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Geometric elasticity · A course with experiments

L3 · Chapter 7

Kresling folds and their stress-free shapes

A Kresling cell is two polygons joined by triangles, and Yasuda et al. (2017) replace its paper panels with springs along the creases. Each crease length depends on both the height and the twist, so more than one shape can put every crease back at its natural length. Working this out, a second such shape with every crease at its natural length exists at a twist of π−2θ0\pi-2\theta_{0} whenever the height formula gives a real number. Such a cell is bistable; without that shape it springs back, or collapses until its plates touch. At θ0=90∘\theta_{0}=90^{\circ} the two shapes merge and the first push meets no stiffness.

A tube of springs

Fold a strip of paper into a ring of triangles and close it, and you get a tube that shortens as it twists. Yasuda et al. call it triangulated cylindrical origami. Paper panels warp and wear as the tube folds, so they keep only the creases: nn valley creases of length aa and nn mountain creases of length bb, each a linear spring of stiffness kk, between two nn-sided polygons on a circle of radius RR.

The cell has two coordinates: the compression uu from the starting height h0h_{0}, and the extra twist ϕ\phi of the top plate from the starting angle θ0\theta_{0}. Each crease is the hypotenuse of a right triangle whose legs are the height and a chord of the circle, which gives Eq. 1 of the paper’s Methods. The stored energy is the sum of the spring energies.

a2=(h0−u)2+4R2sin⁡2α,b2=(h0−u)2+4R2sin⁡2β,α,β=θ0+ϕ2∓π2n,U=nk2[(a−a0)2+(b−b0)2]a^{2}=(h_{0}-u)^{2}+4R^{2}\sin^{2}\alpha,\quad b^{2}=(h_{0}-u)^{2}+4R^{2}\sin^{2}\beta,\quad \alpha,\beta=\frac{\theta_{0}+\phi}{2}\mp\frac{\pi}{2n},\quad U=\frac{nk}{2}\bigl[(a-a_{0})^{2}+(b-b_{0})^{2}\bigr]

The second stress-free shape

A shape with no stored energy needs a=a0a=a_{0} and b=b0b=b_{0} at once. Subtracting the two crease formulas removes the height: b2−a2=4R2sin⁡(α+β)sin⁡(π/n)b^{2}-a^{2}=4R^{2}\sin(\alpha+\beta)\sin(\pi/n). This must keep its starting value, so sin⁡(α+β)\sin(\alpha+\beta) must too. Besides the start, the only way is α+β=π−θ0\alpha+\beta=\pi-\theta_{0}, a total twist of π−θ0\pi-\theta_{0} and an extra twist ϕ2=π−2θ0\phi_{2}=\pi-2\theta_{0}.

Putting this twist back into either crease gives the height H2H_{2}. The shape exists only if H22>0H_{2}^{2}>0. Zero is the lowest energy the springs can have, so the shape is then a second minimum: the cell is bistable, and geometry alone holds it shut.

ϕ2=π−2θ0,H22=h02+4R2[sin⁡2(θ02−π2n)−cos⁡2(θ02+π2n)]\phi_{2}=\pi-2\theta_{0},\qquad H_{2}^{2}=h_{0}^{2}+4R^{2}\Bigl[\sin^{2}\Bigl(\frac{\theta_{0}}{2}-\frac{\pi}{2n}\Bigr)-\cos^{2}\Bigl(\frac{\theta_{0}}{2}+\frac{\pi}{2n}\Bigr)\Bigr]

Three kinds of cell

Yasuda et al. build cells with R=90 mmR=90\ \mathrm{mm} and n=5n=5. With (h0,θ0)=(150 mm,40∘)(h_{0},\theta_{0})=(150\ \mathrm{mm},40^{\circ}) their energy along the valley has two minima; with (90 mm,46∘)(90\ \mathrm{mm},46^{\circ}) it only rises. The formula agrees: the first has H2=49.2 mmH_{2}=49.2\ \mathrm{mm}, the second H22=−10108 mm2H_{2}^{2}=-10108\ \mathrm{mm^{2}}, so no second shape.

