Materials Science · Architected materials and metamaterials

Auxetic behaviour and mechanical lattices

Derive how corner-connected rotating units widen under tension, calculate the stiffness supplied by their hinges, and distinguish finite strain ratios from directional elastic properties and stability.

Useful background: Composites, fibres and interfaces; Elasticity, plasticity and tensile strength. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Interpret negative Poisson ratio through measured transverse strain.
Derive repeat-cell dimensions and transverse response for a stated rotating-square mechanism.
Use hinge strain energy to obtain a local positive force–displacement slope while recognising ideal-mechanism and locking limits.
Interpret a finite strain record without replacing interval or current-configuration definitions by one permanent Poisson ratio.
Separate anisotropic directional ratios from the bounds for a stable three-dimensional isotropic elastic solid, and assess what a finite test actually establishes.

Some structures widen when stretched

Imagine a patterned sheet whose internal blocks rotate as its ends are pulled apart. The sheet can widen while each block changes very little in size. The negative Poisson ratio belongs to the coupled motion of the architecture; the material making up each block need not itself be auxetic. “Geometry, architecture and effective response” introduces this separation of local and averaged behaviour.

For a simple small-strain uniaxial description, Poisson ratio ν is minus transverse strain divided by axial strain. Engineering axial strain is the length change divided by initial length; transverse strain is the width change divided by initial width. An auxetic response has negative ν: axial tension accompanies transverse expansion. Rotating or re-entrant geometric units can produce this effective behaviour.

Define x as the loading direction and y as the transverse direction. For a rectangular gauge with reference dimensions L₀ and W₀, engineering strains are ε_x=(L−L₀)/L₀ and ε_y=(W−W₀)/W₀. Write ν_{x→y}=−ε_y/ε_x for a stated secant comparison. Its numerator and denominator use the same reference state. The notation names the loading direction first; changing that direction is a new test.

A material Poisson ratio in a linear regime is defined under uniaxial stress, with the relevant transverse macroscopic stresses zero. A width held fixed by grips is a different boundary condition: the observed transverse strain can be zero because a lateral reaction is present. In a finite cellular coupon, a measured gauge ratio can also depend on cut cells, end effects and how local rotations are averaged. “Deformation and relative strain” explains why actual gauge separation matters.

Derive one connected rotation mechanism

Consider an ideal planar periodic array of identical rigid squares of side a. Nearest neighbours meet at corner hinges and rotate in opposite senses. Their centres form a rectangular grid. Each square is oriented either +φ or −φ to the x direction, in a checkerboard arrangement. The relative rotation between neighbours changes by 2δφ when their absolute orientations change by ±δφ. This sign pattern and corner connectivity are part of the model, not details that can be omitted from a sketch.

The horizontal centre spacing is p_x=a cosφ+a sinφ. These are the projections of the two square sides onto x; the corresponding vertical spacing is identical, p_y=a sinφ+a cosφ. A repeating cell spans two centre spacings in each direction and contains four squares after sharing boundary pieces. Thus L(φ)=W(φ)=2a(cosφ+sinφ). The dashed repeat box in the first figure cuts some squares; it is not the outer boundary of a gripped coupon.

Use the opening branch 0<φ<π/4, or 0°<φ<45°. Increasing φ increases both dimensions because dL/dφ=dW/dφ=J=2a(cosφ−sinφ)>0, with the derivative taken using radians. No square needs to stretch. Since L=W throughout this ideal motion, their fractional changes are equal: a finite secant from one model state to another gives ν_{x→y}=−1, and the current-configuration differential ratio −(dW/W)/(dL/L) is also −1.

This is a kinematic result, not yet a stiffness or failure prediction. A real ligament has finite width and can bend, stretch, yield, fracture or buckle out of plane. Contact may restrict rotation. Changing the connection scheme or allowing deformation of the blocks changes the permitted motion. A re-entrant lattice can instead widen by opening inward-pointing ribs; that is another mechanism and does not inherit the square formula.

Quantitative ideal rigid-square model, with the same drawing scale in both states. Each square has side 10.0 millimetres. Nearest neighbours rotate oppositely and meet at orange corner hinges; blue and violet only distinguish rotation sense. A dashed two-by-two centre-spacing periodic repeat contains four squares after sharing boundary pieces. Its dimensions are L_x=L_y=2a(cos φ+sin φ). From absolute orientation φ=15 degrees to 30 degrees, each repeat dimension rises from 24.494897 to 27.320508 millimetres, an engineering extension of 11.5355 percent from the first state. Every individual square remains rigid. The visible three-by-three patches extend beyond the periodic boxes; the dashed boxes are not external fixtures or coupon edges. This geometry alone supplies no force, stability or survival prediction.

