oct 22 '18 at 22:27
• ^Precisely. This. – Jonathan R Swift Oct 22 '18 at 22:30
• If you are only considering statics, you can't "apply a force $W$." You can only apply a force which "increases gradually from $0$ up to $W$". On the other hand if you are considering dynamics as well, the "rest of the energy" goes into the kinetic energy of the structure, and it overshoots its equilibrium position and oscillates about it. The kinetic energy is eventually dissipated, either by converting to heat (internal damping within the structure) or by transporting it away to infinity as travelling "sound waves" in the surrounding material (air, water, or whatever.) – alephzero Oct 22 '18 at 22:47

Say we consider a piece of flat steel bar, length L, And a force P pulling at the right end. So the actual force acting on that end is

Total Force = P-kx.

At the balance state P = kx when x grows from zero to x, but at the beginning of stretching the bar

X = 0 , therefore kx = 0 and at the end of stretch, kx = P

So the Total force varies linearly from $$P_{initial}\space to\space P_{ final}= P\space to--> \space(P- kx), \space \space or\space P \space to -->\space 0$$

This is a right angle triangle starting at P on the left side and tapering down to zero at right at distance of x.

And the strain on the bar is the average force times the displacement which is

$$U = \frac {P-0 } {2 }\space\times x$$

$$U = \frac {P\times x }{ 2}$$

$$U = \frac {W }{2 }$$

So as we see there is no 'hidden half' of work, all work gone into stretching bar has added up to its final strain.