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Supplementary Information for the paper A characteristic lengthscale causesanomalous size effects and boundary programmability in mechanical metamaterials

Corentin Coulais,1, 2, 3 Chris Kettenis,2 and Martin van Hecke1, 2

1AMOLF, Science Park 104, 1098 XG Amsterdam, The Netherlands2Huygens-Kamerlingh Onnes Lab, Universiteit Leiden, PObox 9504, 2300 RA Leiden, The Netherlands

3Van der Waals Zeeman Institute, Institute of Physics, Universiteit van Amsterdam,Science Park 904, 1098 XH Amsterdam, The Netherlands

A characteristic length scale causes anomaloussize effects and boundary programmability in

mechanical metamaterials

© 2017 Macmillan Publishers Limited, part of Springer Nature. All rights reserved.

SUPPLEMENTARY INFORMATIONDOI: 10.1038/NPHYS4269

NATURE PHYSICS | www.nature.com/naturephysics 1

http://dx.doi.org/10.1038/nphys4269

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INTRODUCTION

This document contains Supplementary Materials for the main text in the following order : (I.) SupplementaryFigures S1-S4 ; (II.) Supplementary Text for the theoretical description of the rotating-squares meta-chain ; (III.)Supplementary Text for the theoretical description of the topological metamaterial.

I. SUPPLEMENTARY FIGURES

Supplementary Figure S1. Experimental determination of the stiffness and of the rotational rate. (a) CompressiveForce F vs. compressive displacement δ for a meta-chain of size N = 14 (black curve). The stiffness k is measured from thecoefficients of a 2nd order polynomial fit (red line) to the data. (b) Rotation θ(1) of the bottom n = 1 square vs. displacementδ (black curve). The rotational rate ω(1) is determined from a linear fit to the data (red line).





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Supplementary Figure S2. Anomalous stiffness size dependence in 2D and 3D mechanical metamaterials. (a) 2DMetamaterial based on the 2D rotating square mechanism (see Fig. 1 of the main text) under textured boundary conditions. (b)Stiffnesses ke and ko vs. system size N obtained by numerical simulations. The geometric design and method of simulations arethe same as for the meta-chain shown in Fig. 2. (c) 3D Metamaterial [1] under “checkerboard” textured boundary conditions.(d) Stiffnesses ke and ko vs. system size N obtained by numerical simulations. The geometric design and simulation methodfollow those described in [1], with the exception of the value of the struts width, w, which is chosen here twice as small in orderto exhibit a relatively large value of np, yet with a geometry that can realistically be fabricated.

Supplementary Figure S3. FEM Simulation protocol to characterize the hinges. (a) Bending torsional stiffness Cb. (b)Stretching stiffness kj . (c) Shear torsional stiffness Cs. The nominal applied relative displacement has a magnitude 3× 10−4 L.





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Supplementary Figure S4. Hybrid dressed mechanism. To check the applicability of the hybrid model (which we derivein detail in the Supplementary Information, ’model BSS’), we have determined the experimentally relevant values of α and βusing finite element simulations of the hinges, and solved the model using these numerical values. (a) FEM determination of thebending, stretching and shear stiffness Cb, kj and Cs. The magnitude of the imposed displacements was 3×10−4 L, and the colorencodes the ratio of local over imposed displacements (blue : 0, red 1). For the hinge parameters used here (/L = w/L = 0.1),we find Cb = 1.62× 101 N.mm, kj = 3.14× 101 N/mm, Cs = 1.82× 101 N.mm, leading to α = 3.38× 101 and β = 1.39× 102.(b-c) Corresponding stiffnesses ko and ke and rotational rates ω are in excellent agreement with our experimental data. Inparticular, we find that the length scales n∗ = 2.35 and np = 6.9 are in good agreement with the experimentally measured ones(displayed in Fig. 2 of the main text).