A third outcome sits outside the formula. With no second shape the energy may still fall all the way down the valley, and then the cell collapses until its plates meet. The springs alone would let the plates pass through each other into a mirror cell, which no real tube can do. The Lab therefore adds an assumption of its own, outside the paper’s truss: a stiff contact that keeps the plates 0.1 R, 9 mm, apart.

A start with no stiffness

Hold the twist free and push. The cell’s first resistance is the axial stiffness at the start, the ratio of the energy’s Hessian determinant to its twist entry. Writing the crease derivatives out, the numerator is proportional to (sin⁡2α0−sin⁡2β0)2(\sin2\alpha_{0}-\sin2\beta_{0})^{2}, which vanishes when α0+β0=π/2\alpha_{0}+\beta_{0}=\pi/2, that is at θ0=π/2\theta_{0}=\pi/2, whatever h0h_{0}, RR, nn or kk. Yasuda et al. state that this zero-stiffness mode does not depend on kk or h0h_{0} and give the proof in their supplement.

The second shape explains it. At θ0=90∘\theta_{0}=90^{\circ}, ϕ2=0\phi_{2}=0 and H2=h0H_{2}=h_{0}: the two stress-free shapes coincide, and the energy near the start grows as the fourth power of the compression. At 92∘92^{\circ} they are 3.2 mm and 4∘4^{\circ} apart, and the stiffness is 6.86×10−4 nk6.86\times10^{-4}\,nk, 0.112 % of the monostable cell’s 0.610 nk and 0.080 % of the bistable cell’s 0.853 nk. The paper reports 0.26 % and 0.18 % for its prototypes, which are built with a calibration angle of 9.7∘9.7^{\circ}; both pairs of numbers are tiny.

kaxialnk=(aubϕ−aϕbu)2aϕ2+bϕ2 ∝ (sin⁡2α0−sin⁡2β0)2=0atθ0=π2\frac{k_{\mathrm{axial}}}{nk}=\frac{\bigl(a_{u}b_{\phi}-a_{\phi}b_{u}\bigr)^{2}}{a_{\phi}^{2}+b_{\phi}^{2}}\ \propto\ \bigl(\sin2\alpha_{0}-\sin2\beta_{0}\bigr)^{2}=0\quad\text{at}\quad\theta_{0}=\frac{\pi}{2}

Folding until a sheet meets itself

Diamant and Witten (2011) meet the same ending in a different sheet. Their film floating on a liquid buckles into a fold that deepens as it is pushed. Its exact shape keeps working past the point where the fold touches itself, at a pressure of about 0.040 and an end displacement of about 5.6 in their units, but beyond it the solution passes through itself and is unphysical.

In both cases the elastic energy alone does not know that a sheet cannot cross itself. A model has to add that rule, and where the motion ends then depends on it.

Worked example

The cell that clicks shut

Take h0=150 mmh_{0}=150\ \mathrm{mm}, θ0=40∘\theta_{0}=40^{\circ}, n=5n=5, R=90 mmR=90\ \mathrm{mm}. Then π/2n=18∘\pi/2n=18^{\circ}, and the creases start at a0=(1502+32400sin⁡22∘)1/2=150.13 mma_{0}=(150^{2}+32400\sin^{2}2^{\circ})^{1/2}=150.13\ \mathrm{mm} and b0=(1502+32400sin⁡238∘)1/2=186.50 mmb_{0}=(150^{2}+32400\sin^{2}38^{\circ})^{1/2}=186.50\ \mathrm{mm}.

