Textile Testing of Fabrics: Mechanical, Surface and Comfort Properties
2. Stiffness, shear, drape and crease recovery
Stiffness (cantilever method). A strip is pushed forward over the edge of a horizontal platform until its tip drops to a plane inclined at 41.5° to the horizontal. The overhang length l gives the bending length c = l ÷ 2, and the flexural rigidity per unit width is G = W c³, where W is the weight per unit area (N/m², using g = 9.81 m/s²) and c is in metres, giving G in N·m. The cube means a small change in bending length is a large change in stiffness, and stiffness rises with the weight of the fabric at a given bending length. For l = 4.0 cm and 120 g/m²: c = 2.0 cm, G = 0.12 × 9.81 × 0.02³ = 9.4 × 10⁻⁶ N·m.
Shear. In-plane shear is the angular distortion of the grid of yarns (trellising). The shear rigidity is G_s = shear force per unit width ÷ shear angle (in radians or degrees). A low G_s lets a fabric conform to a double-curved surface without wrinkling, so it is important for tailoring and draping, and the hysteresis of the shear loop measures friction between yarns. Drape is the ability to hang in graceful folds. In the Cusick drape tester a circular specimen (for example 30 cm in diameter) rests over a smaller supporting disc (18 cm); the shadow of the draped ring is traced and its area A_d found. Drape coefficient F = (A_d − A_s) ÷ (A_D − A_s) × 100%, where A_D is the area of the specimen and A_s of the disc. A low F means a limp, more drapeable fabric, and F = 100% means a perfectly stiff one.
Crease recovery. A specimen is folded under a stated load for a stated time, released, and allowed to recover; the crease recovery angle is the angle between the two arms after recovery. Taking 180° as full recovery of a fold (flat) the percentage recovery is angle ÷ 180 × 100, and warp and weft angles are often added. Resilient fibres (wool, polyester) and resin-finished cotton recover well, and a fibre of poor elastic recovery such as untreated cotton or flax recovers poorly. Wet crease recovery is measured after wetting, for wash-and-wear fabrics.
3. Tensile, tear and bursting tests
Tensile testing of fabric. In the strip test a specimen of stated width (for example 50 mm) and gauge length is gripped and extended at a constant rate to break, and the result is the breaking force and extension, quoted per specimen width because thickness is ill-defined. In the grab test only the central part of a wider specimen is gripped, so neighbouring yarns also carry load. For woven cloth, the strip strength is about the number of yarns across the width × the single-yarn breaking force × a utilisation factor below 1, since yarns are crimped and break at unequal strains. For 140 yarns of 3.2 N and utilisation 0.85, the strip force is 380.8 N. Knitted fabrics extend far more at break.
Tear strength. The tongue (trouser) tear pulls the two legs of a slit specimen apart, so the tear advances and the force is recorded as the yarns break in turn; the Elmendorf tester uses a falling pendulum, and the energy it loses in tearing a stated length gives the average force, F = E ÷ d. Wing-rip and trapezoid tests are variants. Only a few yarns are loaded at a time, so tear force is far below strip strength. Yarns can slide and bunch together to share the load, so a loosely woven or less heavily finished cloth often tears harder than a tight one of the same yarn.
Bursting strength. A circular area of fabric is clamped and inflated by a rubber diaphragm (hydraulic or pneumatic), or pushed by a steel ball, until it bursts, giving a biaxial test: the result is the bursting pressure or force, and it suits knitted fabrics, nonwovens and parachute cloth that are loaded in all directions. For a dome of spherical shape the membrane tension per unit width is N = P R ÷ 2, where R = (a² + h²) ÷ (2h) for clamp radius a and dome height h. At P = 300 kPa, a = 12.5 mm and h = 8 mm: R = 13.77 mm and N = 300 × 10³ × 0.01377 ÷ 2 = 2065 N/m (2.07 N/mm).
4. Pilling and abrasion resistance
Pilling is the formation of small entangled balls of fibre (pills) on the surface of fabric in wear. The sequence is fuzz formation (fibre ends migrate out), entanglement into balls, anchoring of the pill by fibres still in the cloth, and wear-off of the pill when the anchor breaks. Strong, abrasion-resistant fibres (polyester, nylon) hold on to their pills and so show more lasting pilling, whereas cotton and wool pills break away quickly. Yarn hairiness, low twist, short staple and a loose structure favour pilling.
Pilling is tested by the Martindale method (specimen rubbed in a Lissajous path against itself or a standard abradant), by the ICI pilling box or random-tumble method (specimens tumbled in a lined drum), after a stated number of cycles, and graded visually against photographic standards, commonly from 5 (no pilling) to 1 (very severe). Abrasion resistance is tested by rubbing under a stated pressure by the Martindale instrument (end point a stated breakdown of threads, or the number of rubs) or the Taber instrument (rotating specimen under abrasive wheels, result as mass loss or cycles). It is reported as cycles to the end point, per cent mass or thickness loss or loss of strength. More cycles to the end point means better resistance, and results from different instruments and abradants are not interchangeable.
