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I am a Ph.D. student. I am using Quantum Espresso to calculate the mechanical properties ( in-plane stiffness ) (undergo uniaxial strain). I calculated the strain energy of each applied strain. The problem is how can I employ the data to get in-plane stiffness. I fitted the curve into a second-order polynomial by using Excel as shown in the figure. What is the next step after this figure? how can I calculate the in-plane stiffness?

I hope you can help me. best wishes. enter image description here

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  • $\begingroup$ What was the area of the supercell/unit cell of graphene that you used in your calculation? $\endgroup$
    – Abdul Muhaymin
    Jun 5 at 7:24
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    $\begingroup$ What is the y axis and its unit? In my post (mattermodeling.stackexchange.com/questions/10809), I write the equation related to the elastic stiffness, $(E-E_{0})/A_{0}=\frac{1}{2}C_{j}\epsilon_{j}^{2}$. You can see that this formula appears in many articles( pubs.acs.org/doi/10.1021/acsomega.0c01676, link.springer.com/article/10.1007/s11426-009-0244-3). What you need to do is obtain energies per area, fit these with the strain and investigate the 2nd order coefficient. $\endgroup$
    – Patche
    Jun 7 at 10:00
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    $\begingroup$ @ahmedhassan Okay. So now you have $x = \epsilon$ and $y=E-E_{0}$, right? The relationship between $x$ and $y$ would be $y/A_{0}=\frac{1}{2}Cx^{2}$ according to the above formula. Indeed, your data show the quadratic relationship. And the above equation can be written as $y=\frac{1}{2}CA_{0}x^{2}$. So, the number of $C$ will be approximately $C\approx 2\times(460.05)/ A_{0}$. $\endgroup$
    – Patche
    Jun 7 at 15:48
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    $\begingroup$ @ahmedhassan I think you can neglect it. The function also can be written as y = 460.05(x-0.00045)^2 -0.00005. The displacement 0.00045 is much smaller than the order of strain $x=\sim 0.01$. Thus, $y \approx 460.05 x^{2}$. This means that the undeformed structure (at $x=0$) is indeed a local minimum. $\endgroup$
    – Patche
    Jun 9 at 1:06
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    $\begingroup$ @Patche please write an answer, since comments are "temporary post-it notes" and when they get deleted no one is informed of it. $\endgroup$
    – Nike Dattani
    Jun 11 at 18:43

1 Answer 1

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The elastic stiffness $C_{j}$ with respect to each $j$ direction mediates the energy $E$ at a deformed structure and the mechanical strain $\epsilon_{j}$ as follows [Mir, S. H. et al. ACS Omega. 2020, 5, 24 & L. Tang at al. Sci. China, Ser. B: Chem. 2009, 52, 10].

$$ \frac{E-E_{0}}{A_{0}}=\frac{1}{2}C_{j}\epsilon_{j}^{2} $$ Here, $E_{0}$ is the energy of the structure at the local minimum, and $A_{0}$ is the surface area.

The figure of the original post presents $x$ as $\epsilon_{j}$ and $y$ as $E-E_{0}$, and the data exhibit a quadratic behavior that aligns with the above equation.

Neglecting the lower-order terms, the fitting function can be expressed as $y/A_{0} \approx \frac{1}{2}C_{j}x^{2}$, or $y=\frac{1}{2}C_{j}A_{0}x^{2}$. Therefore, the number of $C_{j}$ will be $C\approx 2\times(460.05)/ A_{0}$.

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  • $\begingroup$ Thank you, Mr. Patche, I really appreciate that $\endgroup$
    – ahmed hassan
    Jun 14 at 4:07

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