Geometry, architecture and effective response

Geometry can change the apparent material

A lattice or cellular solid uses arranged struts, plates or shells. Bending, stretching, rotation and local instability can give an effective bulk response unlike a dense block of its constituent. The architecture is a physical part of the material system.

Imagine two open structures made from the same elastic solid. Under one load, the first mostly lengthens its members; the second bends them. Their external displacement can differ greatly even when the solid’s Young’s modulus E_s is unchanged. Voids do not soften the atomic bonds. They change the amount of material and the paths through which it stores elastic energy.

Topology means which members join which nodes; geometry also includes their directions, lengths, sections and joint shape. A stable, triangulated load path can resist a motion through axial member forces. A jointed mechanism can instead move through rotations with almost no axial extension. Real joints may resist that motion by bending. Counting connections alone does not prove rigidity in every direction, and a drawing without specified joints and supports is not a stiffness calculation.

This lesson develops an initial, quasi-static, small-strain mechanical response. It treats a direction and boundary condition at a time. Auxetic deformation, waves and reconfiguration are separate mechanisms addressed in the following lessons; an initial modulus does not determine their complete behaviour.

Convert a specimen response into a defined average

Let a prismatic specimen have initial gross cross-sectional area A₀ and gauge length H₀. Gross means the full external envelope, including the voids inside it. For a uniform axial test, define nominal averaged stress σ̄_z = F/A₀ and engineering axial strain ε̄_z = ΔH/H₀. In a reversible linear range, the apparent normal slope is E_app = Δσ̄_z/Δε̄_z = (H₀/A₀)(ΔF/ΔH). Its units are Pa. Specimen stiffness ΔF/ΔH instead has units N/m and depends on external dimensions.

An effective Young’s modulus in direction z requires the corresponding uniaxial stress condition: transverse macroscopic stresses are zero, so the structure can contract or expand laterally. If lateral strain is blocked, the measured normal slope is a constrained modulus, not that same Young’s modulus. End friction, grips and the displacement gauge can change an apparent test slope. Machine travel that includes fixture compliance is not automatically specimen extension.

Define relative density φ = ρ*/ρ_s = V_solid/V_gross for a single constituent of uniform density ρ_s, negligible pore-gas mass and a specified gross volume. The apparent density ρ* includes the void volume; skeletal density does not. Added skins, heavy joints, trapped material or a second constituent must be included consistently. A strut’s internal porosity can also change its own effective E_s, rather than merely the open architecture’s φ.

A spatial average is not the stress in every member. The gross axial load divided by gross area can be small while a thin member carries a much larger local stress. Likewise, an averaged strain can combine member stretch, bending and rotation. An effective description replaces that detailed field only for the stated response and scale.

Derive the different cost of stretching and bending

Use a family of three-dimensional open networks with fixed topology and a fixed number of slender members per repeating volume. Let member length scale with ℓ, square section width be t, and cell volume scale with ℓ³. Neglect the overlapping joint volume in this slender-member approximation. Then solid volume scales as t²ℓ, so φ = C_ρ(t/ℓ)², where C_ρ is a dimensionless geometric constant. This relation changes if the topology, section shape or amount of joint material changes.

An axial elastic member has stiffness k_a = E_s t²/ℓ. A uniform slender cantilever with the same square section has transverse tip stiffness k_b = 3E_sI/ℓ³, where I = t⁴/12 is the second moment of area about the bending axis. The cantilever is clamped at one end and loaded by a transverse force at its otherwise free end; its tip can rotate. Different end conditions change the prefactor. Both expressions assume linear elastic material behaviour, small deflection and no dominant shear deformation.

To obtain a scaling rather than a falsely exact cell solution, suppose the relevant member displacement scales with the macroscopic strain times ℓ. Axial elastic energy per member then scales as k_a(ε̄ℓ)², or E_s t²ℓ ε̄². Divide by cell volume ℓ³ and compare with an effective energy density proportional to E* ε̄². The result is E*_stretch/E_s proportional to (t/ℓ)², hence proportional to φ.

For bending, the corresponding energy scales as k_b(ε̄ℓ)², or E_s t⁴ ε̄²/ℓ. Dividing by ℓ³ gives E*_bend/E_s proportional to (t/ℓ)⁴, hence proportional to φ². Bending can therefore lose stiffness faster as a fixed three-dimensional slender architecture is made lighter. The constants still depend on directions, nodes, geometry and boundary conditions; these powers alone do not calculate a particular lattice’s modulus.

