Stress-strain curves are drawn with strain on X-axis and stress on Y-axis. Usually the quantities placed on X-axis are independent quantities whereas the one placed on Y-axis are dependent.
So which one is independent and why?
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$\begingroup$ For the test equipment, say an Instron, you control the strain and measure the stress. For most engineering designs, you expect some stress and determine the strain. Fortunately, for linear elastic response, one implies the other. $\endgroup$ – Jon Custer Nov 20 '19 at 14:22
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$\begingroup$ Depending on how the boundary conditions are imposed, either the stress or the strain can be the independent variable. $\endgroup$ – Chet Miller Nov 21 '19 at 22:48
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I think there is no correct answer to this question. There are cases where strain occurs without stress (i.e. heating of an unrestrained bar) and others where stress occurs without strain (i.e. heating of a fully restrained bar).
If you are asking about uniaxial "engineering" stress-strain experimental curves, then the stress axis is obtained from reading the actuator force sensor and dividing by the initial area of the test bar. Therefore, we could say that this stress is independent of strain.
On the other hand, if you are looking at a "true" stress-strain curve; the actuator force is not divided by the initial area of the bar but by the current area of the bar. Therefore, the stress is a function of the strain.
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$\begingroup$ I'm not sure why mentioning that the actuator force travels at the speed of light is relevant here. Presumably whatever measures your strain also does so with a similar speed. In either case, although the measurement of the applied stress may be able to be reported at the speed of light, the force you measure doesn't even actually propagate the stress at that speed; so I don't see how it has a bearing on what occurs first. $\endgroup$ – JMac Nov 20 '19 at 15:27
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$\begingroup$ If I am not crazy, the question was edited while I was answering, originally it asked about which one ocurrs first. Otherwise you are right, it is not relevant. $\endgroup$ – nodarkside Nov 20 '19 at 15:29
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$\begingroup$ Even then though... I don't see how it actually shows which occurs first. The speed at which you can sense the applied force doesn't really tell you much about the stress propagation or strain propagation. Why wouldn't they occur simultaneously? $\endgroup$ – JMac Nov 20 '19 at 15:33
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$\begingroup$ I didn't edited the text much (just added a hidden text). $\endgroup$ – user238497 Nov 20 '19 at 15:44
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$\begingroup$ Ok, I edited the answer in order to avoid confusion about which one occurs first or not. Otherwise we will end up asking what is time. I hope the answer helps. $\endgroup$ – nodarkside Nov 20 '19 at 15:48
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So, note that here I'll use Einstein's convention on repeated indices with the usual tensor notation.
So, at the beginning, we all know that the forces $f_i$ are obtained from the stress tensor as follows
\begin{equation}
f_i=\frac{\partial\sigma_{ik}}{\partial x_k}
\end{equation}
Therefore, by the nature of the strain tensor $u_{ik}$ we have that the variation of work is
\begin{equation}
\delta W=-\sigma_{ik}\mathrm{d}u_{ik}
\end{equation}
Don't forget to do your calculations in order to reach this.
So, after some equation juggling you'll get to this result
\begin{equation}
\begin{aligned}
\mathrm{d}U&=T\mathrm{d}S+\sigma_{ik}\mathrm{d}u_{ik}\\
\mathrm{d}F&=-S\mathrm{d}T+\sigma_{ik}\mathrm{d}u_{ik}
\end{aligned}
\end{equation}
Therefore
\begin{equation}
\sigma_{ik}=\left(\frac{\partial F}{\partial u_{ik}}\right)_T=\left(\frac{\partial U}{\partial u_{ik}}\right)_S\tag{1}
\end{equation}
Then, jumping some passages (do the math) we can write the free energy as follows
\begin{equation}
\mathrm{d}F=Ku_{ll}\delta_{ik}\mathrm{d}u_{ik}+2\mu\left(u_{ik}-\frac{1}{3}\delta_{ik}u_{ll}\right)\mathrm{d}u_{ik}
\end{equation}
This should pop up if you have done the actual right math
Now you have enough to continue by yourself (note that it's only a simple derivation, and keep in mind that it's valid if and only if Hooke's law is valid in that case) [Use (1)].
source: Course on Theoretical Physics, Vol. 7: Theory of Elasticity, L. D. Landau && E. M. Lifshits, Chapter I, USSR Academy of Sciences.
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$\begingroup$ It does, if he does the derivation. EDIT: Might add some tags in order to help with that if it's not clear enough $\endgroup$ – Birrabenzina Nov 20 '19 at 16:24
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It is possible to impose a known stress and the strain be the result. For example: the strain at the rope of a plumb line with a known weight.
Or the opposite, as when a guitar wire is strained by a known displacement, and the stress can be calculated from the pitch.
But for 3D situations some displacements and some stresses, (or only some displacements) acting on a body are known. The remaining stresses and strains are calculated from the relation between stress and strain tensor. Normally they have to be estimated by finite elements method.
In a rolling mill for example, the imposed reduction of the thickness of a billet results in a stress tensor in the material, which is a function of the friction with the rolls. Depending on that tensor, the end product will elongate more (spreading less) or spreading more (elongating less).
answered Nov 20 '19 at 19:36
