Fibre Structure and Properties: Investigation Methods, Morphology, Mechanics, Moisture and Structure–Property Links
2. X-ray diffraction and birefringence: lattice, crystal size and orientation
Wide-angle X-ray diffraction (WAXD) obeys Bragg's law nλ = 2d sinθ, where λ is the wavelength (Cu Kα radiation has λ = 0.154 nm), d the lattice-plane spacing and θ the glancing angle, so peaks at larger 2θ come from smaller spacings. A powder-like equatorial scan of a fibre gives sharp crystalline peaks on a broad amorphous halo; crystallinity by XRD is the crystalline peak area divided by the total area. Crystallite size is estimated from peak broadening with the Scherrer equation L = Kλ/(β cosθ), where β is the full width at half maximum in radians (instrumental broadening removed), θ is half the peak 2θ, and K ≈ 0.9. Worked example: a peak at 2θ = 25.6° gives d = 0.154/(2 sin 12.8°) = 0.348 nm; if β = 0.80° = 0.01396 rad then L = 0.9 × 0.154/(0.01396 × cos 12.8°) = 10.2 nm. Broader peaks mean smaller or more imperfect crystals.
Orientation by XRD comes from the azimuthal intensity spread of a crystalline reflection. A narrow arc means well-oriented crystals; Hermans' function f = (3⟨cos²φ⟩ − 1)/2 equals 1 for perfect alignment with the fibre axis, 0 for random and −0.5 for perpendicular. A quick estimate is f = (180° − H)/180°, with H the azimuthal half-width in degrees. Small-angle X-ray scattering (SAXS) looks at much larger spacings: a meridional peak reveals the long period of alternating crystalline and amorphous layers (tens of nanometres).
Birefringence Δn = n∥ − n⊥ is the difference of refractive indices for light polarised parallel and perpendicular to the fibre axis, and measures the overall orientation of crystalline plus amorphous chains. In a polarising microscope a compensator measures the retardation Γ (path difference) of a fibre of thickness t, and Δn = Γ/t. The average orientation function is f = Δn/Δn_max, where Δn_max is the birefringence of perfectly oriented chains of that polymer. Worked example: Γ = 1.30 µm, t = 18 µm, Δn_max = 0.22: Δn = 0.0722 and f = 0.0722/0.22 = 0.33. A sonic modulus (sound speed through the fibre) is a third orientation measure, reflecting the overall average orientation of the chains.
3. Optical and electron microscopy, and infrared spectroscopy
- Optical microscopy gives the longitudinal appearance (scales, convolutions, striations, crimp) and, on sections cut by microtome or Hardy's plate technique, the cross-section shape and fibre diameter. Polarised light adds birefringence and orientation; its resolution is limited by wavelength to about 0.2 µm.
- Scanning electron microscopy (SEM) scans a focused beam over a surface and uses secondary electrons: great depth of field, magnification to tens of thousands, ideal for surface topography, damage, fracture ends and finish deposits. Non-conducting fibres are coated (gold or carbon) to avoid charging.
- Transmission electron microscopy (TEM) passes electrons through ultrathin sections (tens of nanometres, stained with a heavy-metal agent to give contrast) and reveals internal fibrils, lamellae and the arrangement of crystalline and amorphous domains, with nanometre resolution and electron diffraction. The beam damages organic polymers, so imaging uses low dose.
- Infrared spectroscopy identifies functional groups by their absorption bands: ester C=O near 1720 cm⁻¹ (PET), amide I near 1640 cm⁻¹ and N–H stretch near 3300 cm⁻¹ (nylon, wool, silk), nitrile C≡N near 2240 cm⁻¹ (acrylic), broad O–H near 3300 cm⁻¹ (cellulose). Some bands are sensitive to conformation, so band ratios track crystallinity. With polarised IR the dichroic ratio R = A∥/A⊥ gives the orientation of the group whose transition moment is parallel or perpendicular to the chain. Beer–Lambert: A = εcl. Attenuated total reflectance (ATR) allows the surface of a fibre to be examined without preparing a film.