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II. MATHEMATICAL DESCRIPTION OF THE META-CHAIN

To capture the physics of metamaterials based on the rotating squares mechanism [2], we develop three closelyrelated models. In all, we consider infinitely rigid squares connected by flexible filaments of rest length which act ashinges (Fig. S5a). The degree of sophistication in the energetics of these filaments specifies the particular model.a. Filament Energetics The most general model results when we allow for bending, stretching and shear (model

BSS - section IIA 1). Here, the pure bending of each hinge is governed by a torsional stiffness Cb (Fig. S5b) ; thestretching of each hinge is governed by a linear stiffness kj (Fig. S5c) ; and the pure shear of each hinge is governed bya shear stiffness Cs (Fig. S5d). We note that these stiffnesses are independent and follow from the detailed geometryof the hinge filaments [3, 4]. We will also consider simplified models which only allow for bending and stretching(model BS - section IIA 2), only allow for bending and shearing (model BS’ - section IIA 3) or bending only (modelB - section IIA 4). In all cases, we focus on small deviations from the equilibrium state and linearised models.

b. Coordinates A priori, each square is characterized by three coordinates, such as their centre of mass Xn +

un (where Xn is their unforced equilibrium position and un captures their response to external forcing) and theirorientation w.r.t. the horizontal, which we take as θn+π/4 (Fig. S5e). To simplify the expressions for the energetics ofthe connecting filaments, we in addition introduce the auxiliary filament characteristics εn (the filament length is givenby (1 + εn) where is the rest length), and ψn (the angle of the filament w.r.t. the horizontal) as shown in Fig. S5e.Using these coordinates, we express the bending energy as Cb/2(θn − θn+1)

2, the stretching energy as kj/2(εn)2 and

the shear energy as Cs/2(

θn+θn+1

2 − ψn

)2

— note that latter form is consistent with rotational invariance.

Supplementary Figure S5. Models based on rigid squares connected by flexible filaments. (a) Undeformed state indicatingfilaments of rest length . (b) Pure bending caused by counter rotating squares. (c) Pure stretch caused by horizontal squaremotion. (d) Pure shear caused by vertical square motion. (e) Coordinate system and filament characteristics. The location and

orientation of each square is specified by Xn + un and θn ; from these, we can deduce the filament characteristics εn, ψn. (f)The most general boundary conditions are when a total force F is applied at the middle hinge — represented by forces F/2that are acting on the lower corner of the upper outer squares, and upper corner of the lower outer squares — and forces F ′/2are exerted at the upper corner of the upper outer squares, and lower corner of the lower outer squares. The compression ischaracterized by δ and δ′ as indicated.





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c. Determination of stiffnesses In order to capture a specific hinge geometry, the three elastic coefficients Cb, kjand Cs can be calibrated by finite element simulations (See e.g. Supplementary Figures S3-S4 and Methods) :Pure bending can be implemented via counter rotation : θn = −θn+1 while εn = ψn = 0.Pure stretching can be implemented via relative horizontal motion : εn = 0 while θn = θn+1 = ψn = 0.Pure shear can be implemented via relative vertical motion : ψn = 0 while θn = θn+1 = εn = 0.

d. Boundary conditions In the majority of cases, we consider a single force F acting on the left and right verticalhinges, where we tacitly assume that we can ignore bending of these hinges, as is small ; in the absence of bending,this is equivalent to forces of strength F/2 acting on both the lower corner of the top outer squares and the top cornerof the lower outer squares (Fig. S5f). As we discuss below, we will also consider generalised boundary conditions,where we in addition force the top of the top outer squares and bottom of the bottom outer squares by a force F ′/2(Fig. S5f). Associated with these forces are the relative compressive displacements δ and δ′ (positive δ, δ′ correspondto compression).