The force comes from resisting the motion

An ideal frictionless pin with no rotational resistance supplies the geometry but no positive elastic resistance along this motion. The free mechanism can change φ without storing hinge energy. To describe a locally restoring response, add an explicitly stated hinge model about a stress-free orientation φ₀. Treat the following motion as quasi-static, with elastic hinges and no dissipative work; inertia is omitted. This does not turn the constituent modulus into the lattice modulus.

Let one hinge store U_h=½k_h(δψ)² for a small relative-angle change δψ, with positive k_h. Angles are inserted in radians; the radian is dimensionless in SI, so k_h has units N·m in this energy expression. Each repeat contains four squares with four corners each, and each hinge joins two corners: eight hinges belong to a repeat. For the uniform rotation mode, δψ=2δφ in magnitude. Summing their energies gives U_cell=8×½k_h(2δφ)²=16k_h(δφ)²=½κ(δφ)², where κ=32k_h.

For a small motion about φ₀, ΔL≈J₀δφ with J₀=2a(cosφ₀−sinφ₀). Apply an axial resultant F_x to opposite repeat faces while the transverse resultant is zero. Virtual work gives F_x dL=dU_cell. To first order, F_x≈κδφ/J₀, so the axial stiffness is k_x=F_x/ΔL=κ/J₀². It is positive when κ>0. The transverse expansion does not make the input work negative.

If the array has a stipulated uniform out-of-plane thickness h, gross axial area is hW₀. Its local nominal axial coefficient is E_app=k_xL₀/(hW₀). For this square repeat L₀=W₀, hence E_app=κ/(hJ₀²). This is an initial uniaxial response of the stipulated rotation mode, not a complete constitutive tensor or a property of the square material. Positive hinge energy establishes a restoring tendency along this mode near the unloaded reference, not stability against every possible disturbance.

As k_h tends to zero, this mode’s stiffness tends to zero while its geometric ratio remains −1. As φ₀ approaches 45°, J₀ tends to zero and the first-order conversion from angle to axial extension becomes singular. A divergent k_x from that formula is not unlimited practical stiffness: block deformation, finite ligament geometry, contact and another mode can become decisive. Beyond the stated opening branch, increased rotation no longer means increased L.

The range and direction matter

Measure strains in the relevant directions and deformation regime. A structure can be auxetic over one interval but not all strains, or in one direction but not every direction. Negative Poisson ratio does not mean negative energy consumption or unlimited stability.

A finite record needs its strain measure as well as its interval. For a common original reference, define the descriptive interval ratio ν_int^(0)=−(ΔW/W₀)/(ΔL/L₀). It differs from a cumulative secant −[(W−W₀)/W₀]/[(L−L₀)/L₀]. Neither five sampled points nor straight joins between them establish an unmeasured instantaneous derivative. At an already deformed state, a differential using current dimensions is instead −(dW/W)/(dL/L); do not silently switch denominators.

The table supplies an invented separate finite lattice gauge, not results of the rigid-square model or a performed simulation. Reference dimensions are L₀=100.0 mm and W₀=50.0 mm. It is pulled quasi-statically along x with transverse sides free of applied traction; the central gauge is tracked at the listed states. No force, unloading, repeat scatter or uncertainty budget is supplied. Read nominal geometric observations without assigning a unique mechanism, an elastic modulus or a bulk isotropic property.

At 2% axial engineering strain, transverse strain is +1% and the cumulative ratio is −0.5, matching the retained sign example. From 3% to 4%, however, width decreases from 50.55 to 50.45 mm. This interval has ν_int^(0)=+0.2 even though the final cumulative secant is −0.225. The gauge remains wider than its starting width while it contracts over the last interval. A headline negative ratio can conceal that change.

Invented separate finite-gauge observations. These are supplied length/width readings, not data for the rotating-square mechanism or a named material; transverse traction is zero on the sides, but finite gauge/end effects have not been eliminated.
State	L (mm)	W (mm)	ε_x (%)	ε_y (%)
Reference	100.0	50.00	0	0
1	101.0	50.30	1	0.60
2	102.0	50.50	2	1.00
3	103.0	50.55	3	1.10
4	104.0	50.45	4	0.90

Invented separate finite-gauge record, not a fit to the rigid-square model. With common original references L₀=100.0 millimetres and W₀=50.0 millimetres, axial engineering strain percentages 0,1,2,3,4 accompany transverse percentages 0,0.60,1.00,1.10,0.90. The top graph marks these five supplied observations and joins adjacent points only as a reading guide. The last interval contracts transversely although final width remains greater than its original value. Lower bars use minus the interval transverse engineering-strain change divided by the axial change, both with original references: −0.60,−0.40,−0.10,+0.20 across intervals 0–1,1–2,2–3,3–4. Negative bars extend downward, the final positive bar upward from zero. The final cumulative ratio is −0.225. Finite chords or these sampled intervals do not establish instantaneous derivatives, an elastic modulus, a unique microscopic mechanism or specimen stability.

