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Materials Science · Composites, fibres and interfaces

Natural and engineered composite examples

Useful background: Units, loads, stress and strain; Atoms, bonding, structure and defects; Processing, structure and useful properties. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

  • Compare natural and engineered composites without assuming identical scales or mechanisms.
  • Distinguish constituent organisation inside a solid wall from the larger cellular structure in wood and bone.
  • Explain why resisting relative sliding can spread deformation in a nacre-inspired architecture, without making the strongest possible interface a universal aim.
  • Use a common-volume loading–unloading comparison to separate input, returned and nonreturned work.
  • Evaluate what a matched manufactured comparison supports and which biological or engineering claims remain untested.
01

Hierarchy appears in familiar materials

Wood combines oriented cellulose-based structures with a surrounding matrix; bone combines organic and mineral constituents in a hierarchy. Engineered fibre composites also arrange contrasting constituents. Their resemblance invites questions about organisation, interfaces and damage control.

Hierarchy means that organised building blocks form larger organised structures. The important question is what changes between those levels. A cellulose-rich fibril, a solid cell wall and a piece of wood containing cell spaces have different geometries and different volume denominators. The property of one level cannot simply be assigned to the next.

Constituents and composite architecture introduces matrix, reinforcement and interface. Here we use those roles to compare particular natural structures with manufactured analogues. An interface transmits contact forces between constituents; its ability to resist, slip or separate affects the available deformation path. This lesson does not assume that every natural composite has the same mechanism or an optimal design for every human use.

02

Wood has a composite wall inside a cellular structure

In a wood cell wall, cellulose-rich fibrils are embedded in a surrounding lignin- and hemicellulose-rich material. Fibril orientation differs between wall layers. At the larger scale, walls enclose elongated cell spaces, called lumina. Thus the surrounding material of a fibril and the space inside a cell are different things; the lumen is not the polymer matrix.

Along the grain, much deformation loads the aligned walls axially. Across the grain, wall bending and the cell arrangement become important. Rays and the orientation of growth rings also distinguish radial from tangential loading. This explains why copying only the cell-wall ingredients does not reproduce the bulk directional response. A manufactured aligned-fibre solid and a cellular wood specimen do not automatically share a modulus or failure mode.

Moisture is part of the state. Water can occupy cell spaces and interact within walls; changing wall moisture can change dimensions and response. Record conditioning and grain direction instead of treating a dry test as a universal wood value. The following accounting example is invented and specifies its own volume and solid density; it is not a measured wood species.

Take a dry cellular teaching specimen with complete volume V = 10.0 cm³ and dry mass 6.00 g. Stipulate an effective solid-wall density of 1.50 g/cm³ and negligible gas mass. Its wall volume is mass/density = 4.00 cm³; the remaining 6.00 cm³ is cell space. The wall fraction of complete volume is 0.400. If cellulose occupies a stipulated 0.500 of that wall volume, its complete-specimen volume is 2.00 cm³ and its bulk fraction is 0.200, not 0.500.

The multiplication 0.400 × 0.500 follows from nested volume accounts: cellulose volume / complete volume = (wall volume / complete volume) × (cellulose volume / wall volume). It is not a modulus estimate or a way to infer cellulose from dry mass without the supplied density and wall composition. This additive-volume illustration ignores unresolved wall porosity and treats all specified wall constituents as filling its solid volume.

03

Bone changes both the constituents and the larger architecture

Bone provides a different mineral–organic hierarchy. Its solid tissue includes collagen-rich organic material, apatite mineral and water. Collagen is a protein; cellulose is a polysaccharide. Similar words such as fibre or hierarchy do not make these constituents chemically interchangeable. At larger scales, compact cortical tissue and the more open trabecular arrangement organise mineralised tissue differently.

Trabeculae are small strut- and plate-like elements whose connectivity and orientation help determine how the tissue carries load. Their deformation and the response inside their solid material are distinct levels. A specimen containing more pore space can therefore behave differently even when a local piece of its mineralised tissue has similar composition.

Living bone can also remodel through cellular activity; a manufactured lattice does not acquire that ability by copying the pore pattern. Hydration, location, direction and biological history belong with a bone observation. This comparison establishes structural questions, not a medical diagnosis, implant qualification or an exercise prescription. Hierarchy in materials made by living systems gives the related biological context.

04

Nacre suggests a contact mechanism to reproduce and test

Nacre, the inner pearly layer of many mollusc shells, contains overlapping aragonite mineral tablets with thin organic regions between them. Aragonite is a crystalline form of calcium carbonate. Neighbouring tablets can transmit tensile loading through tangential contact forces across their overlaps. The staggered arrangement makes that contact path central to the response.