The bracket in H22H_{2}^{2} is sin⁡22∘−cos⁡238∘=0.00122−0.62096=−0.61974\sin^{2}2^{\circ}-\cos^{2}38^{\circ}=0.00122-0.62096=-0.61974, so H22=22500−20080=2420 mm2H_{2}^{2}=22500-20080=2420\ \mathrm{mm^{2}}, H2=49.2 mmH_{2}=49.2\ \mathrm{mm}, and the top plate turns a further 180∘−80∘=100∘180^{\circ}-80^{\circ}=100^{\circ}. At the total twist of 140∘140^{\circ}, 32400sin⁡252∘+2420=22539=a0232400\sin^{2}52^{\circ}+2420=22539=a_{0}^{2} and 32400sin⁡288∘+2420=34781=b0232400\sin^{2}88^{\circ}+2420=34781=b_{0}^{2}. In the Lab the gripper lets go at a compression of 0.233 h0=34.9 mm0.233\,h_{0}=34.9\ \mathrm{mm}, where holding would need a pull, and the cell settles at 49.2 mm and 100∘100^{\circ}.

H2=(22500−32400×0.61974)1/2=49.2 mm,ϕ2=100∘H_{2}=\bigl(22500-32400\times0.61974\bigr)^{1/2}=49.2\ \mathrm{mm},\qquad \phi_{2}=100^{\circ}

An experiment to think through

Predict, change, explain

  1. Before you press: will a cell with h0=140 mmh_{0}=140\ \mathrm{mm} and θ0=40∘\theta_{0}=40^{\circ} click shut, spring back or fold flat? Use the condition H22>0H_{2}^{2}>0.
  2. In the Kresling Lab play “Bistable cell clicks shut”, “Monostable cell springs back” and “Almost no stiffness at first”. Then set the height to 140 mm with the twist at 40° and press the top plate yourself.
  3. Explain each ending with H22H_{2}^{2}, and why the 92° cell barely resists the first push although it is stiff further in.

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 θ0=40∘\theta_{0}=40^{\circ}, n=5n=5 and R=90 mmR=90\ \mathrm{mm}, what is the shortest cell that is bistable?

Read an explained answer

Bistability needs h02>32400×0.61974=20080 mm2h_{0}^{2}>32400\times0.61974=20080\ \mathrm{mm^{2}}, so h0>141.7 mmh_{0}>141.7\ \mathrm{mm}. The 145 mm preset is just above it, with H2=30.7 mmH_{2}=30.7\ \mathrm{mm}; at 140 mm there is no second shape and the Lab cell collapses onto its contact.

2. Show that the (90 mm, 46°) cell has no second stress-free shape.

Read an explained answer

π/2n=18∘\pi/2n=18^{\circ}, so the bracket is sin⁡25∘−cos⁡241∘=0.0076−0.5696=−0.5620\sin^{2}5^{\circ}-\cos^{2}41^{\circ}=0.0076-0.5696=-0.5620 and H22=8100−18208=−10108 mm2H_{2}^{2}=8100-18208=-10108\ \mathrm{mm^{2}}. No real height restores both creases, so the energy rises along the valley and the cell springs back.

3. Keep (150 mm,40∘)(150\ \mathrm{mm},40^{\circ}) but use hexagons, n=6n=6. Is the cell still bistable?

Read an explained answer

Now π/2n=15∘\pi/2n=15^{\circ} and the bracket is sin⁡25∘−cos⁡235∘=−0.6634\sin^{2}5^{\circ}-\cos^{2}35^{\circ}=-0.6634, so H22=22500−21495=1005 mm2H_{2}^{2}=22500-21495=1005\ \mathrm{mm^{2}} and H2=31.7 mmH_{2}=31.7\ \mathrm{mm}. Still bistable, but it folds flatter, because the bistability threshold rises to 146.6 mm.

Sources to read alongside

  1. 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.

  2. Yasuda, Tachi, Lee & Yang, Origami-based tunable truss structures for non-volatile mechanical memory operation, Nature Communications 8, 962 (2017), arXiv:1702.05402v1

    Main text and Methods read: the truss model with its two crease lengths and the stored energy (Methods, Eqs. 1 and 2), the four prototypes (90 mm, 46°), (150 mm, 40°), (140 mm, 92°) and (119 mm, 0°) with a polygon radius of 90 mm and a calibration angle of 9.7°, the zero-stiffness cell with a right-angle starting twist, the stiffness ratios 0.26 % and 0.18 %, and springs of 3.32 kN/m. The supplementary information was not read.

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