Apply an elastic bound only to its stated class

For a stable three-dimensional isotropic linear elastic solid with positive finite bulk modulus K and shear modulus G, ν=(3K−2G)/[2(3K+G)]. Positivity gives −1<ν<1/2. A negative ν is allowed when 3K<2G; it does not require a negative K, a negative G or negative stored energy. For example, K=2 MPa and G=6 MPa give ν=−0.25 and E=9KG/(3K+G)=9 MPa.

The ideal planar rotating-square mechanism is not a finite, fully deformable three-dimensional isotropic solid. It constrains deformation of the blocks and does not provide a finite response to every shear mode. Its exact ν=−1 is an ideal geometric limit, not a violation of the strict bound for the different class just defined. A finite-flexure sheet requires its own directional stiffness and shear evidence.

Anisotropy also changes the interpretation. For an orthotropic plane-stress elastic model in its principal x/y axes, use ε_x=σ_x/E_x−(ν_{y→x}/E_y)σ_y and ε_y=−(ν_{x→y}/E_x)σ_x+σ_y/E_y. Orthotropic means these axes are symmetry directions. In a reversible model with a strain-energy function, the compliance matrix is symmetric: ν_{x→y}/E_x=ν_{y→x}/E_y. The two directional Poisson ratios need not be equal.

The stored energy density of the normal response is u_n=½(σ_x ε_x+σ_y ε_y). Substituting the compliance relations and completing the square gives 2u_n=(σ_x−ν_{x→y}σ_y)²/E_x+[1/E_y−(ν_{x→y})²/E_x]σ_y². Reciprocity makes the second coefficient equal to [1−ν_{x→y}ν_{y→x}]/E_y. This expression tests every combination of the two normal stresses, rather than just a favourable uniaxial example.

For this normal-strain block to have positive strain energy under every nonzero in-plane normal stress, require E_x>0, E_y>0 and 1−ν_{x→y}ν_{y→x}>0, in addition to a positive in-plane shear modulus. These are conditions for the stated finite orthotropic model, not evidence that a photographed lattice satisfies it. They allow a directional ratio below −1 when the directional moduli differ appropriately. A sign calculation alone tests none of these complete stiffness conditions.

Test the deformation mode as well as the ratio

Finite perforated-rubber specimens have shown an early ligament-bending regime, relative unit rotations and later hinge stretching as extension approaches its geometric limit. A particular triangular layout also moved out of plane before full deployment, whereas a two-dimensional model restricted it to in-plane motion. Comparing deformed shapes and the force–displacement history can therefore reveal a missing mode even when a simple in-plane mechanism is attractive. These observations concern specific finite-flexure designs, not the ideal square array derived here.

The useful test connects several observations: where the gauge lies relative to cells and grips; axial and transverse displacements; local unit/hinge rotations; any out-of-plane motion; force and unloading behaviour; and the dimensions, imperfections and rate/temperature conditions. A repeated negative gauge ratio is relevant evidence, but it does not alone establish uniform fields, isotropy, reversible response or a stable deployed state. “Loading direction, geometry and conditions” and “Elastic response and elastic modulus” develop these distinctions.

Some architectures have more than one unloaded equilibrium configuration; others return when the force is removed. That is a question about their energy landscape and constraints. It cannot be decided from negative ν alone or from a single loaded image. “Metastability and persistence” distinguishes local stability and finite retention. In the present positive quadratic hinge model, only the specified local restoring mode has been calculated; no bistability, fatigue life, toughness benefit or service allowable follows.

Worked examples

Work through the case

An idealised test gives axial strain +0.02 and transverse strain +0.01. Find ν.

ν = −εtransverse/εaxial.
ν = −0.01/0.02.

ν = −0.5 as a secant effective strain ratio for this small deformation. An incremental ratio would use changes in strain over a specified interval.

Calculate a finite opening without stretching the squares

For the ideal connected-square model, let a=10.0 mm. Change φ from 15° to 30° on the opening branch. Find the two repeat dimensions and their engineering strains referred to 15°. Does the resulting ratio mean that the squares themselves stretched by that amount?