4. Thermal methods: DSC, DMA, TMA and TGA
| Method | Measures | Reads from a fibre |
|---|---|---|
| DSC (differential scanning calorimetry) | Heat flow against temperature | Tg as a step in heat capacity, cold-crystallisation exotherm, melting endotherm (Tm and ΔHm), crystallinity |
| DMA (dynamic mechanical analysis) | Storage modulus E′, loss modulus E″, tan δ = E″/E′ under oscillating strain | Tg (tan δ peak or E′ drop), secondary transitions, modulus against temperature and frequency |
| TMA (thermomechanical analysis) | Dimension change under small constant load against temperature | Thermal expansion, shrinkage and shrinkage force, softening |
| TGA (thermogravimetric analysis) | Mass against temperature, in nitrogen or air | Moisture and volatile loss, onset of decomposition, residue (char), thermal stability |
Crystallinity by DSC is Xc = (ΔHm − ΔHcc)/ΔHm°, where ΔHm is the measured melting enthalpy, ΔHcc the enthalpy of any cold-crystallisation exotherm on the same run (crystals formed during the scan, not originally present), and ΔHm° the melting enthalpy of 100% crystalline polymer. Worked example: ΔHm = 48 J/g, ΔHcc = 10 J/g, ΔHm° = 140 J/g: Xc = 38/140 = 0.271, or 27.1%. A fibre that crystallised during heating therefore shows a lower true initial crystallinity than ΔHm alone suggests.
5. Structure and morphology of man-made fibres
Man-made fibres share the fibrillar picture: oriented microfibrils with crystalline blocks alternating with amorphous segments along the axis, joined by tie molecules. Spinning route and drawing decide the cross-section and the radial structure.
| Fibre | Cross-section and structure | Why |
|---|---|---|
| PET, nylon (melt spun) | Round, smooth, fairly uniform; spinneret shape can be changed (trilobal for lustre and bulk) | Solidifies by cooling, no solvent loss |
| Viscose | Serrated with skin-core structure; skin oriented and dense, core weaker | Fast skin formation, then core shrinks |
| Polynosic, lyocell | Round, little or no skin, more uniform and more crystalline and oriented than viscose | Slow regeneration and stretch |
| Acrylic | Dry spun: dog-bone; wet spun: round or bean; lateral order only, no true 3-D crystal | Bulky nitrile dipoles and comonomer disturb packing |
| Para-aramid | Skin-core, radially pleated sheets; highly oriented, extended chains | Liquid-crystalline dope aligned in the air gap |
| UHMWPE, carbon | UHMWPE: nearly fully extended-chain crystals; carbon: graphite-like layers aligned along the axis | Ultra-drawing; carbonisation |
6. Mechanical properties and moisture sorption
The stress–strain curve of a fibre is plotted as force per linear density (specific stress) against extension. Tenacity is breaking force divided by linear density, in N/tex, cN/tex or gf/den; initial modulus is the slope near the origin; breaking extension is the strain at break; work of rupture is the area under the curve. Elastic recovery, creep (extension rising under steady load) and stress relaxation (stress falling at fixed extension) describe time dependence. Conversions: 1 gf/den = 8.83 cN/tex, and since stress = tenacity × density, 1 N/tex × 1 g/cm³ = 1 GPa. Worked example: breaking force 9.0 cN on a 2.0 dtex (0.2 tex) polyester fibre of density 1.38 g/cm³: tenacity = 9.0/0.2 = 45 cN/tex = 0.45 N/tex; stress = 0.45 × 1.38 = 0.621 GPa = 621 MPa. The reason to quote tenacity is that fibres are not exactly cylinders and the area is hard to measure.
Moisture sorption. Water is taken up by polar groups (hydroxyl, amide) in the amorphous regions; crystals are largely inaccessible. The regain–humidity isotherm is S-shaped: steep at low humidity (water bound to sites), flatter in the middle, and rising again near saturation (multilayer and condensed water). The desorption curve lies above the absorption curve (hysteresis), so regain at a given humidity is higher when approached from the wet side. Absorption releases the heat of sorption, swells the fibre mostly across its diameter, and plasticises the polymer, lowering Tg. Textiles are tested at a standard atmosphere of 65% relative humidity at 20 °C (or 27 °C in tropical practice). Regain falls with crystallinity and rises with the number of hydrophilic groups; this is why cotton, wool and viscose absorb readily and polyester and polypropylene barely do.
7. Fibre structure–property correlation
| Structural change | Effect on properties |
|---|---|
| Higher orientation (more drawing) | Tenacity and initial modulus up, breaking extension down, birefringence up |
| Higher crystallinity | Density, modulus and dimensional stability up; moisture regain, dye uptake and swelling down |
| Higher molecular weight | Strength, toughness and fatigue resistance up (to a limit); melt viscosity up |
| Oriented amorphous regions | Hot-water shrinkage and shrinkage force up; removed by heat setting |
| Coarser fibre (higher tex) | Bending rigidity rises roughly as tex², so fabric handle becomes stiffer |
| Hydrophilic groups, skin–core, crimp | Regain, comfort and static behaviour; cohesion, bulk and resilience |
A first model for modulus is the parallel (iso-strain) rule of mixtures, E = xcEc + (1 − xc)Ea, where Ec and Ea are the crystalline and amorphous moduli and xc the volume fraction; it is an upper bound, since real fibres have crystals and amorphous chains partly in series. The high strengths of aramid, UHMWPE and carbon follow from the same ideas: chains extended and aligned, with few defects, so strength approaches a fraction of the theoretical value for covalent bonds, while conventional textile fibres reach only a small fraction of it.