A. Models and equations of motion

In this section, we consider the case where F ′ is zero, so that a single force F acts on the left and right verticalhinges ; as discussed above, this is equivalent to forces of strength F/2 acting on both the lower corner of the topouter squares and the top corner of the lower outer squares.a. Kinematic constraints In Fig. S5f we show an example of the configurations of 2 × N chain in a deformed

state. Clearly, in lowest order, the vertical filaments do not experience shear or stretching, and we can simplify thekinematics by assuming that the lower corners of the top squares are all aligned with the x-axis, with the state ofthe lower squares following by mirror symmetry. In this case, θn, εn and ψn are no longer independent, and in linearorder we find :

ψn = −L(θn + θn+1)

2. (1)

In addition, we denote the deviations of the x-coordinates of the centre-of-mass of the top row of squares from theirequilibrium positions (Xn = n(L+ )) by un, where in leading order :

un+1 − un = εn . (2)

Finally, we connect the end-to-end displacement δ of the system to the internal variables via a constraint c = 0, wherec is of the form :

c = δ + uN−u1 +L

2(θN − θ1). (3)

1. Model BSS

If we include bending, stretching and shear, the elastic energy of the system can be written as

E = 2N−1∑n=1

(Cb

2(θn−θn+1)

2 +kj2(εn)

2 +Cs

2

(θn + θn+1

2−ψn

)2)

+N∑

n=1

Cb

2(2θn)

2, (4)

where the first sum corresponds to the energy of the two rows of horizontal hinges, which can both bend, stretch andshear, and the second sum to the vertical hinges, which only undergo bending. Implementing the geometric constraints,by using Eqs. (1-2) to eliminate ψn and εn, we obtain :

E = 2

N−1∑n=1

(Cb

2(θn−θn+1)

2 +kj2(un+1 − un)

2 +Cs

8

(1 +

L

)2

(θn+θn+1)2

)+

N∑n=1

Cb

2(2θn)

2 . (5)

This shows that bending and shear, in leading order, respectively correspond to antiparallel and parallel squarerotations, and also indicate how the effective coupling constant of the parallel rotations, (Cs/8)(1 + L/)2, dependson the filament shear coefficient Cs.This elastic energy has to be minimized in the presence of the geometrical constraint c = 0. We introduce the

Lagrange function :

L = E + Fc , (6)





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where F is the Lagrange multiplier corresponding to the compressive force. Mechanical equilibria are found at statio-nary points of the Lagrange function, which are given when the partial derivatives of L with respect to θn, un and Fare zero.Taking the derivatives with respect to u1, un (for n ∈ [2, N−1]) and uN yields :

u2 − u1 = −F/(2kj) , (7a)

2un − un−1 − un+1 = 0 , for n ∈ [2, N−1] (7b)

uN − uN−1 = −F/(2kj) , (7c)

from which we immediately deduce that un − u1 = −(n − 1)F/(2kj), such that the stretch εn = −F/(2kj) isindependent of the discrete coordinate n. Taking the derivative with respect to F , and using that uN − u1 = −(N −1)F/(2kj) as found above, we obtain that

(N − 1)F = 2kjδ + Lkj(θN−θ1) . (8)

Finally, taking the derivatives with respect to θ1, θn (for n ∈ [2, N−1]) and θN yields :

Cs

4Cb

(1+

L

)2

(θ1+θ2) = (θ2−3θ1) +L

4CbF (9a)

Cs

4Cb

(1+

L

)2

(θn−1+2θn+θn+1) = (θn−1−4θn+θn+1) for n ∈ [2, N−1] (9b)

Cs

4Cb

(1+

L

)2

(θN−1+θN ) = (θN−1−3θN )− L

4CbF , (9c)

Equations (8-9) can be non-dimensionalised by defining the dimensionless force F := F L4Cb

and dimensionless

displacement δ := 2δL , as well as the dimensionless parameters α := Cs

4Cb(1 + L

)2 and β :=

kjL2

4Cbwhich represent the

relative cost of shear and stretch to bending, respectively (see Eqs. (1) of the main text) :

(N − 1)β−1F = (θN−θ1) + δ. (10a)

α(θ1+θ2) = (θ2−3θ1) + F (10b)

α(θn−1+2θn+θn+1) = (θn−1−4θn+θn+1) for n ∈ [2, N−1] (10c)

α(θN−1+θN ) = (θN−1−3θN )− F (10d)