At 15°, L₀=W₀=20(cos15°+sin15°)=24.494897 mm. The dimension is a two-spacing repeat, not the side length of a single square.
At 30°, L=W=20(cos30°+sin30°)=27.320508 mm. The change is 2.825611 mm in both directions. The geometry is exact within this rigid-unit model.
Each engineering strain is 2.825611/24.494897=0.115355, or about 11.5355%. The finite secant ratio is therefore −1. This is not a small material strain assigned to every ligament.
Every square still has side 10.0 mm. The shape change comes from hinge rotations and changing void geometry. Whether real finite hinges survive that motion or remain in plane requires further mechanics and evidence.

Both repeat dimensions increase from about 24.495 to 27.321 mm, giving equal 11.5355% engineering strains and ν=−1 in this ideal mechanism.

A negative Poisson ratio with positive work

At φ₀=15° use a=10.0 mm, h=1.00 mm and per-hinge k_h=6.25×10⁻⁴ N·m in the stated eight-hinge model. For a small δφ=0.0100 rad, calculate initial k_x, E_app and first-order axial extension/force. Check stored work.

Convert a=0.0100 m and h=0.00100 m. J₀=2(0.0100)(cos15°−sin15°)=0.0141421 m per unit radian. The aggregate angular coefficient is κ=32k_h=0.0200 N·m.
k_x=κ/J₀²=0.0200/0.000200=100 N/m=0.100 N/mm. E_app=k_x/h=100000 Pa=0.100 MPa because L₀=W₀. Specimen/cell stiffness and apparent modulus have different units.
First-order ΔL≈J₀δφ=0.000141421 m=0.141421 mm, with ε_x≈0.00577350. The transverse fractional extension is equal, while F_x≈k_xΔL=0.0141421 N.
U_cell=½κ(δφ)²=1.00×10⁻⁶ J. The linear external work ½F_xΔL gives the same positive result. Gross volume hL₀W₀=6.00×10⁻⁷ m³ gives energy density 1.66667 J/m³, also ½E_app ε_x².
These are initial linearised force/extension values. Evaluating the exact geometric length change gives 0.140194 mm instead; finite rotation introduces a higher-order correction. Positive quadratic hinge energy gives local resistance along the stipulated mode, not an all-mode stability or failure proof.

k_x=100 N/m and E_app=0.100 MPa. First-order extension is 0.141421 mm at force 0.0141421 N, storing positive work 1.00 μJ.

An auxetic headline can miss a late contraction

Use the separate finite-gauge table. Calculate the cumulative secant at state 4 and the common-original-reference interval ratios over 0–1, 1–2, 2–3 and 3–4. What does the final interval change?

At state 4, ε_x=4/100=0.0400 and ε_y=0.45/50=0.00900. The cumulative secant is −0.00900/0.0400=−0.225.
Each interval increases L by 1.00 mm, so ΔL/L₀=0.0100. Width changes are +0.30, +0.20, +0.05 and −0.10 mm. Divide each by W₀=50.0 mm before forming the ratios.
The resulting ν_int^(0) values are −0.60, −0.40, −0.10 and +0.20. The last width change is a contraction even though W at state 4 exceeds its original value.
The finite intervals are supplied observations, not a smooth measured derivative. Without force/unloading, uncertainty and local-mode evidence, neither an elastic tensor nor the cause of the sign change has been established.

Final cumulative ratio −0.225; common-reference interval ratios −0.60, −0.40, −0.10, +0.20. The last interval is nominally non-auxetic in this specific gauge measure.

A directional ratio below minus one can be admissible

A separate invented orthotropic plane-stress model has E_x=10.0 MPa, E_y=1.00 MPa, ν_{x→y}=−1.50 and G_xy=2.00 MPa. Find the reciprocal directional ratio and check the stated in-plane positive-energy conditions. What strains follow under σ_x=0.0200 MPa, σ_y=0?

Symmetry requires ν_{y→x}=ν_{x→y}E_y/E_x=−0.150. Substituting −1.50 for both ratios would violate the stated reciprocity when E_x and E_y differ.
E_x,E_y and G_xy are positive. The normal-block product is (−1.50)(−0.150)=0.225<1, satisfying the stated determinant condition. Equivalently, the normal compliance coefficients S_11=0.100,S_22=1.00 and S_12=S_21=0.150 in MPa⁻¹ give determinant 0.0775 MPa⁻²>0.
With only σ_x present, ε_x=0.0200/10.0=0.00200 and ε_y=−(−1.50)(0.0200)/10.0=0.00300. The directional ratio is −1.50, while the stored work density ½σ_x ε_x=20.0 J/m³ is positive.
The 3D isotropic bound does not apply to this specified anisotropic plane-stress model. This is an admissibility check of supplied coefficients, not a measurement of the rotating-square array or a proof that any real coupon is homogeneous and orthotropic.