A crack can turn along an interface instead of cutting straight through a tablet; connected material can also bridge behind an advancing crack. Tablet sliding provides another deformation path. An interface that offers almost no resistance would let tablets separate with little work, while an interface or constraint that forces early tablet breakage can remove an extended sliding path. Neither the weakest nor the strongest possible contact is a universal target.

In a 2011 abalone study, local displacement measurements connected tablet waviness with sideways expansion and increasing resistance during sliding. A larger ABS–epoxy model incorporated a related geometry and constraint to investigate that mechanism. The transfer was a particular sliding/contact process, not a claim that synthetic polymer tablets had become natural mineral tablets or that every shell mechanism had been reproduced.

Separate interface-fracture experiments on three nacre types found that local interface resistance and whole-composite fracture response were not interchangeable. A deflected crack, distributed sliding and bridging change the path and surrounding deformation. This is why a rough interface picture or a large force peak alone is insufficient. The next fictional records ask what work measurements and local observations can jointly support.

Wall, cell space and tablet overlapOriginal schematic, not a micrograph or scale comparison. Six dark lumina lie within pale connected solid walls in a selected wood cross-section. An enlarged selected wall region shows eight oblique cellulose-rich fibrils in a surrounding matrix. A separate nacre-inspired drawing contains fourteen overlapping mineral tablets in three staggered rows with an organic region between them. Opposite arrows at adjacent rows indicate possible relative sliding, not applied-force magnitudes. Neither drawing measures constituent fractions or predicts stiffness.Hierarchy changes the questionSchematic geometry · no common scale or fraction measurementWood: cell spaces and the composite wallAcross-grain sectionSelected wall detailCellulose-rich fibrilsSurrounding matrixNacre-inspired overlap: a different contact pathMineral tablets overlap across the organic regions.Possible sliding is different from an open, unloaded gap.Natural and manufactured constituents remain distinct.
Original schematic, not a micrograph or scale comparison. Six dark lumina lie within pale connected solid walls in a selected wood cross-section. An enlarged selected wall region shows eight oblique cellulose-rich fibrils in a surrounding matrix. A separate nacre-inspired drawing contains fourteen overlapping mineral tablets in three staggered rows with an organic region between them. Opposite arrows at adjacent rows indicate possible relative sliding, not applied-force magnitudes. Neither drawing measures constituent fractions or predicts stiffness.
05

Compare a response path rather than only its tallest point

Consider two original invented manufactured coupons, S and H. They have matched constituent amounts, complete initial gauge volume V₀ = 100 mm³, dimensions, tablet arrangement, conditioning, loading rate and measurement method. S has a contact state that slides at nearly constant resistance after an initial rise. H has a modified contact state with increasing resistance during the observed sliding interval. These descriptions are supplied observations, not predictions from chemistry or a drawing.

Apply a slow tensile displacement programme from δ = 0 to 1.00 mm, then unload to zero force. Here F is the measured axial force and δ is the extension between specimen gauge marks; fixture motion has been excluded. Both coupons remain connected to that prescribed maximum. Treat the table as exact hypothetical input and join successive points by straight lines. During loading H is at 30, 50, 70 and 90 N at δ = 0.25, 0.50, 0.75 and 1.00 mm. S stays at 30 N after 0.25 mm.

A force acting through a displacement increment supplies work. For one straight segment, work is the trapezoid area ½(F₁ + F₂)(δ₂ − δ₁). With N and mm, the result is N·mm; 1 N·mm = 0.001 J = 1 mJ. Add the four loading areas to obtain W_in. A high force over a short distance and a lower force over a long distance can therefore give different work rankings.

The specified unloading path returns mechanical work to the external loading system. S unloads linearly from (1.00 mm, 30 N) to (0.900 mm, 0); H unloads from (1.00 mm, 90 N) to (0.700 mm, 0). The positive returned amount W_return is the triangle area under each unloading branch, even though the signed work done on the specimen during unloading is negative. Define W_nonreturned = W_in − W_return.

This nonreturned work can include dissipation and energy stored in changed internal structure or damage; the record does not separate those contributions. It is not automatically frictional heat, fracture energy per new crack area or a valid critical toughness. Both coupons are still connected at the selected maximum, so this is not work to complete fracture. The zero-force residual extensions also do not prove that all delayed recovery has ended.

