Yarn Structure and Properties: Geometry, Twist, Strength and Irregularity

This chapter answers the last part of Section XE9.2, “Yarn Manufacture, Yarn Structure and Properties”, of XE9 Textile Engineering and Fibre Science: “Influence of fibre geometry, fibre configuration and fibre orientation in yarn; Fibre packing density of yarn; Yarn diameter; Yarn twist and its relation to yarn strength; Helical arrangement of fibres in yarns; Yarn contraction; Fibre migration in yarns; Stress-strain relation in yarn; Mass irregularity of yarn; Structure-property relationship in ring, compact, rotor, air-jet and friction spun yarns.” It teaches how the arrangement of fibres in a yarn decides its density, diameter, strength, extensibility and evenness. The numericals are the number of fibres in a section and the limit irregularity, packing fraction, yarn diameter, surface helix angle, strength utilisation, and twist contraction with its effect on linear density.

1. Fibre geometry, configuration and orientation

Fibre geometry (length, fineness, cross-section, crimp, surface friction, bending and torsional rigidity) sets what a yarn can be. Longer fibres are held by friction over a greater length, so they slip less and give stronger yarn at a given twist. Finer fibres put more fibres in the section: n = yarn tex ÷ fibre tex, so a 20 tex yarn of 0.15 tex fibres has 133 fibres in the section. More fibres gives a more even yarn and permits finer counts; coarse fibres limit the finest count that can be spun, since a yarn needs a minimum number of fibres to hold together. Fibre rigidity matters because a stiff fibre resists the bending needed to follow a helix.

Configuration is the shape of a fibre in the yarn: straight, hooked at the leading or trailing end, folded, looped or buckled. A hook or fold shortens the effective length, so a fibre of length 30 mm folded in half carries load like a 15 mm fibre. Carding leaves many hooks, and each drawing and combing passage straightens and parallelises the fibres, so the product gets more orientation. Orientation is the angle θ between a fibre and the yarn axis. A fibre at θ occupies 1/cos θ of yarn length per unit yarn length, and carries only cos θ of its force along the axis, so the orientation coefficient (the mean of cos θ) is a measure of how well a fibre is used; poor orientation also means fewer fibres of useful length bridging each slip plane.

2. Packing density and yarn diameter

The packing fraction φ is the fraction of the yarn volume that is fibre material. Per metre of yarn, fibre volume = tex ÷ ρ (mm³, with tex in mg/m and ρ in g/cm³, which equals mg/mm³), and the yarn volume is (π d²/4) × 1000 mm³, so

φ = tex ÷ [ρ × (π d²/4) × 1000], d in mm, tex in g/km, ρ in g/cm³. Hence d = 0.03568 × √[tex ÷ (ρ φ)] mm. Example 1: 20 tex, d = 0.18 mm, ρ = 1.50 gives φ = 20 ÷ 38.17 = 0.52. Example 2: 30 tex polyester (ρ = 1.38) with φ = 0.60 has d = 0.03568 × √(30 ÷ 0.828) = 0.215 mm. At a fixed φ the diameter goes as √tex and as 1/√ρ.

  • Circular rods cannot fill the space: the densest hexagonal arrangement of equal circles has φ = π ÷ (2√3) = 0.9069, and real yarns lie well below this because of voids, crimp, hairs and irregular cross-sections.
  • Twist raises φ at first (the yarn is squeezed in) and then flattens, so the diameter at fixed tex falls and levels off. Rotor and friction yarns are less densely packed than ring yarns; compact yarns are more densely packed than ordinary ring yarns.
⚠️ Trap
The diameter in the formula is the diameter of an equivalent circular yarn with no hairs. A measured optical diameter is larger because it includes protruding fibres, and it will give a φ that is too low.

3. Helical model, twist and strength

In the ideal helical model every fibre lies on a coaxial cylinder as a helix. For T turns per unit length at radius r, one turn advances 1/T along the axis and goes 2πr round, so tan θ = 2π r T. The helix angle is zero on the axis and largest at the surface, where tan α = π d T. For d = 0.17 mm and T = 800 turns/m = 0.8 turns/mm, tan α = π × 0.17 × 0.8 = 0.427, so α = 23°. Substituting d from above and T = TM ÷ √tex per cm, tan α = 0.01121 × TM ÷ √(ρ φ), where TM is the tex-system twist multiplier (turns/cm × √tex): the surface angle depends on TM, not on the count.

