I'm reading this very cool paper on the formation of wrinkles in elastic materials. The key result of the paper is a set of scaling laws for the amplitude and wavelength of wrinkles based on the physical properties of the material, and an explanation for why wrinkles of a characteristic wavelength emerge from the balance of bending and stretching energies.

I'm trying to follow their derivation of these scaling laws from formulation of an energy functional. I can get it all except for the physical derivation of a key ingredient: the condition of inextensibility (Eqn. 2 in their paper).

Let me set up the problem: Suppose you have a thin elastic sheet of length $L$ and width $W\ll L$. Now suppose you clamp down two ends and you subject it to a tension force along its length (the $x$ direction). At a critical strain in the material, wrinkles will develop along the $x$ direction. Parametrize the height of the wrinkles as $\zeta(x,y)$ and work in the limit of small deformation $\zeta$ (and small derivatives of $\zeta$). One can write down bending and stretching energy functionals dependent on $\zeta$, and in addition the boundary condition given by the geometry of the problem is the so-called inextensibility condition, which the authors write as


where recall that the $y$ direction is transverse to the applied tension (and the wrinkles), and $\Delta(x)$ is the "imposed compressive transverse displacement" in their words. I feel reasonably confident that this means $\Delta(x)=W-W'$ i.e. the difference between $W$, the relaxed width of the sheet, and $W'$, the current width of the sheet after tension is applied (ignoring the formation of wrinkles). N.B. I'm pretty sure there's a typo in the limits of integration in their paper, which I have corrected here.

My question:

Can someone explain the inextensibility condition to me physically, describe how it applies to this problem, and demonstrate why the mathematical formulation above is correct?

I know that this is an important condition in many situations in continuum mechanics, so I'd like to understand it in general, as well as in this particular context. Wikipedia doesn't seem to have a good article on it.

My thoughts so far:

My physical interpretation of the inextensibility condition is that the width of the sheet at a given point must be conserved in the wrinkling process, so that if you measure the arc length of the sheet in the transverse direction along all the wrinkles, you should get the same width that you would have without any wrinkles ($W'$ in the paragraph above). If I write out that expression mathematically, I get

$$W'=\int_0^W\sqrt{1+(\partial_y\zeta)^2}dy \approx \int_0^W\left(1+\frac{1}{2}(\partial_y\zeta)^2\right)dy $$

Since $W' = W-\Delta(x)$, I can re-write $W' = \int_0^W\left(1-\frac{\Delta(x)}{W}\right)dy$. Equating the two, simplify to obtain


Which differs by a sign from the stated result. The sign that they have there is crucial, since the only way my integrand can be zero is if both components are trivially zero. It is possible that by definition $\Delta(x)<0$, but this seems a very strange convention, and it is not honored later in the paper.

Thus, I am left to believe that either (1) I made a sign error (hopefully not!), (2) the sign of $\Delta(x)$ has a strange convention, or (3) my understanding of the inextensibility condition was incorrect. I'd appreciate any help you have!

  • $\begingroup$ If I understand it correctly, the "inextensibility" condition simply states that because of the "fixed" boundary conditions at the left and right ends of the sheet (see figure 1 of that paper), the length of the sheet in the in the x-direction is fixed, and cannot change. $\endgroup$ Oct 22 '14 at 18:09

If inextensibility means that the arc length of the wrinkled sheet is equal to the unwrinkled sheet width, then the arc length must be $W$ and not $W'$. Also, the upper bound of integration over the y-axis is $W'$ (and not $W$), because there is no peace of the plate present between $W'$ and $W'\!\!+\Delta(x)$ in the deformed situation. This leads to $$W\approx\int_0^{W'}\left(1+\frac{1}{2}(\partial_y\zeta)^2\right)dy$$ Integrating up to the upper bound $W$ means you include an extra peace of $\Delta(x)$ which has to be accounted for. $$\int_0^{W'}\left(1+\frac{1}{2}(\partial_y\zeta)^2\right)dy=\int_0^{W}\left(1+\frac{1}{2}(\partial_y\zeta)^2\right)dy-\Delta(x)$$ Equating and simplifying this just as you did then leads to the condition of inextensibility as proposed in the paper. $$\int_0^W\left[\frac{1}{2}(\partial_y\zeta)^2-\frac{\Delta(x)}{W}\right]dy=0$$


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