Invented loading records at common V₀ = 100 mm³; exact straight segments between points. Both specimens are then unloaded along the separately specified branches.
Gauge extension δ (mm)Force S (N)Force H (N)
000
0.25030.030.0
0.50030.050.0
0.75030.070.0
1.00030.090.0
06

Use the work account with local evidence and the same basis

S takes W_in = 3.75 + 7.50 + 7.50 + 7.50 = 26.25 mJ. Its unloading triangle returns ½ × 30.0 × 0.100 = 1.50 mJ, leaving 24.75 mJ nonreturned. H takes 3.75 + 10.0 + 15.0 + 20.0 = 48.75 mJ. It returns ½ × 90.0 × 0.300 = 13.50 mJ, leaving 35.25 mJ. H takes more input work but also returns more; treating all 48.75 mJ as nonreturned would overstate this latter quantity.

At the common initial volume, W_nonreturned/V₀ is 0.2475 mJ/mm³ for S and 0.3525 mJ/mm³ for H. Since 1 mm³ = 10⁻⁹ m³, 1 mJ/mm³ = 1 MJ/m³. The ratio is 35.25/24.75 = 1.42424, a 42.42% increase for this protocol and basis. It is not a claim of 42.42% greater intrinsic fracture resistance, impact protection or useful life.

Suppose separate local images show sliding concentrated in one region of S but spread across several overlaps in H, with sideways expansion resisted by the surrounding arrangement. That observation is consistent with a distributed sliding mechanism. The global curves alone do not locate sliding or distinguish it from polymer deformation or damage. A further matched control that changes lateral constraint while retaining the interface state would help test whether the proposed contact hardening depends on that constraint.

Matching complete specimen volumes makes this particular work comparison direct. If another coupon instead had 200 mm³ and 60.0 mJ nonreturned, its total would be larger than H’s, but its volume-normalised value would be 0.300 MJ/m³, below H’s 0.3525 MJ/m³. The basis can reverse a ranking. Scaling a geometry can also change flaws, contact roughness and constraints, so the chosen normalisation does not establish scale independence.

Calculated work for the invented common-maximum loading–unloading protocol; quantities are not named-material test results or critical fracture properties.
QuantityS: near-constant sliding resistanceH: increasing sliding resistance
Input work (mJ)26.2548.75
Returned work (mJ)1.5013.50
Nonreturned work (mJ)24.7535.25
Nonreturned / original volume (MJ/m³)0.24750.3525
Input, returned and nonreturned workOriginal invented force–gauge-extension records at common initial specimen volume one hundred cubic millimetres. The solid S loading path reaches thirty newtons at zero point two five millimetre and remains at thirty to one millimetre. The long-dashed H path shares the initial segment and then reaches fifty, seventy and ninety newtons at zero point five, zero point seven five and one millimetre. Short-dashed unloading returns S to zero force at zero point nine millimetre and H at zero point seven millimetre. Loading work is twenty-six point two five and forty-eight point seven five millijoules; returned work is one point five and thirteen point five millijoules; nonreturned work is twenty-four point seven five and thirty-five point two five millijoules. Lower bars split each input into nonreturned and hatched returned amounts. These records do not establish critical fracture resistance or isolate a local mechanism.Record the complete mechanical pathInvented coupons · common initial volume V₀ = 100 mm³Force F (N)02550751000.000.250.500.751.00Specimen gauge extension δ (mm)Load toward 1.00 mm; unload toward residual extension.S loadingH loadingS unloadingH unloadingSplit input work (mJ) on one common bar scaleS26.25H48.75NonreturnedReturnedThe nonreturned part is not automatically frictional heat.
Original invented force–gauge-extension records at common initial specimen volume one hundred cubic millimetres. The solid S loading path reaches thirty newtons at zero point two five millimetre and remains at thirty to one millimetre. The long-dashed H path shares the initial segment and then reaches fifty, seventy and ninety newtons at zero point five, zero point seven five and one millimetre. Short-dashed unloading returns S to zero force at zero point nine millimetre and H at zero point seven millimetre. Loading work is twenty-six point two five and forty-eight point seven five millijoules; returned work is one point five and thirteen point five millijoules; nonreturned work is twenty-four point seven five and thirty-five point two five millijoules. Lower bars split each input into nonreturned and hatched returned amounts. These records do not establish critical fracture resistance or isolate a local mechanism.
07

An analogy is a research guide

Natural materials vary with growth, moisture, orientation and history. Copying one visual pattern does not automatically reproduce its properties. Identify the mechanism and scale to transfer, then test the manufactured analogue under matched conditions.

A useful transfer statement names the proposed process, its scale, loading direction, contact and boundary conditions, and the observation that would distinguish it from an alternative. For the manufactured coupons, that means recording local tablet motion together with the global path, not relabelling every nonreturned millijoule as one mechanism.