Numerical solution. — This system of equations is linear and can be solved analytically for each value of N ,yielding solutions of the form F = kδ, where the dimensionless stiffness is

k =1

(N − 1)β−1 + PN (α). (11)

However, the expressions PN (α) take a different algebraic form for every value of N and become impractically largefor large N . Therefore, we solved the equations numerically for each value of N , α and β in order to calculate thestiffness and rotational field (Fig. 3 of the main text and Supplementary Figure S4). The peak length scale np is

determined by using the location of the maximum of a quadratic fit to the numerically estimated stiffness ke vs. N inthe vicinity of the maximum value of ke. The characteristic length scale n∗ is determined by using the decay lengthof an exponential fit to the rotational field of a meta-chain of length N = 1000.Continuum limit of the bulk equation. — In order to obtain the scaling of for the characteristic length n∗ with

α, it is worthwhile to consider Eq. (10c) in the continuum limit. Assuming that the envelope of the counter-rotating

field has small gradients, we introduce a continuum field θ(x), such that θn = θ(x), and θn±1 = −θ(x ± b), where

b = L+ is the distance between two neighboring squares. Using the Taylor expansions of θ(x ± b), we find that

θn±1 = −θ(x)∓ b∂xθ(x)− b2

2 ∂xxθ(x). Substituting this into Eq. (10c), we find :

α− 1

6∂xxθ(x)− θ(x) = 0 . (12)

As a result, solutions for the continuum (staggered) field θn are linear combinations of exp(n/n∗), exp(−n/n∗),where n∗ =

√(α− 1)/6. We note that the scaling of n∗ with α is consistent with the numerical solution of Eqs. (10)

discussed in the main text (See Fig. 3a-inset and Fig. S6a) [6].





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Approximate solutions for the mechanical equilibrium of finite meta-chains can be obtained by substituting θn =A exp (−n/n∗) +B exp (n/n∗) in the boundary equations Eqs. (10b) and (10d). After a few algebraic manipulations,we find

θn =

F[(α+ 3)(ΓN−1 + 1)− (α− 1)Γ(ΓN−3 + 1)

]−1 (ΓN−n + Γn−1

)if N is even

F[(α+ 3)(ΓN−1 − 1)− (α− 1)Γ(ΓN−3 − 1)

]−1 (ΓN−n − Γn−1

)if N is odd

, (13)

where we defined Γ := exp(−1/√α−16 ) for convenience.

2. Bending and Stretch : Model BS

The BSS-model can be simplified by preventing shear interactions, to only include stretch and rotations. However,the absence of shear basically leads to a separation of stretching and rotations, and introduces an extra constraint :θn+1 = −θn — without shear there is no decay of the rotational field. We thus can write the the rotation asθn = (−1)n−1Ω and the energy reads

E = 2

N−1∑n=1

(kj2(un+1 − un)

2

)+ 2(3N − 2)CbΩ

2 , (14)

which, once combined with the constraint c = 0 yields

(3N − 2)Ω =1 + (−1)N

2F , (15a)

(N − 1)β−1F = −Ω(1 + (−1)N ) + δ. (15b)

In this simple case, the force-displacement relation F (δ) can be explicitly calculated and reads

F =δ

(N − 1)β−1 + (1+(−1)N )2

2(3N−2)

(16)

3. Model BS’

Alternatively, the BSS model can be simplified by preventing stretch interactions, and by only including bendingand shear. The energy then becomes

E = 2N−1∑n=1

(Cb

2(θn−θn+1)

2 +Cs

2

(θn + θn+1

2−ψn

)2)

+N∑

n=1

Cb

2(2θn)

2, (17)

and can be combined with the constraint c = 0 in the Lagrangian framework to obtain the following non-dimensionalequations

0 = (θN−θ1) + δ. (18a)

α(θ1+θ2) = (θ2−3θ1) + F (18b)

α(θn−1+2θn+θn+1) = (θn−1−4θn+θn+1) for n ∈ [2, N−1] (18c)

α(θN−1+θN ) = (θN−1−3θN )− F . (18d)

