A clamped strip buckles into an arch at the Euler load

A clamped strip whose ends are pushed together first shortens by compressing. Once the compressive force reaches the Euler load

4\pi^{2}B/L^{2}

, the straight strip can no longer hold it, and every further bit of shortening goes into an arch while the force stays fixed. In units of the stretchability this happens at

d/S=4\pi^{2}\approx39.5

. The deep arch of the Lab sits at 7.6 times that confinement and rises by 8.9 % of its length.

Sandbox Physics · 2026-10-30

Where the extra length goes

Lengths are measured in units of the strip length $L$ , so the clamps sit at $x=\pm\tfrac12$ and hold $w=0$ and $w'=0$ there. The ends are pushed together by $d$ , and $w(x)$ is the sideways deflection, positive upwards.

A deflected strip is longer than the straight line between its ends. A short piece of it has length $\sqrt{1+w'^{2}}\,dx\approx(1+\tfrac12w'^{2})\,dx$ when the slope is small. So the shape alone takes up $\tfrac12\int w'^{2}dx$ of the shortening, and only the rest compresses the middle surface. Hooke’s law turns that leftover strain into a compressive force, written as $\tau^{2}$ in units of $B/L^{2}$ ; this is Eq. 2 of Pandey et al. (2014) integrated along the strip.

S\,\tau^{2}=d-\frac12\int_{-1/2}^{1/2}w'^{2}\,dx

The load at which the straight strip gives way

A strip at rest under a compressive force obeys the static beam equation: the fourth derivative comes from bending, the second from the force acting on the tilted pieces. It is Eq. 1 of Pandey et al. with the time derivatives and the indenter removed.

Try the clamped shape $w=\tfrac12w_{0}(1+\cos2\pi x)$ . It satisfies $w=w'=0$ at $x=\pm\tfrac12$ , and inserting it gives $\tfrac12w_{0}\cos2\pi x\,[(2\pi)^{4}-\tau^{2}(2\pi)^{2}]=0$ . A nonzero arch can therefore exist only when $\tau^{2}=4\pi^{2}$ , the Euler load of a clamped strip, $4\pi^{2}B/L^{2}$ in ordinary units.

While the strip is straight, the whole shortening compresses it, so $\tau^{2}=d/S$ . It stays straight as long as $d/S<4\pi^{2}$ ; Pandey et al. read $d-4\pi^{2}S$ as the shortening left after the compression has done what it can.

\frac{d^{4}w}{dx^{4}}+\tau^{2}\frac{d^{2}w}{dx^{2}}=0,\qquad w=\tfrac12w_{0}(1+\cos2\pi x)\;\Rightarrow\;\tau^{2}=4\pi^{2}

How high the arch rises

Past the threshold the force stays at the Euler value, so Hooke’s law fixes the compression at $4\pi^{2}S$ and the arch must take up the rest. With $\tfrac12\int w'^{2}dx=\pi^{2}w_{0}^{2}/4$ for the arch shape, $4\pi^{2}S=d-\pi^{2}w_{0}^{2}/4$ gives the rise below.

The energy of chapter 1 tells the same story. The arch costs $4\pi^{2}d-8\pi^{4}S$ in units of $B/L$ , and the straight strip $d^{2}/2S$ . Their difference is a perfect square divided by $2S$ , so the arch is cheaper whenever it exists, and the two meet at the threshold.

w_{0}=\frac{2}{\pi}\bigl(d-4\pi^{2}S\bigr)^{1/2},\qquad U_{\text{straight}}-U_{\text{arch}}=\frac{\bigl(d-4\pi^{2}S\bigr)^{2}}{2S}

Worked example

The three arches of the Lab

The deep-arch preset has $S=7.5\times10^{-5}$ and $d=0.0225$ , so $d/S=300$ . The Euler compression is $4\pi^{2}S=2.96\times10^{-3}$ , only 13 % of the shortening. The rise is $w_{0}=\tfrac{2}{\pi}(0.0225-0.00296)^{1/2}=0.0890$ , so a strip 100 mm long arches up by 8.9 mm. Its energy, 0.830 in units of $B/L$ , is 4.07 times below the straight strip’s 3.375.