The assumptions matter. An extruded thin-wall honeycomb has a different solid-volume scaling: for fixed in-plane shape, φ can scale with t/ℓ rather than its square. Its in-plane bending modulus can consequently scale with φ³. Closed faces, plate or shell deformation, contact, a mechanism or dense joints can also change the argument. “Bending dominated” is a mechanism in a specified model, not a universal exponent for every architecture.

Calculate one stretching network without changing its constituent

Now calculate a more explicit ideal model. Assume a stable periodic stretching network whose nodes follow the imposed homogeneous small strain: this is an affine motion, with no additional internal relaxation. There are six equally weighted axial-member direction families: the two face diagonals in each of the xy, xz and yz planes. Each family contains one sixth of the solid volume. Direction vectors, up to reversal, are (1,1,0), (1,−1,0), (1,0,1), (1,0,−1), (0,1,1) and (0,1,−1), each divided by √2.

This direction model idealises the axial load paths of an octet-type network. Its stability and connectivity are stipulated; the list of directions alone does not establish them. All members have the same E_s, joints contribute negligible volume and compliance, there is no prestress, and only normal strains with zero shear are considered here. Member bending, buckling and local material nonlinearity are excluded from this initial response.

For a unit direction n, the axial member strain is ε_m = n_x²ε̄_x + n_y²ε̄_y + n_z²ε̄_z when shear is zero. Thus the two xy families each have strain (ε̄_x + ε̄_y)/2; similarly for xz and yz. Weight each member’s elastic energy density, E_s ε_m²/2, by its fraction φ/6 of gross volume. The resulting gross energy density is u = (E_sφ/24)[(ε̄_x + ε̄_y)² + (ε̄_x + ε̄_z)² + (ε̄_y + ε̄_z)²]. Void space stores no elastic energy in this model.

The macroscopic normal stress conjugate to ε̄_x is the derivative of u with respect to that strain: σ̄_x = (E_sφ/12)(2ε̄_x + ε̄_y + ε̄_z). The y and z formulas follow by exchanging indices. This use of energy keeps the member and macroscopic work consistent. It is not an arithmetic average of the constituent modulus and zero.

Under uniaxial z stress with free lateral response, set σ̄_x = σ̄_y = 0. The two equations give ε̄_x = ε̄_y = −ε̄_z/3. Substituting into σ̄_z gives σ̄_z = (E_sφ/9)ε̄_z, so E*_z = E_sφ/9. The factor 1/9 comes from these directions and this stress condition. It is not the factor for every stretching lattice or every loading direction.

If lateral strain is instead blocked, ε̄_x = ε̄_y = 0. The same energy gives σ̄_z = (E_sφ/6)ε̄_z, a constrained normal coefficient M*_z = E_sφ/6. Lateral reaction stresses are then nonzero. M*_z/E*_z = 1.5: the measured normal slope can rise by 50% purely through restraint. Cubic-looking symmetry does not justify using isotropic formulas to infer every shear or off-axis property; only the stated normal response has been calculated.

The local strains also differ from the average. In the free-lateral uniaxial case, the xy families have strain −ε̄_z/3 and the xz and yz families have +ε̄_z/3. Some members are therefore in compression even during macroscopic tension. This is one reason why a positive averaged tensile modulus cannot exclude a local buckling limit.

Compare the models at a stated density and response

For the table and graph, take an invented constituent with E_s = 90.0 GPa and ρ_s = 4500 kg/m³. Compare the free-lateral affine model E*_z/E_s = φ/9, its constrained coefficient M*_z/E_s = φ/6, and an illustrative bending model E*_b/E_s = C_bφ² with C_b stipulated as 1.00. The bending coefficient is supplied for this teaching comparison, not measured here or fixed by all foams. The plotted interval is φ = 0.010 to 0.100, with each model assumed to remain in its stated initial elastic regime.

At φ = 0.050, apparent density is 225 kg/m³. The free-lateral stretching modulus is 500 MPa, the constrained coefficient is 750 MPa, and the bending-model modulus is 225 MPa. Comparing 750 with 225 as if both were uniaxial Young’s moduli would mix boundary conditions. At matched free lateral conditions, the supplied stretching model is about 2.22 times as stiff as the bending model at this φ.

Within these models, halving φ halves E*_z but quarters E*_b. The specific stiffness E*/ρ* therefore remains E_s/(9ρ_s) for the stretching model and equals C_b(E_s/ρ_s)φ for the bending model. This is an axial material comparison, not a guarantee of minimum component mass: a beam, panel or component with constrained space has its own geometry and load requirements.