4. Bending only : Model B

In the simplest approach, we only consider bending interactions, and the model has only one degree of freedom(counter rotations) that carry a finite energy penalty. We can deduce this model B from model BS by taking the limitwhere β → ∞, which yields :

(3N − 2)Ω =1 + (−1)N

2F , (19a)

0 = −Ω(1 + (−1)N ) + δ. (19b)





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and

F =2(3N − 2)

(1 + (−1)N )2δ (20)

B. Asymptotics for np

We show in Fig. 3b-inset of the main text as well as in Fig. S6bc below that np/√β collapses when plotted as

function of α/β (in the limit where both α and β are large) ; here we rationalize this scaling form. In the regime whereα/β 1, Cs kj and Cs Cb, and we therefore can assume that the physics is set by a balance between stretchingand bending, with shearing forbidden. Model BS describes this situation, and from Eq. (16) we find that

ke =1

(N − 1)β−1 + 4/(6N − 4). (21)

To find the scaling of np, we require ∂N ke = 0, which yields β(6β − (2− 3N)2) = 0, or

np =√2/3

√β , (22)

which accurately describes the plateau of np for large α/β (See Fig. S6c and Inset of the Figure 3b of the main text).In the regime where α/β 1, kj Cs Cb, and we therefore can assume that the physics is set by a balance

between shear and bending, with stretching forbidden. This is the situation described in model BS’. By substitutingthe continuum limit solutions (Eq. (13)) into Eq. (18a), we can analytically express the ratio F /δ, i.e. the stiffnesses

ke =(α+ 3)

(ΓN−1 + 1

)− (α− 1)Γ

(ΓN−3 + 1

)2 (ΓN−1 + 1)

(23)

and

ko =(α+ 3)

(ΓN−1 − 1

)− (α− 1)Γ

(ΓN−3 − 1

)2 (ΓN−1 − 1)

. (24)

To find the scaling of np, we require ∂N ke = 0, which for α 1, yields in leading order to :

np =√2/3

√α , (25)

which describes the scaling of of np for small α/β (See Fig. S6c and Inset of the Figure 3b of the main text).

Supplementary Figure S6. Scaling of the lengthscales as function of the hinge constitutive parameters α and β.(a) Characteristic lengthscale n∗ vs. α for all values of β showing that n∗ is independent of β and scales as

√α for large α. (b)

Peak lengthscale np vs. α for various values of β ranging from 101 (orange) to 105 (black). (c) Rescaled version of panel (b),where np/

√β vs. α/β collapses on a master curve, indicating two asymptotic regimes : np ∼

√α for α/β 1, and np ∼

√β

for α/β 1.





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C. Hybrid mechanism model for the meta-chain with complex boundary conditions

Here we consider generalized boundary conditions, where we also force the top of the top outer squares and bottomof the bottom outer squares by a force F ′/2 (Fig. S5f). Associated with these forces are the relative compressivedisplacements δ and δ′ (positive δ, δ′ correspond to compression). We can express these as

δ = u1 − uN +L

2(θ1 − θN ) , (26)

δ′ = u1 − uN +L

2(θN − θ1) , (27)

which leads to two constraints

c = δ − u1 + uN − L

2(θ1 − θN ) , (28)

c′ = δ′ − u1 + uN − L

2(θN − θ1) , (29)

and a Lagrangian

Lh = E + Fc+ F ′c′ . (30)

The equations of motion of this coupled system follow analogously to the F ′ = 0 model BSS. Taking the derivatives∂Lh/∂ui yields again a linear gradient in un :

un − u1 = − (n− 1)(F + F ′)

2kj. (31)

Taking the derivatives ∂Lh/∂F and ∂Lh/∂F′ and using the above expression (31) for uN − u1 yields :

(N − 1)(F + F ′) = 2kjδ + Lkj(θN − θ1) , (32)

(N − 1)(F + F ′) = 2kjδ′ + Lkj(θ1 − θN ) , (33)

which generalizes Eq. (8), and supports the intuitive picture that F + F ′ plays the role of the compressive force.Taking the derivatives ∂Lh/∂θi yields the equations of motion :