A printed lattice resembling trabecular bone has copied some architecture, not collagen–mineral chemistry, living remodelling or clinical suitability. A layered polymer sample resembling nacre has copied some geometry, not automatically the same sliding resistance, crack path or hydration sensitivity. Compare the actual natural and manufactured specimens under the intended conditions; unrelated tests with similar units are not enough.

The separate mixture-estimate, load-transfer and fracture lessons develop quantitative models for their stated geometries and measurements. Here the central skill is to identify which constituent, architectural or contact feature the evidence actually supports. A successful analogue can reproduce one useful mechanism while remaining a different material with different limitations.

Worked examples

Work through the case

A manufactured layered material resembles a natural shell. Has shell-like toughness been established?

  1. Identify the proposed toughening mechanism.
  2. Check constituent and interface behaviour.
  3. Measure fracture response in the manufactured architecture.
  4. For a nacre-inspired claim, look for the proposed overlap sliding, constraint and crack path, rather than only a brick-shaped pattern. Record hydration, scale and the actual fracture measurement.

No. Visual similarity alone does not establish the mechanism or property.

Carry the volume denominator through a hierarchy

Use the invented dry cellular specimen: complete volume 10.0 cm³, dry mass 6.00 g, solid-wall density 1.50 g/cm³, and cellulose occupying 0.500 of the solid wall volume. Find wall, cell-space and bulk cellulose fractions.

  1. Wall volume = 6.00/1.50 = 4.00 cm³. Complete volume includes the cell spaces, so their volume is 10.0−4.00=6.00 cm³.
  2. Wall and cell-space fractions are 0.400 and 0.600. The stipulated wall model resolves no additional pores inside that solid volume.
  3. Cellulose volume =0.500×4.00=2.00 cm³, giving complete-volume fraction 2.00/10.0=0.200.
  4. The remaining wall material occupies 2.00 cm³. The volume account is 2.00+2.00+6.00=10.0 cm³; no stiffness or force fraction follows from it.

Wall 40.0%, cell space 60.0%, cellulose 20.0% of complete volume. Cellulose remains 50.0% on the different wall-only basis.

Subtract returned work before comparing the residual account

For S and H, use the given straight loading and unloading paths. Compare input and nonreturned work at the common 100 mm³ volume. Does the result establish intrinsic nacre-like toughness?

  1. S loading has first triangle 3.75 mJ and three rectangles 7.50 mJ each, total 26.25 mJ. H has trapezoids 3.75, 10.0, 15.0 and 20.0 mJ, total 48.75 mJ.
  2. The S unloading triangle is ½×30×(1.00−0.900)=1.50 mJ. H returns ½×90×(1.00−0.700)=13.50 mJ.
  3. Subtract to obtain 24.75 and 35.25 mJ nonreturned. Divide by 100 mm³ to obtain 0.2475 and 0.3525 MJ/m³.
  4. H/S=1.42424 for this measure. Imaging and changed-constraint controls are still needed to attribute the difference; no precracked fracture measurement or new-crack-area account was supplied.

H takes 48.75 mJ input but has 35.25 mJ nonreturned, versus 24.75 mJ for S. This is a conditional protocol comparison, not proof of intrinsic shell-like toughness.

A changed return path changes the conclusion

Keep H’s loading data, but a new independently observed unloading branch reaches zero force at δ=0.500 mm. What nonreturned work follows? Can the old 42.42% contrast still be used?

  1. The loading area stays 48.75 mJ because that path is unchanged.
  2. The new returned triangle is ½×90×(1.00−0.500)=22.50 mJ.
  3. Nonreturned work becomes 48.75−22.50=26.25 mJ, or 0.2625 MJ/m³ at 100 mm³.
  4. Relative to S’s 24.75 mJ, the increase is 1.50/24.75=6.06%. Changed recovery cannot be ignored merely because the loading peak is unchanged.

26.25 mJ nonreturned and a 6.06% contrast with S. The former 42.42% applies only to the original complete path.

Try the reasoning

  1. Why report wood’s grain direction?

    Show the answer and reasoning

    Its architecture is directional.

    Longitudinal and transverse responses differ.

  2. Can an analogy replace a property test?

    Show the answer and reasoning

    No.

    It proposes a mechanism to investigate.

  3. A new dry cellular model has complete volume 20.0 cm³ and wall volume 6.00 cm³. Cellulose occupies 0.600 of its wall. Find the complete cellulose fraction. Has wall-only fraction 60% established 60% of the whole body?