Key takeaways
- Density-gradient column gives density; two-phase crystallinity: volume fraction (ρ − ρa)/(ρc − ρa), mass fraction ρc(ρ − ρa)/[ρ(ρc − ρa)]. Density, XRD and DSC crystallinities differ because they define order differently.
- XRD: Bragg nλ = 2d sinθ, Scherrer L = Kλ/(β cosθ) with K ≈ 0.9, azimuthal width gives crystal orientation (Hermans f); birefringence Δn = Γ/t and f = Δn/Δn_max give overall orientation.
- Optical microscopy shows shape and sections, SEM surfaces and fractures, TEM internal fibrils on stained thin sections; IR identifies groups and, with polarisation, orientation via the dichroic ratio.
- DSC reads Tg, cold crystallisation and melting with Xc = (ΔHm − ΔHcc)/ΔHm°; DMA gives E′, E″, tan δ; TMA gives expansion and shrinkage; TGA gives mass loss and decomposition.
- Tenacity = force/tex; 1 N/tex × 1 g/cm³ = 1 GPa; 1 gf/den = 8.83 cN/tex. Moisture is taken up by amorphous polar sites, with S-shaped isotherm, hysteresis, heat of sorption and plasticisation.
- Orientation raises strength and modulus and cuts extension; crystallinity raises density and stability but cuts regain and dye uptake; spinning route fixes cross-section (viscose skin-core, dry-spun acrylic dog-bone, polynosic and lyocell round).
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In a density-gradient column used for fibre density, a fibre sample comes to rest at a particular height because
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Answer: C — the liquid density at that level equals the fibre density
The fibre sinks or rises until the liquid at its level has the same density as the fibre, where buoyancy balances weight; the height is read against glass floats of known density. The fibre must not dissolve or swell in the liquids, and crystallinity is then calculated from the density through a two-phase model rather than read directly.The crystallinity of the same polyester fibre is found to be 44% from density (two-phase model) and noticeably lower from DSC. The most sound explanation is that
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Answer: B — density assumes fixed phase densities while DSC counts melting enthalpy, so imperfect crystals and intermediate order are counted differently
Each method defines crystallinity operationally: density through assumed ρc and ρa, DSC through enthalpy relative to a 100% crystalline reference (and affected by cold crystallisation or recrystallisation during the scan). Differences are expected for real fibres with imperfect crystals and intermediate order. DSC measures melting of the crystals, not the amorphous phase, and no instrument fault is needed.Which statement correctly contrasts birefringence with the azimuthal X-ray method for orientation?
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Answer: A — X-ray azimuthal spread gives crystal orientation, birefringence gives an average over all chains
A crystalline reflection sees only the lattice planes of the crystals, so the azimuthal width reports crystal orientation; birefringence responds to all chain segments and therefore reports an average. The two can differ, notably after heat setting. Birefringence is, in fact, most commonly measured on drawn fibres.In polarised infrared spectroscopy, the dichroic ratio R = A∥/A⊥ of an absorption band of a fibre is used to find
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Answer: D — the alignment of chain segments carrying the group
A band absorbs most when the electric vector lies along the transition moment of the group, so the ratio of absorbances parallel and perpendicular to the fibre axis measures how well those groups, and hence the chain segments, are aligned. Molecular weight, Tg and crystal density are not found from dichroism.A thermogravimetric run in nitrogen on a cotton fibre shows a mass loss of about 7% up to 120 °C, then essentially no change until a sharp loss at a much higher temperature. The first loss is mainly
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Answer: B — loss of sorbed moisture
Mass lost near and below 100 to 120 °C is water driven out of the hydrophilic fibre (a regain of about 7 to 8% fits cotton). Cellulose decomposition occurs at far higher temperature, which is the later sharp loss; crystallisation and glass transitions involve no mass change.Which pair of fibre and typical morphology is correct?
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Answer: C — Dry-spun acrylic: dog-bone cross-section
In dry spinning a skin forms first as solvent evaporates, then the core shrinks as the rest of the solvent leaves, leaving a dog-bone shape. Melt spinning gives smooth round sections, polynosic has little or no skin, and lyocell is round and uniform.A polyester fibre is drawn to a higher draw ratio. Which set of changes is expected?