Cs

4Cb

(1+

L

)2

(θ1+θ2) = (θ2−3θ1) +L

4Cb(F − F ′) (34a)

Cs

4Cb

(1+

L

)2

(θn−1+2θn+θn+1) = (θn−1−4θn+θn+1) for n ∈ [2, N−1] (34b)

Cs

4Cb

(1+

L

)2

(θN−1+θN ) = (θN−1−3θN )− L

4Cb(F − F ′), (34c)

which are straightforward extensions of the equations of motion of the BSS model (Eqs. (10b -10d), where now(F − F ′)L, instead of FL, exerts the torque.

1. Effective Stiffness

As we are dealing with two independent forces (F and F ′) and two compressive displacements (δ and δ′), there isno single definition of the effective stiffness of the ensuing structure. Physically this implies that different patternedboundaries may lead to different effective stiffnesses. A simple example of a patterned boundary that allows to tunethe relative strength of F and F ′, consider the case shown in Fig. S7, where a single global force Fg is coupled to thesquares via bars and pins. The relation between the global force Fg and the forces F and F ′ is given by

F = (1− µ)Fg , (35)

F ′ = µFg . (36)

The changes in chain length are now naturally given by δg, which is related to δ and δ′ as

δg = (1− µ)δ + µδ′ . (37)

These relations for Fg and δg allow to define the effective stiffness kg := Fg/δg, which we use to produce the np curvein Fig. 4b of the main text. We note that we recover the F ′ = 0 limit considered earlier for µ = 0. In addition we notethat in the main text we define F ′ = λF , so that λ = µ/(µ− 1).





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Supplementary Figure S7. Physical implementation of the hybrid boundary conditions. In order to realise the mecha-nical loading (F, F ′) shown in Fig. S5f, we propose a rigid hinge mechanism (depicted in red) that applies a load Fg located atthe red points located on the outermost squares. The crucial parameter is the dimensionless length µ, which allows to controla hybrid force/torque boundary condition. For µ = 0, only the central vertices are loaded (equivalent to F = Fg and F ′ = 0).For µ = 0.5, central, bottommost and topmost vertices are equally loaded (equivalent to F ′ = F ) and for µ = 1, only thebottommost and topmost vertices are loaded (equivalent to F = 0 and F ′ = Fg). Negative values of µ, or µ > 1, can be realizedby extended bars and hinges. This concrete realization provides a definite relation between the global force Fg and the forcesF and F ′, as well as a definite relation between the global compression δg and δ and δ′.

D. Equations governing the dynamic mechanical equilibrium

The lengthscale n∗ discussed above can also be uncovered via a vibrational analysis. To this end, we introduce thedynamic Lagrangian

L = E −K (38)

where the elastic energy E is as defined in the BSS model (Eq. (5)) and the kinetic energy is

K =N∑

n=1

mL2

12θ2n +mu2

n. (39)

Here ˙ denotes the time derivative and m the mass of the squares. Applying the Euler-Lagrange theorem on Eq. (38)leads to the equations

0 = θ1 + ω20 ((α+ 3)θ1 + (α− 1)θ2) (40a)

0 = θn + ω20 ((2α+ 4)θn + (α− 1)(θn−1 + θn+1)) for 1 < n < N (40b)

0 = θN + ω20 ((α+ 3)θN + (α− 1)θN−1) (40c)

and

0 = u1 +β

6ω20 (u1 − u2) (41a)

0 = un +β

6ω20 (2un − un−1 − un+1) for 1 < n < N (41b)

0 = uN +β

6ω20 (uN − uN−1) (41c)

where ω20 = 24Cb

mL2 is the characteristic rotational frequency.Infinite meta-chain : Band Diagram. — In the limit of infinite system size, these equations admit plane waves

solutions of the form un = u0 exp(i(ωt−qn)) and θn = θ0 exp(i(ωt−qn)), leading to the following dispersion relations :

(ω

ω0

)2

= 2((α+ 2) + (α− 1) cos q) (42)

