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Geometric elasticity · A course with experiments

L3 · Chapter 6

The thickness where a disc buckles

The energy of a non-Euclidean plate is hES+h3EBhE_{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 hC=0.163h_{C}=0.163 for the Lab’s disc at K=1K=1. The flat disc in fact gives way earlier, at hB=0.299h_{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.

Two prices that scale differently

Efrati, Sharon and Kupferman (2009) reduce the elastic energy of a thin non-Euclidean plate to two integrals. ESE_{S} measures how far the midsurface lengths miss the target metric and EBE_{B} how much the surface bends. Neither depends on thickness; thickness only weights them, by hh and h3h^{3}.

The two extremes are easy to price. A flat disc cannot bend, so its energy is hES∞hE_{S}^{\infty}, where ES∞E_{S}^{\infty} is the stretching of the best flat state; for the Lab’s disc at K=1K=1 and ν=0.5\nu=0.5 this is 0.01434. The cap of chapter 5 has the target lengths exactly, so its energy is h3EB0h^{3}E_{B0} with EB0=0.541E_{B0}=0.541.

E=∫r0r1(h wS+h3 wB) Φ dr=h ES+h3EB,EB0=23(1+ν)(cos⁡Kr0−cos⁡Kr1)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 hC=(ES∞/EB0)1/2h_{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, hC=(0.01434/0.541)1/2=0.163h_{C}=(0.01434/0.541)^{1/2}=0.163.

Both prices grow with curvature: ES∞E_{S}^{\infty} as K2K^{2} for gentle growth (the computed ratio is 0.0135) and EB0E_{B0} as KK. So hCh_{C} grows as K\sqrt{K}: stronger growth bends thicker discs.

hC=(ES∞EB0)1/2=(0.014340.541)1/2=0.163h_{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 h2h^{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 hh and negative, the bending term proportional to h3h^{3} and positive, so the flat disc becomes unstable when hh falls below hBh_{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 hCh_{C}. Like hCh_{C}, it scales as K\sqrt{K}: hB/Kh_{B}/\sqrt{K} is 0.297 at K=0.5K=0.5 and 0.299 at 1.

δ2E=h Qσ[ζ]+h3QB[ζ],hB2=max⁡ζ−Qσ[ζ]QB[ζ]≈0.2992(K=1, ν=0.5)\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=0.04h=0.04 to 0.01, the stretching per unit thickness falls 32-fold, from 4.88×10−54.88\times10^{-5} to 1.50×10−61.50\times10^{-6}, and the bending per unit thickness 13-fold, from 6.25×10−46.25\times10^{-4} to 4.66×10−54.66\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 qq costs bending Bq4Bq^{4} and lifts liquid against gravity at a cost ρg\rho g, both per unit squared amplitude, while the push releases Pq2Pq^{2}.

The film buckles first at the qq that minimizes P(q)=Bq2+ρg/q2P(q)=Bq^{2}+\rho g/q^{2}, giving their wavelength λ=2π(B/ρg)1/4\lambda=2\pi(B/\rho g)^{1/4} and threshold P=2(Bρg)1/2P=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)=Bq2+ρgq2 ⇒ q4=ρgB,λ=2π(Bρg)1/4,Pmin⁡=2 (Bρg)1/2P(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=0.26h=0.26, K=1K=1, ν=0.5\nu=0.5. Kept flat it would cost 0.26×0.01434=3.73×10−30.26\times0.01434=3.73\times10^{-3}. Bent into the full cap it would cost 0.263×0.541=9.52×10−30.26^{3}\times0.541=9.52\times10^{-3}, 2.55 times more. Since 0.26<hB=0.2990.26<h_{B}=0.299, the flat disc is unstable all the same.

The Lab’s equilibrium has stretching content 0.00801 and bending content 0.0800, so its energy is 0.26×0.00801+0.263×0.0800=3.49×10−30.26\times0.00801+0.26^{3}\times0.0800=3.49\times10^{-3}, 6.5 % below the flat disc. Stretching holds 59.7 % of it and the dome rises 0.313, 0.577 of the cap. During growth this disc leaves the plane only at KB=0.760K_{B}=0.760, where hB(K)h_{B}(K) has climbed to 0.26.