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Answer: D — Higher birefringence, higher tenacity and modulus, lower breaking extension
Drawing aligns chains along the axis, which raises birefringence, tenacity and modulus and removes the slack that gave extension. The other combinations contradict that, for example higher extension together with higher orientation.A fibre is brought to 65% relative humidity once from a drier state (absorption) and once from a wetter state (desorption). The moisture regain at 65% relative humidity is
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Answer: A — higher after desorption than after absorption (hysteresis)
Sorption isotherms of hygroscopic fibres show hysteresis: the desorption branch lies above the absorption branch, so regain is higher when approached from the wet side. This is why test specimens are pre-conditioned from the dry side to a defined state. Equilibrium here is not memoryless because internal bonds re-form differently on drying and wetting.Which statements about X-ray diffraction of fibres are correct?
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Answer: A — Bragg's law nλ = 2d sinθ links the glancing angle to the lattice spacing; B — Broader diffraction peaks indicate smaller or more imperfect crystallites
Bragg's law and Scherrer broadening are standard results. A narrow arc means the crystals are well aligned, not poorly; and at fixed wavelength d = λ/(2 sinθ), so a larger angle means a smaller spacing.Other factors being equal, an increase in the crystallinity of a fibre is expected to cause which of the following?
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Answer: A — A rise in density; B — A fall in moisture regain; C — A slower rate of dye uptake
Crystals are denser, exclude water and dye molecules, and lock chains in place, so density rises while regain and dye uptake rate fall and shrinkage tends to fall. Greater shrinkage is associated with oriented amorphous regions that are free to relax, not with more crystals.Which statements about moisture sorption by textile fibres are correct?
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Answer: A — Water is held mainly at polar sites in the amorphous regions; B — Heat is released when a dry fibre absorbs moisture; C — Absorbed water plasticises the fibre and lowers its Tg
Sorption occurs at hydrophilic sites in the amorphous phase, is exothermic (heat of sorption), and plasticises the polymer. Polyester has few accessible polar groups and a regain of well under 1%, far below viscose with its many accessible hydroxyl groups.A PET fibre has density 1.385 g/cm³. Assume a two-phase structure with crystalline density 1.455 g/cm³ and amorphous density 1.335 g/cm³. The crystalline mass fraction, as a percentage to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 43.8
Mass fraction xc = ρc(ρ − ρa)/[ρ(ρc − ρa)] = 1.455 × (1.385 − 1.335)/[1.385 × (1.455 − 1.335)] = 1.455 × 0.050/(1.385 × 0.120) = 0.07275/0.1662 = 0.4377, i.e. 43.8%. The volume fraction (0.05/0.12 = 41.7%) is the usual wrong answer when mass fraction is asked.A diffraction peak of a polymer fibre is at 2θ = 25.6° with a full width at half maximum of 0.80° (instrumental broadening already removed). Using the Scherrer equation L = Kλ/(β cosθ) with K = 0.9, λ = 0.154 nm, and β in radians, the apparent crystallite size in nm, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 10.2
θ = 12.8°, cos θ = 0.9751. β = 0.80 × π/180 = 0.013963 rad. L = 0.9 × 0.154/(0.013963 × 0.9751) = 0.1386/0.013615 = 10.18 nm, so 10.2 nm. Using the full 2θ in place of θ in the cosine, or β in degrees, are the usual slips.A drawn fibre of thickness 18 µm gives a retardation of 1.30 µm in a polarising microscope. The maximum birefringence for perfect orientation of this polymer is 0.22. The average orientation function f = Δn/Δn_max, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.33
Δn = Γ/t = 1.30/18 = 0.0722 (both in µm). f = 0.0722/0.22 = 0.328, so 0.33.In a DSC run on a polyester fibre the melting endotherm is 48 J/g and a cold-crystallisation exotherm on the same run is 10 J/g. The melting enthalpy of 100% crystalline polymer is taken as 140 J/g. The initial crystallinity Xc = (ΔHm − ΔHcc)/ΔHm°, as a percentage to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 27.1
Xc = (48 − 10)/140 = 38/140 = 0.2714, i.e. 27.1%. Ignoring the cold-crystallisation exotherm would give 48/140 = 34.3%, which overstates the crystallinity the fibre had before heating.A single polyester fibre of fineness 2.0 dtex breaks at a force of 9.0 cN. Take the fibre density as 1.38 g/cm³ and the cross-section as uniform. The breaking stress in MPa, to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 621
2.0 dtex = 0.20 tex = 0.20 g per 1000 m = 2.0 × 10⁻⁷ kg/m. Area = (2.0 × 10⁻⁷)/1380 = 1.449 × 10⁻¹⁰ m². Force = 9.0 cN = 0.090 N. Stress = 0.090/1.449 × 10⁻¹⁰ = 6.21 × 10⁸ Pa = 621 MPa. Check: tenacity 45 cN/tex = 0.45 N/tex, and 0.45 × 1.38 = 0.621 GPa.