ν_{y→x}=−0.150. The stipulated finite in-plane model passes its positive-energy checks and gives strains +0.00200 and +0.00300 under the specified uniaxial stress.

Try the reasoning

Would transverse contraction under axial tension give negative ν?
Show the answer and reasoning
No.
The minus sign then produces a positive ratio.
Can geometry produce auxetic response with ordinary constituents?
Show the answer and reasoning
Yes.
Rotation or bending can change effective strains.
In the same local square model, double every k_h while keeping a,h and φ₀ fixed. What happens to the initial modulus, force at a fixed small extension, and geometric Poisson ratio?
Show the answer and reasoning
The modulus and force at that extension both double; the kinematic ratio remains −1.
κ=32k_h and k_x=κ/J₀² change with hinge resistance. The equality L(φ)=W(φ) is geometric and has not changed. This does not describe a real material treatment that changes only hinge resistance.
A calculation takes the initial hinge formula increasingly close to φ₀=45° and reports unlimited useful stiffness. Which step becomes singular, and what information is missing?
Show the answer and reasoning
J₀=dL/dφ tends to zero, so the linear conversion ΔL≈J₀δφ becomes singular. The extrapolated stiffness is not an established useful property.
Check exact geometry, contact, finite block/hinge deformation, out-of-plane modes and the load/constraint-dependent stability. The stated opening branch and local approximation cannot qualify arbitrary finite motions near its limit.
For the table’s 3–4 interval, use the dimensions at state 3 as a new reference: L_r=103.0 mm,W_r=50.55 mm. Calculate that finite interval ratio. Is it an independently measured instantaneous Poisson ratio?
Show the answer and reasoning
−[(−0.10)/50.55]/[1.00/103.0]=+0.203759, rather than +0.200 with the original reference. It is another finite ratio, not an instantaneous derivative.
The strain denominators changed. Their positive scale factors preserve the sign here, while changing the numerical value. State the reference and strain measure; five points alone do not supply continuous local derivatives.
The same patterned sheet is prevented from changing width during an axial test, and a report finds −ε_y/ε_x=0. Does this prove that the freely responding architecture is not auxetic?
Show the answer and reasoning
No. Width restraint changes the boundary condition and can supply a transverse reaction.
A material Poisson ratio assumes the relevant transverse stress is zero. Determine the strain field in an appropriate free-transverse gauge and account for finite end effects; a constrained zero ratio is not the same test.
A claimed stable 3D isotropic linear elastic material has E>0 and ν=−1.20. Can these values be accepted in that class simply because some anisotropic models have ratios below −1?
Show the answer and reasoning
No. They violate the stated positive-finite-modulus 3D isotropic range. Anisotropic examples do not validate this isotropic claim.
With the isotropic relation G=E/[2(1+ν)], the proposed pair would require G<0. Re-examine class, direction, boundary/strain definition and evidence rather than transferring an anisotropic exception without its assumptions.
For the invented orthotropic model, apply σ_y=0.00200 MPa with σ_x=0. Find ε_y and ε_x, and compare the ratio with the x-loaded case.
Show the answer and reasoning
ε_y=0.00200 and ε_x=0.000300, so ν_{y→x}=−0.150, not −1.50.
Use the directional compliance and zero transverse stress. The two moduli differ by a factor of 10; reciprocity relates the ratios through the moduli rather than forcing them equal.
A finite array widens under tension in one central image, but its force drops and some hinges leave the plane. Does the image alone validate the rigid planar mechanism or show a stable released shape?
Show the answer and reasoning
No. It documents one loaded configuration; additional mode, force and unloading evidence is needed.
Out-of-plane motion violates a planar constraint, and a force drop can have several structural causes. Neither auxetic sign nor an initial hinge model establishes bistability, repeatability, an entire constitutive law or survival after release.

What to carry forward

Use the strain definition and sign convention.
Attach direction and range to auxetic claims.
Derive auxetic response from actual connectivity and permitted motion; a geometric ratio is not a stiffness or strength.
Positive hinge energy can coexist with negative Poisson ratio; ideal zero-resistance or locking limits need separate care.
Keep finite secant, reference-interval and current differential definitions distinct, together with loading direction and transverse restraint.
The bound −1<ν<1/2 belongs to positive-finite-modulus 3D isotropic linear elasticity; directional anisotropic and constrained planar models are different cases.

References

Teaching resources and research supporting this lesson’s explanations and examples.