Eflat=3.73×10−3,Ecap=9.52×10−3,Edome=3.49×10−3E_{\mathrm{flat}}=3.73\times10^{-3},\qquad E_{\mathrm{cap}}=9.52\times10^{-3},\qquad E_{\mathrm{dome}}=3.49\times10^{-3}

An experiment to think through

Predict, change, explain

  1. Set the thickness to 0.15. Will the disc leave the plane before it reaches K=1K=1? Estimate the growth KBK_{B} at which it does, using hB≈0.299Kh_{B}\approx0.299\sqrt{K}.
  2. In the non-Euclidean plate Lab play “Thick disc stays flat”, “Just past the buckling threshold” and “Thin disc grows into a cap”. Then set the thickness to 0.15 and read the growth at which the disc leaves the plane.
  3. Explain each preset with hC=0.163h_{C}=0.163 and hB=0.299h_{B}=0.299, and say why the disc at 0.26 buckles even though the full cap would cost 2.55 times the flat disc.

The disc Lab finds equilibrium shapes of the axisymmetric energy of Efrati, Sharon and Kupferman on 80 radial elements as growth is imposed.

Try explaining it yourself

1. Using hB≈0.299Kh_{B}\approx0.299\sqrt{K}, estimate the growth at which a disc of thickness 0.15 leaves the plane.

Read an explained answer

KB≈(0.15/0.299)2=0.252K_{B}\approx(0.15/0.299)^{2}=0.252. The Lab’s stability calculation gives 0.258; the small gap is because hB/Kh_{B}/\sqrt{K} drifts down at smaller KK, to 0.295 near K=0.26K=0.26.

2. Setting the Poisson ratio to 0 raises hBh_{B} to 0.368. Does that make the disc more or less willing to bend, and does the cap’s bending content change?

Read an explained answer

More willing: discs up to 0.368 thick now buckle. The cap’s bending content falls to 23×1×0.541=0.361\tfrac{2}{3}\times1\times0.541=0.361, two thirds of its value at ν=0.5\nu=0.5. The flat disc’s stretching changes too, so the new threshold has to be recomputed from both terms.

3. Suppose a gel annulus of the Lab’s shape had K=0.11 cm−2K=0.11\ \mathrm{cm^{-2}} and a width of 1/K=3.02 cm1/\sqrt{K}=3.02\ \mathrm{cm}. Which thickness would mark the buckling threshold, and where do Efrati et al.’s 0.6 to 0.75 mm gels sit?

Read an explained answer

With lengths in units of 3.02 cm, K=1K=1, so hB=0.299×3.02 cm=9.0 mmh_{B}=0.299\times3.02\ \mathrm{cm}=9.0\ \mathrm{mm}. The gels are 12 to 15 times thinner than that, so they should be almost isometric to their target. The annulus is a supposition made for this exercise; the paper reports only the gels’ curvature and thickness.

Sources to read alongside

  1. Efrati, Sharon & Kupferman, Elastic theory of unconstrained non-Euclidean plates, Journal of the Mechanics and Physics of Solids 57, 762 (2009), arXiv:0810.2411v1

    Read in full: the definition of a non-Euclidean plate and why it cannot be both flat and strain-free (Sec. 3), the energy as stretching times h plus bending times h cubed, the axisymmetric reference metric and energy (Eqs. 4.1 and 4.2), the buckling thickness of about 0.3, the smaller crossover thickness and the missing equipartition (Sec. 4.2, Figs. 4.1 and 4.2), and the gel discs shaped by uneven shrinking to a Gaussian curvature of 0.11 per square centimetre.

  2. Diamant & Witten, Compression induced folding of a sheet: an integrable system, Physical Review Letters 107, 164302 (2011), arXiv:1107.5505v1

    Read in full: the wrinkle wavelength of a sheet on liquid, the exact fold shapes (Eq. 12), the exactly quadratic fall of the pressure as the sheet is pushed in, and the point where the symmetric fold touches itself, at a pressure of about 0.040 in their units.

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