Uniformly scaling every member and cell length by the same factor keeps t/ℓ, φ and these classical normalized initial moduli unchanged, provided E_s, joint similarity and the model assumptions persist. Merely changing ℓ while leaving t fixed does not. Surface effects, manufacturing resolution, member porosity or a different joint fraction can also spoil geometric similarity. Dynamic resonances and wave wavelengths introduce additional scales and require their own response model.

An apparent crossing of these simple laws outside the displayed interval is not a newly discovered material limit. The slender-member and fixed-geometry assumptions fail as the architecture becomes dense; substituting φ = 1 does not turn either reduced network into the homogeneous solid. Compare valid models and actual observations over their supported regimes.

An initial effective modulus does not remove member limits

A compressive member can lose stability while its constituent remains elastic. For a perfectly straight, uniform, slender pin-ended member of length ℓ, the first Euler load is P_cr = π²E_sI/ℓ². Its local mean compressive stress at that load is P_cr/A. End restraint changes the critical load; crookedness, eccentric loading and joint compliance can change the practical response. This formula is a bifurcation prediction for the ideal member, not an allowable load for a fabricated lattice.

Compare that local limit with local yielding or fracture, not just with the averaged specimen stress. The full structure can also buckle as a component even while individual members remain stable. A topology that is efficient in initial axial stiffness need not maximise damage tolerance, energy absorption, fatigue life or robustness to a missing member. Post-yield localization and redistribution require additional analysis and evidence.

An initial slope and a later collapse are different parts of the evidence. Similar initial stiffness can coexist with different post-yield localization and redistribution. For a real comparison, retain constituent and member-scale tests, actual dimensions and density, joint/defect characterization, local deformation observations and the specimen’s complete reversible or irreversible response. A good match to one initial slope does not validate every field or failure mechanism.

Effective means averaged at a specified scale

An effective modulus relates averaged loading and deformation for a suitable representative structure. It depends on relative density, topology, dimensions and boundaries. The constituent material still supplies its own local strength, stiffness and loss limits.

To replace a resolved architecture by a continuum, choose a region large enough to represent its cells but small compared with the length over which the component-scale field changes. A representative volume element (RVE) is intended to support that averaged response at the chosen precision. Homogenisation connects averaged stress and strain while keeping microscopic and macroscopic work consistent. In a mechanical power statement, the volume average of local stress times local strain rate must match the macroscopic stress–strain-rate work under suitable boundary conditions. Separate averages cannot simply be multiplied without that justification.

A periodic calculation imposes compatible repeating displacement fluctuations and balancing face tractions, the boundary forces per area. The fluctuation is the local departure from the imposed macroscopic linear displacement field. A finite coupon with gripping, cut cells and free edges is a different boundary-value problem. A few-cell modulus can consequently be a specimen response rather than a boundary-insensitive bulk value. Finite-volume estimates can depend on scale and boundary prescription under explicit material and statistical assumptions. No universal minimum number of cells follows for every lattice.

The following separate invented computational-study record compares two microscopic boundary prescriptions on increasing cubic domains, N cells per edge, of the same architecture and constituent. For each prescription the full normal stiffness/compliance response is assumed to have been obtained, and the table reports the resulting uniaxial Young’s coefficient with zero transverse macroscopic stress. A imposes compatible affine boundary displacements; B imposes uniform macroscopic-traction-type boundary loading. The values are supplied illustrative outputs, not simulations performed here and not measurements of the six-family model.

Both columns therefore target the same macroscopic response definition. Their difference illustrates finite-domain boundary sensitivity; it is not the 50% difference between lateral restraint and free contraction derived earlier. The record alone does not establish a variational bound for a real void geometry or prove representative sampling. Details of boundary application, mesh/member convergence, geometric realization and any local instability would still need checking.

Use a declared descriptive screen to read this record: the difference between A and B divided by their midpoint must be at most 5%. At N = 16, the gap is 31.25 MPa and the midpoint 503.125 MPa, giving 6.21%, so the screen is not met. At N = 32, the corresponding result is about 3.12%, so it is met in the supplied record. This screen is a chosen comparison tolerance, not an uncertainty estimate, a statistical confidence level or proof that every field is representative.

An effective-property claim should therefore state the constituent state, topology, relative-density basis, cell dimensions, direction, loading regime, specimen size, boundary prescription and measured output. Changing those conditions is a new transfer question. A scalar modulus is useful because it compresses a justified response, not because the local architecture has ceased to matter.

