--- title: "The thickness where a disc buckles" description: "Stretching scales with thickness, bending with its cube. Find where a growing disc leaves the plane, why a cap is not yet cheaper there, and how energy splits." canonical_url: "https://sandboxphysics.com/stories/geometric-elasticity/the-thickness-where-a-disc-buckles/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/geometric-elasticity/the-thickness-where-a-disc-buckles/" last_modified: "2026-10-30" --- [Geometric elasticity · A course with experiments](https://sandboxphysics.com/stories/geometric-elasticity/) L3 · Chapter 6 # The thickness where a disc buckles The energy of a non-Euclidean plate is $hE_{S}+h^{3}E_{B}$. A flat disc pays only the first term and a perfect cap only the second, so a naive guess puts the change where they are equal, at $h_{C}=0.163$ for the Lab’s disc at $K=1$. The flat disc in fact gives way earlier, at $h_{B}=0.299$, where Efrati et al. report about 0.3. In between, the disc settles on a shallow dome that keeps some stretching and costs less than either extreme. Thinner still, the dome becomes the cap, and the stretching left over falls faster than the bending. Sandbox Physics · 2026-10-30 In this chapter [All eight chapters](https://sandboxphysics.com/stories/geometric-elasticity/) Before you start Chapter 5’s target metric and stretch-free cap; chapter 1’s scaling of stretching with $h$ and bending with $h^{3}$; minimizing a function; and the idea of a stability threshold from chapter 2. ## Two prices that scale differently Efrati, Sharon and Kupferman (2009) reduce the elastic energy of a thin non-Euclidean plate to two integrals. $E_{S}$ measures how far the midsurface lengths miss the target metric and $E_{B}$ how much the surface bends. Neither depends on thickness; thickness only weights them, by $h$ and $h^{3}$. The two extremes are easy to price. A flat disc cannot bend, so its energy is $hE_{S}^{\infty}$, where $E_{S}^{\infty}$ is the stretching of the best flat state; for the Lab’s disc at $K=1$ and $\nu=0.5$ this is 0.01434. The cap of chapter 5 has the target lengths exactly, so its energy is $h^{3}E_{B0}$ with $E_{B0}=0.541$. $$ E=\int_{r_{0}}^{r_{1}}\bigl(h\,w_{S}+h^{3}\,w_{B}\bigr)\,\Phi\,dr=h\,E_{S}+h^{3}E_{B},\qquad E_{B0}=\tfrac{2}{3}(1+\nu)\bigl(\cos\sqrt{K}r_{0}-\cos\sqrt{K}r_{1}\bigr) $$ ## A naive crossover Setting the two prices equal gives $h_{C}=(E_{S}^{\infty}/E_{B0})^{1/2}$. Thicker than this, the flat disc is cheaper than the cap; thinner, the cap is cheaper. With the Lab’s values, $h_{C}=(0.01434/0.541)^{1/2}=0.163$. Both prices grow with curvature: $E_{S}^{\infty}$ as $K^{2}$ for gentle growth (the computed ratio is 0.0135) and $E_{B0}$ as $K$. So $h_{C}$ grows as $\sqrt{K}$: stronger growth bends thicker discs. $$ h_{C}=\Bigl(\frac{E_{S}^{\infty}}{E_{B0}}\Bigr)^{1/2}=\Bigl(\frac{0.01434}{0.541}\Bigr)^{1/2}=0.163 $$ ## Where the flat disc actually gives way The flat disc is not uniformly strained. In Efrati et al.’s gel the rim shrinks more than the centre, so the short outer rings squeeze the inside of the disc. Compression in a plate is resisted by bending, and the resistance grows as $h^{2}$ relative to the stress. Below some thickness the squeezed region buckles out of the plane, just as the strip of chapter 2 did. Testing every small out-of-plane shape of the flat disc gives that threshold. In the energy’s second variation the stress term is proportional to $h$ and negative, the bending term proportional to $h^{3}$ and positive, so the flat disc becomes unstable when $h$ falls below $h_{B}$. On 40, 80 and 160 segments this gives 0.29905, 0.29911 and 0.29912; Efrati et al. report about 0.3 and note that it lies above $h_{C}$. Like $h_{C}$, it scales as $\sqrt{K}$: $h_{B}/\sqrt{K}$ is 0.297 at $K=0.5$ and 0.299 at 1. $$ \delta^{2}E=h\,Q_{\sigma}[\zeta]+h^{3}Q_{B}[\zeta],\qquad h_{B}^{2}=\max_{\zeta}\frac{-Q_{\sigma}[\zeta]}{Q_{B}[\zeta]}\approx0.299^{2}\quad(K=1,\ \nu=0.5) $$ ## A compromise between the two thicknesses Between 0.163 and 0.299 the flat disc is unstable even though the full cap would cost more. The way out is a shallow dome: it bends a little to relieve the worst compression and keeps the rest as stretching. Its energy is below both extremes. As the disc thins further, the dome rises towards the cap. Some stretching remains: near the free edges the dome lowers its bending by stretching a little, in a band that narrows as the disc thins. ## No equal sharing One might expect the two energies to end up comparable, as kinetic and potential energy share equally in many physical systems. Efrati et al. find no such equipartition, and so does the Lab. From $h=0.04$ to 0.01, the stretching per unit thickness falls 32-fold, from $4.88\times10^{-5}$ to $1.50\times10^{-6}$, and the bending per unit thickness 13-fold, from $6.25\times10^{-4}$ to $4.66\times10^{-5}$. The stretching share drops from 7.2 % to 3.1 %. On log axes the two fall with slopes 2.51 and 1.87. A very thin sheet is therefore almost isometric to its target geometry: its shape is chosen by geometry, and the material only fixes the price. ## The same bargain on a floating sheet Diamant and Witten (2011) study another sheet that must choose between bending and a second energy: a film floating on a liquid and pushed from its ends. A ripple of wavenumber $q$ costs bending $Bq^{4}$ and lifts liquid against gravity at a cost $\rho g$, both per unit squared amplitude, while the push releases $Pq^{2}$. The film buckles first at the $q$ that minimizes $P(q)=Bq^{2}+\rho g/q^{2}$, giving their wavelength $\lambda=2\pi(B/\rho g)^{1/4}$ and threshold $P=2(B\rho g)^{1/2}$. As with the disc, a threshold comes from setting a bending cost against another energy, and neither alone fixes the answer. $$ P(q)=Bq^{2}+\frac{\rho g}{q^{2}}\ \Rightarrow\ q^{4}=\frac{\rho g}{B},\qquad \lambda=2\pi\Bigl(\frac{B}{\rho g}\Bigr)^{1/4},\qquad P_{\min}=2\,(B\rho g)^{1/2} $$ Worked example ## The disc just past the threshold Take the preset “Just past the buckling threshold”: $h=0.26$, $K=1$, $\nu=0.5$. Kept flat it would cost $0.26\times0.01434=3.73\times10^{-3}$. Bent into the full cap it would cost $0.26^{3}\times0.541=9.52\times10^{-3}$, 2.55 times more. Since $0.26