Geometric elasticity · A course with experiments

L2 · Chapter 3

Pushing an arch until it snaps

A blade that pushes the middle of an arch down at a set position meets a force that first rises and then falls. A shallow arch stays symmetric and is pressed flat. A deep arch, once $d/S>80.76$ , leans into an S shape at a critical force $129.53\,(d-80.76\,S)^{1/2}$ , after which the force falls in proportion to the midpoint height. Either way the force reaches zero just as the midpoint reaches the line between the clamps. The blade cannot pull, so there the strip leaves it and snaps into the inverted arch.

Sandbox Physics · 2026-10-30

The blade sets the position, not the force

The blade presses the midpoint $x=0$ and holds its height $w_{0}$ at whatever value it is set to. The force $f$ it needs, in units of $B/L^{2}$ and negative when pushing down, is part of the answer. Pandey et al. (2014) call this displacement control. A dead weight is the other extreme: it fixes $f$ and lets $w_{0}$ follow.

At each blade position the strip is in balance, so the static beam equation of chapter 2 holds with the point force added as a delta function. Away from the blade the solution is a sum of $1$ , $x$ , $\cos\tau x$ and $\sin\tau x$ , and the force appears as a jump of $f$ in the third derivative at $x=0$ .

\frac{d^{4}w}{dx^{4}}+\tau^{2}\frac{d^{2}w}{dx^{2}}=f\,\delta(x),\qquad w\Bigl(\pm\tfrac12\Bigr)=w'\Bigl(\pm\tfrac12\Bigr)=0

The symmetric branch

For a shape that is the same on both sides, solve on $0<x<\tfrac12$ with $w'(0)=0$ and $w'''(0^{+})=f/2$ . The equation is linear, so the shape is $f$ times a fixed profile that depends only on $\tau$ . Call the midpoint height per unit force $G(\tau)$ and the slope integral per unit force squared $J(\tau)$ .

Hooke’s law then reads $S\tau^{2}=d-\tfrac12f^{2}J(\tau)$ , which gives the force for each load. As the blade descends, $\tau$ climbs from the Euler value $2\pi$ towards $(d/S)^{1/2}$ , the load of the flat compressed strip. There $f$ and $w_{0}$ vanish together. The force rises from zero and falls back to zero on the way, so it has a peak.

f=-\Bigl[\frac{2\,(d-S\tau^{2})}{J(\tau)}\Bigr]^{1/2},\qquad w_{0}=f\,G(\tau),\qquad 2\pi<\tau<\Bigl(\frac{d}{S}\Bigr)^{1/2}

The twisted branch of a deep arch

The clamped strip has a second Euler shape, odd about the middle: $\sin\tau x-2x\sin(\tau/2)$ . It meets the clamped conditions when $\tan(\tau/2)=\tau/2$ , first at $\tau_{\max}=8.9868$ , so $\tau_{\max}^{2}=80.76$ . This odd shape does not move the midpoint, so the blade cannot hold it back.

If $d/S$ exceeds 80.76, the load reaches $\tau_{\max}$ before the strip is flat. From then on any amount of the odd shape can be added at that same load, and it soaks up the length that Hooke’s law no longer allows the symmetric part to keep. The load stays at $\tau_{\max}$ , so the force is simply $w_{0}/G(\tau_{\max})$ : a straight line through zero. Solving it here gives the slope 207.745 and the coefficients 129.530 and 0.6235, matching Pandey et al.’s 207.75, 129.53 and 0.62.

f=-207.75\,w_{0},\qquad f_{\mathrm{c}}=-129.53\,\bigl(d-80.76\,S\bigr)^{1/2}\quad\text{at}\quad w_{0}=0.62\,\bigl(d-80.76\,S\bigr)^{1/2}

Why it jumps where the force is zero

Both branches end at $w_{0}=0$ with $f=0$ : the flat compressed strip for a shallow arch, the pure odd shape for a deep one. Continuing below the clamp line would, by symmetry, need a force pulling up. The blade can only push, so contact is lost, no balanced shape is left nearby, and the strip falls into the inverted arch. Pandey et al. use this to explain why their arches snap as the midpoint reaches the clamp line, and not at the force peak.

