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Let's say I have a large isotropic body in uniaxial tension, onto which a small rectangular piece of film is applied. Think of a strain gage on a big body, but the 'strain gage' is one single continuous isotropic layer.

Let x- and y-direction be on the surface of the big body and z-direction the normal (piercing trough the strain gage).

In the big body I get $\sigma_x = \sigma_0; \sigma_y=\sigma_z=0$. Therefore

$$\begin{align} \epsilon_x &= \frac{\sigma_x}{E} \\ \epsilon_y = \epsilon_z &= \frac{\sigma_x\nu}{E} \end{align}$$

What is the strain in the small body?

My thinking was: Assuming I have an ideal boundary layer and negligible thickness of the film, the strain should be the same as in the big body. This is consistent with literature for $\epsilon_x$ and $\epsilon_y$. But I keep reading an expression for $\epsilon_z$ in the small body involving $\nu_B$ of the big body and $\nu_s$ of the small body:

$$\epsilon_{z,smallbody}=-\frac{\nu_s(1-\nu_B)}{1-\nu_s}\sigma_x/E$$

Can someone explain how to get this result?

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  • $\begingroup$ Wait, is the literature stating that the $\epsilon_y \neq \epsilon_z$ for the big body in uniaxial tension? $\endgroup$ – Wasabi Feb 1 '17 at 12:27
  • $\begingroup$ No only for small body. I edited the question to make this clearer. $\endgroup$ – JLo Feb 1 '17 at 12:27
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I would like to start by stating that my answer is not exactly the same as the one you've found in the literature. It is almost the same, but one sign is switched. I don't know if you wrote it wrong or if I messed up somewhere (very possible, please correct me if you find my mistake). Regardless, I am so close that I'm sure the mistake is somewhere on this page (and not in the literature, of course). And it is close enough to show you where that equation comes from, which is your actual question.


It is worth noting a key difference between the small and big bodies: the big body is in uniaxial tension, but the small one is under biaxial stress. It is under tension in the x axis but under compression in the y axis. Now, the deformation equations for biaxial stress are:

$$\begin{align} \epsilon_x &= \dfrac{1}{E}(\sigma_x - \nu\sigma_y) \\ \epsilon_y &= \dfrac{1}{E}(-\nu\sigma_x + \sigma_y) \\ \epsilon_z &= \dfrac{-\nu}{E}(\sigma_x + \sigma_y) \\ \end{align}$$

Your small body is under known $\epsilon_x$ and $\epsilon_y$ (equal to those of the big body). From that we can calculate $\sigma_{x,s}$ and $\sigma_{y,s}$. (All variables with index $s$ are for the small body, $b$ for the big body)

$$\begin{align} \epsilon_{x,s} = \epsilon_{x,b} &= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\ \dfrac{\sigma_{x,b}}{E_b} &= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\ \therefore \sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\sigma_{y,s}\\ \epsilon_{y,s} = \epsilon_{y,b} &= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\ \dfrac{\nu_b\sigma_{x,b}}{E_b} &= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\ \therefore \sigma_{y,s} &= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s} \\ \therefore \sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s}\right) \\ \sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} \\ \therefore \sigma_{y,s} &= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\ \sigma_{y,s} &= \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \end{align}$$

Now we just plug these in the $\epsilon_z$ equation. There's a lot to simplify here, so I'll show each of the steps.

$$\begin{align} \epsilon_z &= -\dfrac{\nu_s}{E_s}\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\ \epsilon_z &= -\nu_s\left(\dfrac{\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}(1+\nu_s) + \nu_b\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \nu_b\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \dfrac{(1-\nu_s)\nu_b}{1-\nu_s}\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b + \nu_b-\nu_s\nu_b}{1-\nu_s}\right) \\ \epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_b}{1-\nu_s} \\ \end{align}$$

This is almost identical to the equation you've given from the literature, with the exception that my equation has $1 + \nu_b$, while the literature has $1 - \nu_b$. I can't figure out where I went wrong (or if you wrote it wrong), but this is so close that I believe it should demonstrate where the literature got this equation from.

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  • $\begingroup$ Wow, thanks a lot! The deformation equation for $\epsilon_z$ has a typo - it says $\epsilon_x$. I am still working through the rest of it. I checked the literature again and I copied it correctly (DOI 10.1016/0040-6090(74)90001-7, eq. 18) $\endgroup$ – JLo Feb 1 '17 at 15:02
  • $\begingroup$ Hacked it in Matlab and it gives exactly your solution. So it seems like my literature is wrong or uses a different sign convention or I misunderstood? I will try to figure out. $\endgroup$ – JLo Feb 1 '17 at 15:28
  • $\begingroup$ Just occurred to me that it might be because I defined $\epsilon_{y,b} = \dfrac{\nu\sigma}{E}$, but I think it should be negative (compression). I can't redo the math to see if that's it right now, will try later (or you can try on Matlab). $\endgroup$ – Wasabi Feb 1 '17 at 15:43
  • $\begingroup$ Good point! Checked Matlab and it gives $-\dfrac{1-\nu_b}{1-\nu_s}$ (omitting the other fraction). And it is consistent with literature because I found my error. Will edit the question to reflect the sign. It is also plausible because the term will always give a negative $\epsilon_z$. So I am pretty sure this is now the correct answer. $\endgroup$ – JLo Feb 1 '17 at 16:35

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