From what I understand, when we calculate elastic potential energy per unit volume of a material which extends linearly, we calculate the area under the graph of stress- strain OR strain- stress graph, they both will give the same value. Likewise, if I want to calculate the elastic potential energy of a linear extension, I could get the value from the area under the graph of force-extension/ extension- force graph. However, I am unable to understand why we can’t do the same for a non linear extension. For example in a non linear extension, the areas obtained from stress-strain and strain-stress curve are not the same. I only know that this has something to do with integration, but I still can’t seem to wrap my head around it. So to conclude, my question is 1) why we can’t obtain elastic potential energy per unit volume under a curve of a strain stress graph


If $F=kx$, in which $k$ is a constant, $$\int_{x=0}^X Fdx=\int_{x=0}^X kxdx=\left[\tfrac12\ kx^2\right]^X_0=\tfrac12F_XX$$ in which $F_X=kX$. But $$\int_{F=0}^{F_X} x dF=\int_{x=0}^X x d(kx)=\int_{x=0}^X k x dx=\tfrac12F_XX$$ This equality of the two integrals is wholly dependent on our initial assumption: $F=kx.$ The same argument applies for the stress vs. strain graph.

Less precisely, if the graph of $F$ against $x$ started as $F=kx$ but curved upwards for higher values of $x$, this would make the area between the graph and the x-axis (up to a given value, $X$) higher, but would make the area between the graph and the $F$ axis (up to $F_X$) lower!

  • $\begingroup$ Thank you. However, which area under the graph would be the correct representation of work done? $\endgroup$ Jun 2 '19 at 2:12
  • $\begingroup$ The area between the graph line and the $x$ axis. This follows from the definition of $work.$ $\endgroup$ Jun 2 '19 at 7:49
  • $\begingroup$ Thank you, I think I got it. $\endgroup$ Jun 2 '19 at 14:11
  • $\begingroup$ @YSH, if this answer your question you should considering accepting it. $\endgroup$
    – nicoguaro
    Jun 10 '19 at 14:46

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