    Show the answer and reasoning

    Cellulose occupies 3.60 cm³ and 18.0% of complete volume, not 60%.

    Multiply the conditional fractions: (6/20)×0.600=0.180. The remaining 14.0 cm³ of cell space is part of the complete denominator.

  4. The original 10.0 cm³ teaching specimen gains 0.600 g of water. Its complete volume is independently observed to remain 10.0 cm³. What is the new total mass density? Does dividing its new 6.60 g mass by the dry-wall density identify extra dry wall?

    Show the answer and reasoning

    Total mass density is 0.660 g/cm³=660 kg/m³. The dry-wall mass remains 6.00 g; 6.60/1.50 is not a valid dry-wall volume.

    Water is an added constituent in this stipulated account. Reusing a dry-solid density for total wet mass would misassign water as new solid. Real wood may also swell; unchanged volume is an input here, not a general moisture law.

  5. Coupon Q has 60.0 mJ nonreturned over 200 mm³. Compare its total and volume-normalised values with H. Does one ranking answer every design question?

    Show the answer and reasoning

    Q has more total nonreturned work than H’s 35.25 mJ, but Q’s 0.300 MJ/m³ is below H’s 0.3525 MJ/m³.

    Divide by each complete original volume. A fixed-volume comparison and an available-whole-coupon comparison address different constraints; neither alone establishes fracture resistance or component suitability.

  6. A third invented coupon reaches 100 N linearly at 0.250 mm and breaks there; force then becomes zero without further specimen-extension work. Find its input work. Is its higher peak than H enough to show more work taken?

    Show the answer and reasoning

    Input work is ½×100×0.250=12.50 mJ, less than H’s 48.75 mJ.

    The high peak spans a shorter path. This third coupon fails early whereas H remains connected to 1.00 mm, so retain the different endpoints; the calculation does not assign an intrinsic failure property to either.

  7. H’s global curve rises during loading. A close image shows no tablet sliding, only deformation of the polymer between tablets. Has the intended sliding-hardening mechanism been verified?

    Show the answer and reasoning

    No. A rising global force can arise from a different mechanism.

    The local observation conflicts with the proposed relative sliding during that interval. Reconsider the account and test a discriminating control rather than infer the mechanism solely from the curve shape.

  8. A manufactured porous mineral–polymer lattice matches a selected bone pore image and its initial compression slope. What has that comparison established about living remodelling and use inside a patient?

    Show the answer and reasoning

    Neither living remodelling nor clinical suitability follows.

    Selected geometry and one mechanical response do not supply cellular activity, biological compatibility, degradation history or evidence for the intended medical use. Keep the structural analogy and the untested functions separate.

What to carry forward

  • Use biological examples to ask structural questions.
  • Validate the mechanism in the actual material.
  • Distinguish a constituent fraction within a wall from its fraction in the whole cellular specimen.
  • Wood, bone and nacre use different constituents and structures; hierarchy is not one universal recipe.
  • Compare complete response paths: work put in, work returned and work not returned have different meanings.
  • A plausible manufactured analogue needs local mechanism evidence, matched conditions and a clearly defined comparison basis.

References

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

  1. Massachusetts Institute of Technology3 sources
    1. Natural Honeycombs: Wood · transcribed lecture notesLorna Gibson · MIT OpenCourseWareCourse notes · PDFocw.mit.eduRead source
    2. Trabecular Bone · lectures 10–11 transcribed notes (2015 course)Lorna Gibson · MIT OpenCourseWareCourse notes · PDFocw.mit.eduRead source
    3. Lecture 21: Biomaterials for Organ Replacement · selected bone structureMIT OpenCourseWare · 3.051J/20.340J Materials for Biomedical Applications (2006)Historical university lecture notes · PDFocw.mit.eduRead source
  2. Moisture relations and physical properties of wood · Wood Handbook, chapter 4 (2021)Samuel V. Glass and Samuel L. Zelinka · USDA Forest Service, Forest Products LaboratoryGovernment technical handbook chapter · PDFresearch.fs.usda.govRead source
  3. Tablet-level origin of toughening in abalone shells and translation to synthetic composite materialsHoracio D. Espinosa and colleagues · Nature Communications 2, 173 (2011)Primary research, selected official indexed sections and direct abstractnature.comRead source
  4. The weak interfaces within tough natural composites: Experiments on three types of nacre · accepted author paper (2012)Ahmad Khayer Dastjerdi, Reza Rabiei and François BarthelatPrimary research author paper · PDFbarthelat-lab.mcgill.caRead source