(ω

ω0

)2

=β

3(1− cos q) (43)





12

Supplementary Figure S8. Vibrational analysis of the rotational modes of the meta-chain. (a) Band spectrum ωω0

vs. wave

vector q for α = 34, corresponding to the experimental value (See Extended Data Figure 4). The band-gap is depicted in purpleand the eigenfrequency of the lowest edge mode, shown in panel (b), is represented by a red line. (b) Eigenmode profiles ofthe field θn (dashed red line) and static response (solid black line) for a meta-chain with N = 14 and with α = 34. (c) Decaylengths of the mode with the smallest eigenfrequency (dashed red line) and of the static response n∗ vs. α. The data has beenproduced with a meta-chain of length N = 2000.

Eq. (42) exhibits a band-gap in the limit where the counter-rotating angles θn decay slowly, that is q → π (See

Fig. S8a). In this limit, Eq. (42) can be expanded around q = π as(

ωω0

)2

= 6(1+ α−16 (q− π)2)+O((q− π)4). Hence,

the length scale n∗ =√

α−16 emerges from the dispersion relation, in perfect agreement with the continuum limit

result for n∗ derived above.

Finite meta-chain : Vibrational analysis. — For a finite chain, we solve the system of equations Eqs. (40), and findthat a finite meta-chain with open boundary conditions has two extra modes. One of these modes has an eigenfrequencythat is very close to the bulk spectrum (See Fig. S8a). These eigenmodes are exponentially localised at the edges,qualitatively similar to the static response (See Fig. S8a). However, they exhibit a different—larger—decay lengththan n∗ (See Fig. S8b), which scales differently with α (See Fig. S8c).

To understand the difference between n∗ and the spatial decay of the surface or edge mode, we note that the staticdisplacement field u in response to external forcing F is a linear combination of all the eigenvectors ψi :

u =

N∑i=1

ψi · Fω2i

ψi, (44)

where ω2i are the eigenvalues. In some instances (e.g. in isostatic lattices), the edge mode has a much lower eigenfre-

quency than all the others modes (the surface mode could even be a zero mode) and the previous sum boils down

to u ψ1·Fω2

1ψ1. However, the edge and bulk modes have comparable eigenfrequencies here, and n∗ cannot simply be

related to the decay of a single mode.

III. MATHEMATICAL DESCRIPTION OF THE TOPOLOGICAL METAMATERIAL

In Figure 4 of the main text, we introduce a topological mechanical metamaterial (See Fig. S9a) which was previouslystudied by Coulais et al. [5]. By contrast with the previous study where only the pure mechanism was consideredunder simple boundary conditions, we here focus on the mechanical response of the mechanism-based metamaterialby taking into account the elastic hinge deformations which compete with the mechanism and by using more generic





13

boundary conditions. The system Lagrangian reads

L =1

2C

N∑n=1

δθ2n +1

2k

N∑n=1

α2n +

1

2k

N−1∑n=1

ε2n

+

N−1∑n=1

κn

(2√2 cos(θ) +

√2 ((αn + 1) (− sin (θ + δφn)) + (αn+1 + 1) sin (θ + δφn+1)

− (εn,n+1 + 1) cos (θ − δψn,n+1))− sin(−θ − δθn +

π

4

)− sin

(θ + δθn+1 +

π

4

))

+

N−1∑n=1

λn

(√2 ((αn + 1) (− cos (θ + δφn)) + (αn+1 + 1) cos (θ + δφn+1)

+ (εn,n+1 + 1) sin (θ − δψn,n+1))− cos(−θ − δθn +

π

4

)+ cos

(θ + δθn+1 +

π

4

))

+N∑

n=1

µn (−2 cos(θ) + (αn + 1) cos (θ + δφn) + cos (θ + δθn))

+√2F ((α1 + 1) sin (δφ1 + θ)− sin (δθ1 + θ)) + 2

√2F ′ (α1 + 1) sin (δφ1 + θ)

, (45)

where the quantities δθi, αi, εi,i+1 and δψi,i+1 are internal degrees of freedom of the structure. The quantities λi, µi,κi, F and F ′ are Lagrange multipliers associated to the geometric constraints.