Strength. With n fibres, each of breaking force F and linear density t, all at helix angle θ and breaking together without slip: yarn force = n F cos θ and yarn tex = n t ÷ cos θ, so yarn tenacity = (F ÷ t) cos² θ. At θ = 30° the tenacity is cos² 30° = 0.75 of the fibre tenacity. Real yarns have a strength-utilisation coefficient below 1 for this obliquity and for slip of short fibres, uneven sharing of load, and irregularity.

Twist and strength. At very low twist, fibres are barely bound, so the yarn fails by fibre slip. More twist raises the lateral pressure and the friction that grips fibres, so strength rises. Beyond the critical (optimum) twist, the obliquity cos² θ and the uneven stress on the outer fibres dominate, so strength falls, and the yarn fails by fibre breakage. The optimum is lower for longer and rougher fibres and higher for short or smooth fibres. Extension at break rises with twist up to a limit, but the yarn gets harder and less bulky.

4. Yarn contraction and fibre migration

Twist contraction is the shortening of a strand when it is twisted: C = (L₀ − L) ÷ L₀ × 100, with L₀ the untwisted and L the twisted length. A fibre on the surface helix has a path of L ÷ cos α for a yarn length L, so the surface fibre’s contraction is 1 − cos α, and the contraction rises with twist. Because the same mass now lies in less length, the yarn linear density rises: tex_yarn = tex_strand ÷ (1 − C). A strand of nominal 20 tex that contracts by 4 % becomes 20 ÷ 0.96 = 20.83 tex. The twisting-in of a drafted strand, and the folding of yarns, both give contraction, which is why a delivered count differs a little from the calculated one.

Fibre migration is the movement of a fibre between the core and the surface of the yarn along its length. In ring spinning it arises in the spinning triangle: fibres at the edge of the ribbon travel a longer path and take higher tension than those in the middle, so they are pulled inward, while the lower-tension fibres move outward in turn. A fibre then follows a helix of changing radius, and its ends are bound into the yarn at some points. Without migration every fibre would keep a fixed radius, ends would lie loose on the surface and the yarn would be a bundle of independent layers with poor cohesion; with migration the fibres interlock, so strength, cohesion and hair-binding improve. Migration is measured with tracer (dyed) fibres, as the mean fibre position and the amplitude of the radial excursion. Rotor yarns, formed from a ring of fibres in a groove, show less regular migration and have wrapper fibres on the surface.

5. Stress-strain behaviour of yarn

A staple yarn’s tensile curve has three regions: an initial, nearly linear elastic region, then a yield region in which fibres slip and the structure rearranges and the slope falls, and finally a region leading to break. Twisted yarns extend more and have a lower initial modulus than the fibres, for two reasons.

  • Geometry: for a helix of fixed radius, a yarn strain ε_y stretches the fibre by ε_f = ε_y cos² θ only, since the fibre unwinds a little as the yarn extends. A fibre at 25° is strained to 0.82 of the yarn strain (cos² 25° = 0.82).
  • Structure and fibre: slip between fibres, migration, and the fibre’s own stress-strain curve (for example the yield and the high extensibility of some synthetic fibres) are added to the geometric effect. Higher twist (a larger helix angle) lowers the initial modulus and raises the extension at break, up to a limit.

Yarn strength is also a weak-link property: a long yarn breaks at its weakest place, so the mean breaking force falls and its scatter narrows as the test gauge length is raised. Strengths measured at different gauge lengths cannot be compared, and a thin-place-rich yarn will show a big drop in strength between a short and a long gauge length.

6. Mass irregularity

Irregularity is measured as CV % = (standard deviation ÷ mean) × 100 or U % = (mean deviation ÷ mean) × 100 of the mass per unit length over a short cut length. For a normal distribution U = 0.8 CV (mean deviation = 0.798 σ), so a yarn with U = 10 % has CV = 12.5 %. A perfectly random arrangement of equal fibres has a lower bound, the limit irregularity:

CV_lim (%) = 100 ÷ √n, with n the mean number of fibres in the section; if the fibre linear densities vary with a coefficient of variation CV_t (as a fraction), CV_lim = (100 ÷ √n) × √(1 + CV_t²). A 25 tex yarn spun from 0.125 tex fibres has n = 200 and CV_lim = 7.07 %. The index of irregularity I = CV_actual ÷ CV_lim; real yarns have I above 1, and I shows how much the process, rather than the fibre count, adds. Independent sources add as variances: CV_total² = CV₁² + CV₂², so a yarn with CV 12 % whose limit is 8 % has an extra 8.94 % from drafting, preparation and machine faults. Doubling n independent slivers lowers the CV by √n, and a finer count raises CV_lim since n falls.