The shallow-arch preset has $S=3\times10^{-4}$ and $d=0.018$ : $d/S=60$ and $w_{0}=0.0500$ . The barely buckled preset has $d=0.0135$ : $d/S=45$ , just above 39.5, and $w_{0}=0.0259$ . There the arch saves only 1.5 % of the energy, which is why the next chapters find it the most delicate of the three.

\frac{d}{S}=300:\;w_{0}=0.0890,\qquad\frac{d}{S}=60:\;w_{0}=0.0500,\qquad\frac{d}{S}=45:\;w_{0}=0.0259

A sheet on water buckles differently

A clamped strip buckles at a load set by its own length, and after buckling the load stays put. A sheet floating on liquid buckles at a load set by its bending stiffness and the liquid, $2(BK)^{1/2}$ per unit width with $K=\rho g$ , whatever its length. Diamant and Witten (2011) showed that its pressure then falls exactly as $P=2-\Delta^{2}/16$ in those units while the wrinkles gather into one fold.

This chapter has used the small-slope beam equation. Pandey et al. compare their experiments with a Kirchhoff rod model that allows large slopes; the Lab uses the small-slope form, which is accurate while the rise stays well below the length, as it does in all three presets.

Predict, change, explain

Keep

h/L=0.06

and lower the end shortening from 1.35 % to 1.0 %. Predict

d/S

and whether the strip is still an arch.

Choose “Barely buckled: a slow snap”, then move “End shortening, percent of the length” down to 1.0 % and back up past 1.2 %.

Explain where the arch disappears using the threshold

d/S=4\pi^{2}

and the rise formula, and why the confinement readout changes in proportion to the slider.

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. A strip with $h/L=0.03$ has its ends pushed together by 0.5 % of its length. Does it buckle, and how high is the arch?

Read an explained answer

$d/S=0.005/7.5\times10^{-5}=66.7>39.5$ , so it buckles. The rise is $\tfrac{2}{\pi}(0.005-0.00296)^{1/2}=0.0287$ of the length.

2. Why does the compressive force stop growing once the strip has buckled?

Read an explained answer

The arch shape can balance only the Euler load $\tau^{2}=4\pi^{2}$ . Any extra shortening is taken up by a higher arch through $\tfrac12\int w'^{2}dx$ , so the strain of the middle surface, and with it the force, stays at $4\pi^{2}S$ .

3. Show that the arch energy $4\pi^{2}d-8\pi^{4}S$ equals the bending energy plus the stretching energy of the buckled strip.

Read an explained answer

Bending gives $\pi^{4}w_{0}^{2}=4\pi^{2}(d-4\pi^{2}S)$ after inserting the rise. Stretching gives $(4\pi^{2}S)^{2}/2S=8\pi^{4}S$ . Their sum is $4\pi^{2}d-16\pi^{4}S+8\pi^{4}S=4\pi^{2}d-8\pi^{4}S$ .

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.

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.

Where the extra length goes

The load at which the straight strip gives way

How high the arch rises

The three arches of the Lab

A sheet on water buckles differently

Predict, change, explain

Try explaining it yourself

1. A strip with h/L=0.03h/L=0.03h/L=0.03 has its ends pushed together by 0.5 % of its length. Does it buckle, and how high is the arch?

2. Why does the compressive force stop growing once the strip has buckled?

3. Show that the arch energy 4π2d−8π4S4\pi^{2}d-8\pi^{4}S4π2d−8π4S equals the bending energy plus the stretching energy of the buckled strip.

Sources to read alongside

1. A strip with $h/L=0.03$ has its ends pushed together by 0.5 % of its length. Does it buckle, and how high is the arch?

3. Show that the arch energy $4\pi^{2}d-8\pi^{4}S$ equals the bending energy plus the stretching energy of the buckled strip.