Mechanical equilibria are found at the stationary points of the Lagrangian, therefore follow from the equations∂L/∂δθi = 0, ∂L/∂δαi = 0, and so on. After a few simple algebraic manipulations and substitutions, we find the

Supplementary Figure S9. Sketch of the mechanical metamaterial 2. a. Geometry of the chain constituted of N unit cells,characterized by their initial tilt angle θ and connected in the middle by a torsional spring C (red dots). bc. Geometry of twounit cells in undeformed (b) and deformed (c) configurations. For simplificity, in the text we work the condition a = 1, withoutloss of generality. Adapted from [5].





14

following equations

−2√22F + 2F ′

ksin(2θ) = sin(θ)

(−4

C

kδθ1 +

(4c1c2δθ2 − 4c21δθ1

))

+ (α1 − α2) (sin θ + sin(3θ) + 5 cos(θ) + cos(3θ))

(46a)

0 =4C

kδθn − 4c1c2δ(θn−1 + δθn+1) + 4(c21 + c22)δθn

+ 2((cot θ + 1)(c1 + c2)− 1)αn+1 − 2(2c1 − 3 cot θ − 2)αn−1 − 4 cot(θ)(c1 + c2)αn,(46b)

0 =4C

kδθN − 4c1c2δθN−1 + 4c22δθN + 4c2 cot(θ) (αN−1 − αN ) (46c)

and

−2√2F + 2F ′

kcsc θ =2 cot θ (−c1δθ1 + c2δθ2) + 2 csc2 θα1 − 2α2 cot

2 θ (47a)

0 =2 cot θ (c1δθn−1 − 2(c1 + c2)δθn + c2δθn+1) + 2(1 + 2 cot2 θ)αn − 2 cot2 θ (αn−1 + αn+1) (47b)

0 =2 cot(θ) (c1δθN−1 − c2δθN ) + 2αN csc2(θ)− 2αN−1 cot2(θ), (47c)

with c1 = 12 (sin(2θ) + cos(2θ) + 2) and c2 = 1

2 (− sin(2θ) + cos(2θ) + 2). To produce the results shown in figure 4d-f ofthe main text, we solve these equations numerically for C = 0.1, k = 10, θ = π/16 and varying the ratio between Fand F ′. The hybridisation of the left-localised and right-localised deformation modes occurs because the deformationfields αn and δθn are mixed.

[1] C. Coulais, E. Teomy, K. de Reus, Y. Shokef and M. van Hecke, Combinatorial design of textured mechanical metamaterials,Nature 535, 529-531 (2016).

[2] J. N. Grima and K. E. Evans, Auxetic behavior from rotating squares, J. Mater. Sc. Lett. 19, 1563–1565 (2000).[3] A.R. Day, K.A. Snyder, E.J. Garboczi and M.F. Thorpe, The elastic moduli of a sheet containing circular holes, J. Mech.

Phys. Solids 40, 1031 - 1051 (1992).[4] C. Coulais, Periodic cellular materials with nonlinear elastic homogenized stress-strain response at small strains, Intl. J.

Solids Struct. 97–98, 226–238 (2016).[5] C. Coulais, D. Sounas and A. Alu, Static non-reciprocity in mechanical metamaterials, Nature 542, 461-464 (2017).[6] We point out that n∗ can also be calculated analytically for a semi-infinite media using the Z-transform to solve the Eqs.

for the BSS model. The result is

θn =F

1 +√3√1 + 2α

(−1)n exp− n

n∗Z, (48)

where

n∗Z = log

α− 1

α+ 2−√3√1 + 2α

. (49)

The exact solution is therefore a staggered field modulated by an exponentially decreasing envelope of characteristic lengthn∗Z . In the limit where α 1, we can expand the log and find n∗

Z = ((α− 1)/6)1/2 +O(1), consistent with the continuumscaling.





a characteristic length scale causes anomalous size …10.1038...in the format provided by the...

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