⚠️ Trap
CV_lim = 100 ÷ √n refers to the ideal, with no variation of fibre size, and it is a bound, not a target: no real yarn is better. A yarn with I below 1 is practically never found and usually means the data or n is wrong.

7. Structure-property relations of ring, compact, rotor, air-jet and friction yarns

The table compares typical structures and the properties that follow from them, for the same fibre and count. These are tendencies, not laws, and they change with fibre, count and machine setting.

YarnStructureTypical consequences
RingFibres in a helix with regular migration, good orientationHigh strength, good extension, higher hairiness
CompactRing structure with edge fibres captured, denser packingHigher strength and elongation, lower hairiness, smaller diameter
RotorFibres less oriented and less extended, with wrapper (belt) fibres on the surface; less regular migrationLower strength than ring, better evenness, bulkier, less hairy, stiffer
Air-jetNearly parallel, untwisted core with wrapper fibresLow hairiness, lower strength, firm, harsher handle
FrictionLoosely packed, often core-sheath; poorer orientationCoarse, bulky, lower strength, high production

The reasoning is mechanical: strength is highest when the fibres are parallel and well gripped over their whole length by lateral pressure (the helical yarns), and it falls when fibres are poorly oriented, hooked or only held by a few wrapper turns. Hairiness falls when edge fibres are captured (compact) or wrapped (air-jet).

Key takeaways

  • Fibres in a section n = yarn tex ÷ fibre tex; fibre length, fineness, hooks and orientation (mean cos θ) decide how fully fibre strength is used.
  • φ = tex ÷ [ρ (π d²/4) × 1000] and d = 0.03568 √(tex ÷ ρφ) mm; the maximum hexagonal packing of equal circles is 0.9069.
  • In the helical model tan θ = 2π r T and tan α = π d T at the surface, which depends on TM and not on count; ideal tenacity = fibre tenacity × cos² θ.
  • Strength rises with twist while slip dominates and falls past the critical twist as obliquity dominates; contraction C raises linear density to tex ÷ (1 − C).
  • Migration from tension differences in the spinning triangle binds fibre ends in; yarn strain gives fibre strain ε_y cos² θ, and strength falls as test length rises (weak link).
  • CV_lim = 100 ÷ √n, U = 0.8 CV, I = CV ÷ CV_lim above 1, independent variances add, and doubling n slivers divides CV by √n; yarn type decides orientation, packing, strength and hairiness.

Practice questions (17)

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.

  1. Which set of fibre features generally raises the strength of a staple yarn of a given count and twist?

    1. Coarser, crimpier fibres, which reduce the number in the section, so that fewer fibres share the load
    2. Shorter fibres with many hooks, which lie across the yarn axis, and these slip out under a very small load
    3. Longer, straighter, well-oriented fibres of adequate strength
    4. Fibres of widely varying length, so that all slip planes differ, which spreads the load equally among them
    Show answer

    Answer: C — Longer, straighter, well-oriented fibres of adequate strength

    Long, straight, parallel fibres are gripped along a greater length and carry load along the axis, so fibre strength is better used. Hooks, cross-lying fibres and more coarse fibres do the opposite.

  2. At constant yarn tex, what generally happens to the yarn diameter and packing fraction as the twist is raised from a low to a moderate level?

    1. Diameter falls and packing fraction rises, since the twist squeezes the fibres together
    2. Diameter rises and packing fraction falls, since the yarn untwists internally, because the fibres move apart from one another
    3. Both are unchanged, since diameter depends only on tex and not on how the fibres are arranged in the section
    4. Diameter and packing fraction both rise, since fibres are added although no new fibre enters the yarn at a fixed tex
    Show answer

    Answer: A — Diameter falls and packing fraction rises, since the twist squeezes the fibres together

    Twist binds and compacts the fibres, so φ rises and, because tex and density are fixed, d = 0.03568 √(tex ÷ ρφ) falls, until the structure is fully packed and the change levels off. Diameter is not a function of tex alone.

  3. In a spun yarn spun at very low twist, which failure mode usually governs the breaking of the yarn?

    1. Breakage of all fibres at the same instant, as the load is shared perfectly among all of them
    2. Melting of the fibres by friction at the points where they cross one another
    3. Torsional failure of the fibres themselves, since the twist is highest inside the fibres
    4. Slippage of fibres past one another, since the lateral pressure is small
    Show answer

    Answer: D — Slippage of fibres past one another, since the lateral pressure is small

    With little twist the grip between fibres is weak and they slide out. Fibre breakage governs at higher twist, past the critical twist.

