Geometric elasticity · A course with experiments

L4 · Chapter 8

Reading the snapping-beams paper

Pandey, Moulton, Vella and Holmes, “Dynamics of snapping beams and jumping poppers”, EPL 105, 24001 (2014), is five figures and ten numbered equations, with derivations in a supplement. Its argument has one control parameter, $d/S$ , three regimes of pushing and two time scales, the growth of the snap and the ringing after it. Read it in that order. For each figure, ask whether a point came from the experiment, from Kirchhoff numerics that allow large slopes, or from the shallow-angle formulas, and check the constants you can recompute.

Sandbox Physics · 2026-10-30

What the paper claims

The abstract makes three claims. The stretchability $S=h^{2}/12L^{2}$ decides whether a pushed arch stays symmetric or twists. It also sets how fast the snap grows and how fast the arch rings afterwards. And the ringing of arches and of snapping spherical shells follows the same scaling, a time proportional to $L^{2}/ch$ , with $c$ the sound speed.

Keep the three claims apart while reading. The first is tested by force curves (Fig. 2), the second by high-speed video (Figs. 3 and 4), and the third by the sound of toys (Fig. 5).

Fig. 1 and Fig. 2a: the objects

Fig. 1 is the motivation: a rubber popper, $E\approx25\ \mathrm{MPa}$ and $\rho\approx1200\ \mathrm{kg\,m^{-3}}$ , leaping from a table. Fig. 2a is the real experiment: a polyvinylsiloxane strip, $E=789.49\ \mathrm{kPa}$ , with its ends embedded in cured silicone and pushed at the apex by a razor blade in an Instron microtester with a 0.5 N load cell. The blade sets the position, not the force.

Two strips were used, with $S_{1}=1.912\times10^{-4}$ and $S_{2}=9.706\times10^{-4}$ , that is $h/L=(12S)^{1/2}$ of 0.048 and 0.108. Panels (i) and (ii) show a deep arch twisting and then snapping; (I) and (II) a shallow arch flattening and then snapping.

Eqs. 1 to 3: the model in three lines

Eq. 1 is the dynamic beam equation with the blade as a point force, Eq. 2 the Hooke law that turns end shortening into compression, and Eq. 3 the force on the twisted branch. The paper uses this shallow-angle (Föppl–von Kármán) model for formulas and asymptotics, but compares experiments with numerical solutions of the Kirchhoff rod equations, which allow large slopes. When a dashed curve and a solid curve differ in a figure, that is usually why.

Three constants can be recomputed with a few lines of code: $\tau_{\max}$ from $\tan(\tau/2)=\tau/2$ , the slope 207.75 and the coefficient 129.53. Chapter 3 derives them; this course’s benchmark reproduces them as 8.9868, 207.745 and 129.530.

\tau_{\max}^{2}=80.76,\qquad f=-207.75\,w_{0},\qquad f^{(c)}=-129.53\,\bigl(d-80.76\,S\bigr)^{1/2}

Fig. 2b to 2d: the force curves

Panel b is a symmetric arch, panel c two twisted arches at different $d/S$ , panel d the critical force against $d$ for $S_{1}$ . For panel d, Eq. 3 is a prediction with no fitted parameter: the curve starts at $d=80.76\,S_{1}=0.0154$ and rises as a square root.

Two details matter. Symmetric arches are shown only for the thicker strip, because the paper notes that the beam’s weight becomes important there. And in panels b and c the force and the midpoint height reach zero together. That observation is the paper’s explanation of where the snap happens: the blade cannot pull, so contact is lost at $w_{0}=0$ .

Fig. 3: watching one snap

A camera at 3000 frames per second follows the midpoint from A, the moment of release, to B, the inverted arch, after which it rings like a lightly damped oscillator. The inset plots the absolute midpoint deflection on a logarithmic axis. A straight stretch there means exponential growth; its slope in reciprocal seconds, multiplied by $t^{*}$ , is the dimensionless growth rate $\sigma$ .

When you read such a plot, check that the straight stretch covers more than a factor of a few in amplitude, and that the time axis has not been rescaled already. A fit over too short a stretch says little about the exponent.

Fig. 4: growth rate and ringing as confinement changes

The main panel plots $\sigma$ against the confinement $d/S$ : points are experiments, solid curves numerics, dashed curves the near-threshold Eqs. 6 and 7. Below 80.76 the growth rate climbs with confinement; above it the shallow-angle model gives a constant 24.113, while the Kirchhoff numerics keep a small dependence on $d/S$ .