A dead weight would behave differently: once it exceeds the peak force, no balanced shape can hold it, and the jump starts at the peak. Gomez, Moulton and Vella (2017) study a third case, an arch with one end clamped at an angle, whose inverted shape stops existing at a fold without first becoming unstable. Where the jump starts is set by how the structure is loaded.

Worked example

The deep and shallow arches of the Lab

Deep arch: $S=7.5\times10^{-5}$ , $d=0.0225$ , $d/S=300>80.76$ , so it twists. $d-80.76\,S=0.01644$ and its square root is 0.1282. The twist begins at $|f_{\mathrm{c}}|=129.53\times0.1282=16.61$ with the midpoint at $0.6235\times0.1282=0.0800$ , after the blade has pushed it down from 0.0890 by only 0.0090. Halfway down the straight line, at $w_{0}=0.04$ , the force is $207.75\times0.04=8.31$ .

The Lab, on 48 segments, records a peak of 16.47, 0.8 % below 16.61; on 96 segments the gap falls to 0.44 %. Shallow arch: $d/S=60<80.76$ , so it stays symmetric. Following the symmetric branch gives a peak of only 1.95, and the strip is pressed flat before it snaps.

|f_{\mathrm{c}}|=129.53\,(0.0225-80.76\times7.5\times10^{-5})^{1/2}=16.61,\qquad w_{0}=0.0800

An experiment to think through

Predict, change, explain

Before you watch: will the shallow-arch preset ( $d/S=60$ ) show a straight-line stretch on its force chart? At what midpoint height will each preset leave the blade?
In the arch Lab play “Deep arch twists, then snaps” and then “Shallow arch flattens, then snaps”, watching the force chart. Then open the Kresling Lab, choose “Bistable cell clicks shut” and note where its gripper lets go.
Explain the shape of both force curves using $\tau_{\max}^{2}=80.76$ , and why the strip and the Kresling cell both leave their pusher where the holding force reaches zero.

Press a clamped arch until it snaps Press a folded Kresling tube

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. The Kresling Lab treats each crease as a linear spring in the truss model of Yasuda and colleagues and stops the plates on contact 9 mm apart.

Try explaining it yourself

1. For $h/L=0.03$ , what is the smallest end shortening at which the pushed arch twists?

Read an explained answer

$d=80.76\,S=80.76\times7.5\times10^{-5}=6.06\times10^{-3}$ , about 0.61 % of the length. Below that the arch is pressed flat.

2. Take $h/L=0.03$ and $d=0.03$ . Find the critical force and the midpoint height where the twist begins.

Read an explained answer

$d-80.76\,S=0.02394$ , whose root is 0.1547. So $|f_{\mathrm{c}}|=129.53\times0.1547=20.0$ at $w_{0}=0.6235\times0.1547=0.0965$ . The unloaded rise is $\tfrac{2}{\pi}(0.03-0.00296)^{1/2}=0.105$ .

3. Hang a weight equal to 17 on the deep arch instead of using the blade. Where does it snap?

Read an explained answer

The weight exceeds 16.61, the largest force the arch can carry in a stable shape. Symmetric shapes further down carry more, but each one gives way to the twist, so there is nothing to stop at and the arch jumps straight away. A weight just below 16.61 would stop near $w_{0}=0.080$ , and the slightest extra load there sets it off: under a weight the jump starts at the force peak, while under the blade it starts at $w_{0}=0$ .

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.

The blade sets the position, not the force

The symmetric branch

The twisted branch of a deep arch

Why it jumps where the force is zero

The deep and shallow arches of the Lab

Predict, change, explain

Try explaining it yourself

1. For h/L=0.03h/L=0.03h/L=0.03, what is the smallest end shortening at which the pushed arch twists?

2. Take h/L=0.03h/L=0.03h/L=0.03 and d=0.03d=0.03d=0.03. Find the critical force and the midpoint height where the twist begins.

3. Hang a weight equal to 17 on the deep arch instead of using the blade. Where does it snap?

Sources to read alongside

1. For $h/L=0.03$ , what is the smallest end shortening at which the pushed arch twists?

2. Take $h/L=0.03$ and $d=0.03$ . Find the critical force and the midpoint height where the twist begins.