  4. Other things being equal, which change lowers the twist at which yarn strength is a maximum?

    1. Using shorter fibres of the same fineness, which hold together more easily
    2. Using longer fibres, which are held over a greater length
    3. Using smoother, more slippery fibres, which grip one another better
    4. Using a finer count only, whatever the length of the fibres
    Show answer

    Answer: B — Using longer fibres, which are held over a greater length

    Longer fibres are gripped along a greater length and need less twist to avoid slip, so the optimum is lower. Short or smooth fibres need more twist to avoid slip.

  5. Two yarns of different counts are made from the same fibre with the same packing fraction and density, and both are spun with the same tex-system twist multiplier TM (turns/cm × √tex). How do their surface helix angles compare?

    1. The finer yarn has the larger angle, since it has more turns per cm, as it has the smaller diameter and the same twist
    2. The coarser yarn has the larger angle, since it has the larger diameter, as the number of fibres in the section is the same
    3. They are equal, because tan α = 0.01121 TM ÷ √(ρφ) and does not contain the count
    4. They differ by the ratio of the square roots of the counts and also by the ratio of their diameters
    Show answer

    Answer: C — They are equal, because tan α = 0.01121 TM ÷ √(ρφ) and does not contain the count

    tan α = π d T. The coarser yarn has a larger d, but its turns per cm are lower in the same proportion (T = TM ÷ √tex and d ∝ √tex), so the product is the same. This is why TM is used to compare twist across counts.

  6. What is the main benefit of fibre migration in a ring-spun yarn?

    1. The yarn becomes thinner at every point, since the helix then lies closer to the axis
    2. The fibres cease to bear tension along the yarn axis, since the twist takes the tension off them
    3. The twist contraction falls to zero, since the fibres can no longer shorten
    4. Fibres move between core and surface, so their ends are gripped and cohesion improves
    Show answer

    Answer: D — Fibres move between core and surface, so their ends are gripped and cohesion improves

    Migration makes each fibre follow a helix of varying radius, so no fibre end lies free on the surface for its whole length, and strength and hair-binding improve. It does not remove tension from fibres or eliminate contraction.

  7. Why does a more highly twisted yarn usually show a lower initial modulus than a lightly twisted yarn of the same fibre?

    1. Because a larger helix angle strains the fibre less than the yarn and tilts its force off the axis
    2. Because the fibres are shorter in the highly twisted yarn, so they slip out at a low load
    3. Because the fibres are stronger in the highly twisted yarn, and so stretch less at the same load
    4. Because the diameter of the yarn is smaller, so fewer fibres carry the extension
    Show answer

    Answer: A — Because a larger helix angle strains the fibre less than the yarn and tilts its force off the axis

    A fibre in a helix of angle θ is strained by only ε_y cos² θ and contributes its force along the axis with a factor cos θ, so a larger θ reduces the initial stiffness. The fibres are not shorter or stronger because of twist.

  8. What does an index of irregularity (actual CV divided by the limit CV) of 1.6 for a yarn signify?

    1. The yarn is 1.6 times more regular than the random limit, because the process removes the fibre variation
    2. The yarn is less even than a perfectly random fibre arrangement, as the process adds irregularity
    3. The yarn has 1.6 times the number of fibres in the section and a larger limit irregularity as well
    4. The fibre linear density has a CV of 1.6 %, unrelated to the yarn’s own irregularity
    Show answer

    Answer: B — The yarn is less even than a perfectly random fibre arrangement, as the process adds irregularity

    I = CV_actual ÷ CV_lim, so 1.6 means that the actual CV is 60 % above the ideal random limit, owing to drafting waves, preparation faults and machine faults. A value below 1 would be physically impossible for random fibres.

  9. Every fibre in a yarn lies at the same helix angle of 30° to the axis, all fibres have the same breaking force and linear density, break together and do not slip. The ratio of yarn tenacity to fibre tenacity is

    1. 0.87
    2. 0.58
    3. 0.50
    4. 0.75
    Show answer

    Answer: D — 0.75

    Yarn force = n F cos θ and yarn tex = n t ÷ cos θ, so the tenacity ratio is cos² 30° = 0.75. The 0.87 value is cos 30° only, which forgets that the extra length of the inclined fibres per metre of yarn adds mass.