The inset plots the ringing frequency. It rises from the Euler threshold, reaches 42.38 at $d/S=122.4$ in the Kirchhoff numerics, and stays there. The supplement gives 44.36 for the same plateau in the shallow-angle model. Eq. 8 converts the plateau into the ringing time $t_{r}=0.148\,t^{*}\approx0.513\,L^{2}/ch$ .

\sigma\approx24.113\quad(d/S>80.76),\qquad t_{r}=\frac{2\pi}{\omega}\,t^{*}\approx0.148\,t^{*}\approx0.513\,\frac{L^{2}}{c\,h}

Fig. 5 and Eqs. 9 and 10: from strips to poppers

For a spherical cap, Eq. 9 adds the stretching term of curvature to the beam equation, and Eq. 10 gives its ringing time with a numerically found eigenvalue of 3.196. For most shells the bracket in Eq. 10 is close to one, and with $\nu\approx1/2$ the result is $t_{r}\approx0.133\,t^{*}$ , with the base diameter as the length.

Fig. 5 plots measured ringing times against $t^{*}$ for strips and for shells, the shells’ frequencies taken from the sound of the pop with a microphone. Both sets follow the predicted proportionality. Note what this does and does not show: it tests the ringing after a snap, not how fast a shell snaps. How stretchability affects the snapping growth rate of shells is the open question the authors end with.

Worked example

One point of the paper, recomputed

Take the thinner strip, $S_{1}=1.912\times10^{-4}$ , at $d=0.03$ . Then $d/S_{1}=156.9$ , above 80.76, so the arch twists. Eq. 3 gives $|f^{(c)}|=129.53\,(0.03-0.01544)^{1/2}=15.63$ at $w_{0}=0.6235\times0.12066=0.0752$ . Since $d/S$ is also above 122.4, the snap should grow at about 24.1 and ring at the Kirchhoff plateau, 42.38, so $t_{r}=0.148\,t^{*}$ .

The Lab’s nearest setting is $h/L=0.048$ and 3 % shortening, so $S=1.92\times10^{-4}$ and $d/S=156.25$ , for which Eq. 3 gives 15.59. The Lab’s 48-segment strip records 15.50, 0.6 % low; a growth rate of 21.12 against the damped prediction 21.23; and ringing at 43.9 against 44.05. All of these compare the Lab with the paper’s theory; the measured strips appear only in the paper’s own figures.

\frac{d}{S_{1}}=\frac{0.03}{1.912\times10^{-4}}=156.9,\qquad |f^{(c)}|=129.53\times0.1207=15.63

Where the paper sits today

Gomez, Moulton and Vella (2017) note that estimates based on $t^{*}$ often predict snaps faster than observed, a gap earlier attributed to dissipation. In their strip, clamped at an angle and released just past a fold, the slow snap comes instead from critical slowing down, with a time growing as $\Delta\mu^{-1/4}$ .

This does not contradict Pandey et al. Their strips are released from shapes that are genuinely unstable, with a finite growth rate. Gomez et al.’s strips start next to a shape that has just stopped existing. Before applying either time scale to a new experiment, ask which of the two situations the release resembles.

An experiment to think through

Predict, change, explain

Set the arch Lab to $h/L=0.048$ and 3 % end shortening, close to the paper’s thinner strip. Before playing, predict the largest blade force from Eq. 3 and the growth rate from $\sigma=24.113$ with the damping of 6.
Play one push and read the force, growth and ringing readouts. Then lower the end shortening until the force chart loses its straight-line segment, and note the confinement shown.
Account for each readout using Eq. 3, the damped growth formula and the ringing plateau, and explain why the straight-line segment disappears close to $d/S=80.76$ .

Press a clamped arch until it snaps

The arch Lab solves the clamped beam equation with midline stretching on 48 segments, pushed at the middle by a blade that can only press.

Try explaining it yourself

1. For the thicker strip, $S_{2}=9.706\times10^{-4}$ , which end shortenings give the symmetric regime of Fig. 2b?

Read an explained answer

$4\pi^{2}S_{2}=0.0383$ to $80.76\,S_{2}=0.0784$ , so between 3.8 % and 7.8 % shortening. For $S_{1}$ the same window is 0.75 % to 1.54 %, where the arch is low and, as the paper notes, its weight matters.

2. Suppose a popper has a base diameter of 3 cm and a thickness of 1 mm, with Fig. 1’s $E\approx25\ \mathrm{MPa}$ and $\rho\approx1200\ \mathrm{kg\,m^{-3}}$ . Estimate the frequency of its pop.

Read an explained answer

$t^{*}=0.03^{2}\,(12\times1200/(25\times10^{6}\times10^{-6}))^{1/2}=9\times10^{-4}\times24=0.0216\ \mathrm{s}$ . With $t_{r}\approx0.133\,t^{*}=2.87\ \mathrm{ms}$ , the pop rings near 350 Hz. The diameter and thickness are assumed here, not taken from the paper.