Worked examples

Try the reasoning

1. Does effective modulus describe every individual strut stress? Use the free-lateral six-family case with ε̄_z = 0.000500, E_s = 90.0 GPa and φ = 0.050 to illustrate.

   Show the answer and reasoning

   No. Nominal axial stress is 0.250 MPa, while local member stresses are −15.0 MPa in xy families and +15.0 MPa in xz/yz families.

   Local fields require structural analysis. These values come from the stated affine member-strain model and volume/direction averaging; a generic lattice’s fields cannot be recovered from one scalar modulus alone.

2. Why report cell dimensions? Distinguish uniformly doubling all dimensions from doubling cell length at fixed square-member thickness in the same slender-model family.

   Show the answer and reasoning

   Scale and boundaries influence averaging. Uniform doubling keeps t/ℓ and φ unchanged in the classical model; doubling ℓ alone quarters φ, quarters the stretching modulus and reduces the bending-model modulus to one sixteenth.

   A few-cell specimen may not behave like a large bulk array. The scaling conclusion additionally assumes the same topology, section shape, constituent state and regime; it does not preserve wave frequencies or account for manufacturing size effects.

3. For an axial member and a transverse-tip-loaded cantilever with the same ℓ, t and E_s, use k_a = E_st²/ℓ and k_b = 3E_s(t⁴/12)/ℓ³. Find k_a/k_b when ℓ/t = 50. Do the support conditions represent the same member test?

   Show the answer and reasoning

   k_a/k_b = 4(ℓ/t)² = 10000. No: one test is axial extension; the other is transverse bending with a clamped root and freely rotating tip.

   The ratio isolates how the load path and length dependence can change structural stiffness. It is not a 10000-fold whole-lattice modulus claim; averaging and cell geometry still supply their own factors.

4. An initial normal coefficient is 750 MPa at φ = 0.050 in the ideal model. May a report use it as evidence that the free-lateral Young’s modulus exceeds the predicted 500 MPa?

   Show the answer and reasoning

   Not without checking the boundary condition. Lateral restraint alone predicts a constrained coefficient of 750 MPa with unchanged constituent and architecture.

   Under the same axial strain, restraint requires nonzero transverse reaction stresses. Comparing that coefficient with a free-lateral Young’s modulus confounds two different mechanical responses.

5. The supplied finite-domain study reports A and B at N = 8 as 537.5 and 475 MPa. Calculate gap/midpoint. Does averaging them prove a 506.25 MPa bulk modulus?

   Show the answer and reasoning

   Gap/midpoint = 62.5/506.25 ≈ 0.1234568, or 12.35%. Their midpoint is 506.25 MPa, but it is not thereby proved to be the bulk modulus.

   The two prescriptions still differ substantially under the chosen 5% screen. A midpoint has no special claim to remove boundary effects, sampling limitations or discretization error.

6. In the separate ideal pin-ended-member example, double both ℓ and t while preserving E_s. How do P_cr and P_cr/A change?

   Show the answer and reasoning

   I grows by 2⁴ = 16 and ℓ² by 4, so P_cr grows by 4 to about 4.73741 N. Area also grows by 4, leaving the ideal mean buckling stress about 29.6088 MPa.

   Geometric similarity preserves this classical normalized stress, not the force. It also leaves the member’s slenderness unchanged; material size effects and imperfections were excluded.

7. Two lattices have matched density, initial Young’s modulus and specimen geometry, but one contains a weakly bonded set of nodes. Does the matched initial slope establish identical collapse or fatigue behaviour?

   Show the answer and reasoning

   No. Initial stiffness, joint failure and repeated local damage are distinct responses.

   A defect can initiate localization, change load redistribution or evolve during cycles. Local and complete-response evidence are needed; an initial effective modulus does not supply those mechanisms or a life distribution.

What to carry forward

• Effective stiffness relates a defined average loading and deformation; local member response and specimen stiffness are different quantities.
• Fixed three-dimensional slender-member models can give E*/E_s proportional to φ in stretching and to φ² in bending. Geometry and regime determine the constants and exponents.
• The six-family affine model gives E*_z = E_sφ/9 under uniaxial stress and M*_z = E_sφ/6 under lateral restraint, with the same constituent.
• Local compression, buckling, yielding, defects and component instability remain possible even when an initial effective modulus is known.
• Boundary and size comparisons test an averaging assumption; a few-cell result or a chosen convergence screen is not a universal property.