  10. Which statements about yarn diameter and packing fraction are correct, for a circular yarn of tex T, fibre density ρ and packing fraction φ?

    1. At fixed φ and ρ, the diameter is proportional to √T
    2. At fixed T and φ, a denser fibre gives a smaller diameter
    3. The packing fraction is the ratio of the yarn diameter to the fibre diameter
    4. The packing fraction of a real yarn can exceed the hexagonal close-packing value of circular fibres
    Show answer

    Answer: A — At fixed φ and ρ, the diameter is proportional to √T; B — At fixed T and φ, a denser fibre gives a smaller diameter

    From d = 0.03568 √(T ÷ ρφ), d goes as √T and as 1/√ρ. φ is a volume fraction, not a diameter ratio, and it stays far below 0.9069 for real yarns because of voids, crimp and hairs.

  11. Which statements about twist, contraction and migration are correct?

    1. Twist contraction rises as the twist is increased
    2. Migration along the length binds fibre ends into the yarn
    3. The helix angle of a fibre is largest on the yarn axis
    4. The linear density of a yarn falls when the strand is twisted
    Show answer

    Answer: A — Twist contraction rises as the twist is increased; B — Migration along the length binds fibre ends into the yarn

    Contraction grows with twist and makes the yarn tex higher, not lower. Migration ties the fibres in. In the helical model the angle is zero on the axis and largest at the surface.

  12. Which statements about yarn structure and properties, for the same fibre and count, are correct?

    1. A compact ring yarn typically has lower hairiness than a conventional ring yarn
    2. A rotor yarn has wrapper fibres on its surface, and its fibres are typically less oriented than in a ring yarn
    3. An air-jet yarn has a core of fibres twisted at the same helix angle as a ring yarn
    4. A friction yarn has the densest packing and best orientation of all spun yarns
    Show answer

    Answer: A — A compact ring yarn typically has lower hairiness than a conventional ring yarn; B — A rotor yarn has wrapper fibres on its surface, and its fibres are typically less oriented than in a ring yarn

    Compact spinning captures edge fibres, which lowers hairiness; rotor yarns have belt fibres and poorer orientation. An air-jet core is nearly parallel and untwisted, and friction yarns are loosely packed and poorly oriented.

  13. A 25 tex yarn is spun from fibres of uniform linear density 0.125 tex. For a random arrangement of fibres, CV_lim = 100 ÷ √n, where n is the number of fibres in the cross-section. The limit irregularity is ____ % (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 7.07

    n = 25 ÷ 0.125 = 200. CV_lim = 100 ÷ √200 = 100 ÷ 14.142 = 7.07 %.

  14. A yarn of 20 tex has a diameter of 0.18 mm and a fibre density of 1.50 g/cm³. Using φ = tex ÷ [ρ × (π d²/4) × 1000], with d in mm, tex in g/km and ρ in g/cm³, the packing fraction is ____ (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.52

    π d²/4 = π × 0.0324 ÷ 4 = 0.02545 mm². Multiply by ρ = 1.50 and by 1000 to get 38.17. φ = 20 ÷ 38.17 = 0.524, which is 0.52.

  15. A 30 tex polyester yarn (fibre density 1.38 g/cm³) has a packing fraction of 0.60. Using d = √[4 × tex ÷ (π ρ φ × 1000)] mm, with tex in g/km and ρ in g/cm³, the yarn diameter is ____ micrometres (nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 215

    ρ φ = 1.38 × 0.60 = 0.828. 4 × 30 = 120; π × 0.828 × 1000 = 2601.2. d² = 120 ÷ 2601.2 = 0.04613 mm², so d = 0.2148 mm = 215 micrometres.

  16. In the ideal helical model the helix angle α at the surface of a yarn follows tan α = π d T, with d the yarn diameter and T the turns per unit length in the same length unit. A yarn of diameter 0.17 mm carries 800 turns per metre. The surface helix angle is ____ degrees (nearest whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 23

    T = 800 turns/m = 0.8 turns/mm. tan α = π × 0.17 × 0.8 = 0.4273, so α = arctan 0.4273 = 23.1°, which is 23 degrees.

  17. A drafted strand of nominal linear density 20 tex shortens by 4 % on being twisted into yarn (the length after twisting is 0.96 of the length before). The mass is unchanged. The linear density of the yarn is ____ tex (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 20.83

    The same mass now lies in 0.96 of the length, so tex = 20 ÷ 0.96 = 20.83. Multiplying by 0.96 (19.2) is the usual slip.