3. A reader fits the inset of Fig. 3 and finds a growth rate of 30 in units of $t^{*}$ for an arch with $d/S=200$ . What would you check before calling this a disagreement?

Read an explained answer

First the conversion: the slope in reciprocal seconds must be multiplied by $t^{*}$ , which needs $L$ , $h$ , $E$ and $\rho_{s}$ . Then the fit window: a straight stretch over a small amplitude range does not fix the exponent. Finally the model: 24.113 is the shallow-angle value; Fig. 4 shows the Kirchhoff numerics and the data, and those are what a measured point should be compared with.

4. Gomez et al. (2017) and Pandey et al. (2014) both use $t^{}$ . Why can one predict a snap time near $0.04\,t^{}$ and the other $0.3\,t^{*}$ or more?

Read an explained answer

Pandey et al. release from an unstable equilibrium, so the e-folding time is $t^{*}/24.113=0.041\,t^{*}$ . Gomez et al. release next to a fold, where no growing mode exists; the snap time $0.179\,\Delta\mu^{-1/4}\,t^{*}$ is $0.32\,t^{*}$ at $\Delta\mu=0.1$ and longer closer to the fold.

Sources to read alongside

Pandey, Moulton, Vella & Holmes, Dynamics of snapping beams and jumping poppers, EPL 105, 24001 (2014), arXiv:1310.3703v2
Read in full, with the supplementary information: the dynamic beam equation and the integrated Hooke law (Eqs. 1 and 2), the three ranges of d/S and the force on the twisted branch (Eq. 3), the linear stability problem (Eqs. 4 and 5), the growth rate 24.113 and the near-threshold formulas (Eqs. 6 and 7), the ringing frequencies 42.38 and 44.36, the popper equation (Eq. 9) and Figs. 1 to 5.
Gomez, Moulton & Vella, Critical slowing down in purely elastic “snap-through” instabilities, Nature Physics (2017), arXiv:1609.03809v2
Main text and Methods read: the elastic time scale (Eq. 2), the fold where the inverted arch stops existing, the amplitude equation (Eq. 7) whose solution starts by growing as the square of time, the snap time that grows as the distance to the fold to the power minus one quarter (Eq. 9, Fig. 4) and the PET and steel strip properties. The appended supplementary derivation was not read in detail.

What the paper claims

Fig. 1 and Fig. 2a: the objects

Eqs. 1 to 3: the model in three lines

Fig. 2b to 2d: the force curves

Fig. 3: watching one snap

Fig. 4: growth rate and ringing as confinement changes

Fig. 5 and Eqs. 9 and 10: from strips to poppers

One point of the paper, recomputed

Where the paper sits today

Predict, change, explain

Try explaining it yourself

1. For the thicker strip, S2=9.706×10−4S_{2}=9.706\times10^{-4}S2​=9.706×10−4, which end shortenings give the symmetric regime of Fig. 2b?

2. Suppose a popper has a base diameter of 3 cm and a thickness of 1 mm, with Fig. 1’s E≈25 MPaE\approx25\ \mathrm{MPa}E≈25 MPa and ρ≈1200 kg m−3\rho\approx1200\ \mathrm{kg\,m^{-3}}ρ≈1200 kgm−3. Estimate the frequency of its pop.

3. A reader fits the inset of Fig. 3 and finds a growth rate of 30 in units of t∗t^{*}t∗ for an arch with d/S=200d/S=200d/S=200. What would you check before calling this a disagreement?

4. Gomez et al. (2017) and Pandey et al. (2014) both use t∗t^{*}t∗. Why can one predict a snap time near 0.04 t∗0.04\,t^{*}0.04t∗ and the other 0.3 t∗0.3\,t^{*}0.3t∗ or more?

Sources to read alongside

1. For the thicker strip, $S_{2}=9.706\times10^{-4}$ , which end shortenings give the symmetric regime of Fig. 2b?

2. Suppose a popper has a base diameter of 3 cm and a thickness of 1 mm, with Fig. 1’s $E\approx25\ \mathrm{MPa}$ and $\rho\approx1200\ \mathrm{kg\,m^{-3}}$ . Estimate the frequency of its pop.

3. A reader fits the inset of Fig. 3 and finds a growth rate of 30 in units of $t^{*}$ for an arch with $d/S=200$ . What would you check before calling this a disagreement?

4. Gomez et al. (2017) and Pandey et al. (2014) both use $t^{}$ . Why can one predict a snap time near $0.04\,t^{}$ and the other $0.3\,t^{*}$ or more?