5. Objective evaluation of low-stress mechanical characteristics
Fabric hand (the feel in the fingers) and tailoring behaviour depend on deformations far smaller than those of a breaking test: the strains of wear are small, often below a few per cent. Low-stress mechanical characteristics are therefore measured over small, reversible deformations of tension, shear, bending, compression and surface friction. The Kawabata Evaluation System for Fabrics (KES-F) does this on four instruments, and the FAST (Fabric Assurance by Simple Testing) system does it with simpler tools for the worsted and tailoring industry: a compression meter for thickness, a bending meter, an extension meter and a dimensional-stability test.
| Instrument and mode | Parameters | Meaning |
|---|---|---|
| FB1 tension | LT, WT, RT | Linearity of the load-extension curve, tensile energy and tensile resilience |
| FB1 shear | G, 2HG, 2HG5 | Shear stiffness and shear hysteresis at two shear angles |
| FB2 pure bending | B, 2HB | Bending rigidity per unit width and bending hysteresis |
| FB3 compression | LC, WC, RC | Linearity of compression, compressional energy and compressional resilience |
| FB4 surface | MIU, MMD, SMD | Mean coefficient of friction, its mean deviation and geometrical roughness |
| Also | T, W | Thickness and weight per unit area |
Using the results. The hysteresis terms measure energy not recovered: a high 2HG or 2HB means the fabric has been set by the deformation and will not spring back. Kawabata related the parameters to hand through regression equations that give primary hand values (stiffness, smoothness, and fullness with softness) and a total hand value from them, but the weighting factors are specific to the fabric class they were fitted for, so a hand value is not transferable between, say, suiting and knitwear. The same data feed tailoring and drape predictions, and a bending curvature K and bending moment per unit width M are related by M = B K for small K. For B = 2.0 × 10⁻⁵ N·m and K = 40 m⁻¹ the moment per unit width is 8.0 × 10⁻⁴ N.
6. Air permeability, wetting and wicking
Air permeability is the volume flow of air through unit area of fabric under a stated pressure difference. A circular area of fabric is clamped, a fan draws air through it, the pressure drop is held at the stated value (for example 100 Pa for clothing fabrics) and the flow is read; air permeability = Q ÷ A, in mm/s or L/m²/s. For Q = 0.30 L/s through 20 cm²: 3.0 × 10⁻⁴ ÷ 2.0 × 10⁻³ = 0.15 m/s = 150 mm/s. Flow is controlled by the open area and pore size, so it falls with cover, with calendering and with swelling when wet, and it is not strictly proportional to the pressure drop at larger flows. It governs wind-proofness, breathability and filtration.
Wetting. A liquid wets a solid when the contact angle θ is below 90°, with Young’s relation cos θ = (γ_SV − γ_SL) ÷ γ_LV. Roughness amplifies the tendency: a wetting surface wets more and a repellent one repels more. Tests are the drop (wetting-time) test, the spray test (visual rating of surface wetting) and the hydrostatic head test. A pore of radius r resists penetration of water with θ > 90° up to a pressure Δp = 2 γ |cos θ| ÷ r, so halving the pore radius doubles the pressure the fabric withstands, and the water head is Δp ÷ (ρ g). For r = 25 μm, θ = 110° and γ = 0.072 N/m: Δp = 2 × 0.072 × 0.342 ÷ 25 × 10⁻⁶ = 1970 Pa, a head of about 0.20 m.
Wicking is the spontaneous capillary flow of a liquid along the fabric. For a capillary of radius r the Lucas–Washburn relation, ignoring gravity, is h² = γ r cos θ t ÷ (2 η), so the height rises as √t and falls with viscosity. Gravity limits the final height to 2 γ cos θ ÷ (ρ g r). Larger pores give a faster early rise, while smaller pores and a more hydrophilic surface (smaller θ) give a greater final height. Wicking is measured by the rise of a dyed liquid up a vertical strip against time. For γ = 0.072 N/m, r = 10 μm, cos θ = 0.8, η = 1.0 mPa·s and t = 20 s: h² = 0.072 × 10⁻⁵ × 0.8 × 20 ÷ (2 × 10⁻³) = 5.76 × 10⁻³ m², h = 76 mm.
7. Water-vapour transmission and thermal resistance
Water-vapour transmission. In the cup method a cup holds a desiccant or water, is covered with the fabric and sealed at the edge, and is placed in a controlled atmosphere; the cup’s mass change gives WVT = Δm ÷ (A t), in g/m²/day, where A is the exposed area. For Δm = 0.30 g in 4 h through 25 cm²: 0.30 ÷ 0.0025 = 120 g/m² in 4 h = 720 g/m²/day. The sweating guarded hot plate (a skin model) measures the heat needed to keep a wet plate at skin temperature under the fabric, and gives the evaporative resistance R_et = (p_plate − p_air) ÷ (evaporative heat flux per unit area) in Pa·m²/W, where a lower value means a more breathable fabric. A real fabric’s resistance includes the still-air layer at its surface, so the thickness of that layer, and not only the fabric, matters.
Thermal resistance. A guarded hot plate held at skin temperature T_s under the fabric loses heat flux q (W/m²) to air at T_a; the dry thermal resistance is R_ct = (T_s − T_a) ÷ q, in m²·K/W. The clothing unit is 1 clo = 0.155 m²·K/W (1 tog = 0.1 m²·K/W). Fabric warmth comes mostly from trapped still air, so for a fibrous layer R ≈ t ÷ k_eff, rising with thickness t and falling when the fabric is compressed or wet; layers in series add, R_total = R₁ + R₂ + R_air-gap, and wind lowers the surface resistance. For t = 2.0 mm and k_eff = 0.040 W/m·K: R = 0.002 ÷ 0.040 = 0.050 m²·K/W = 0.32 clo.
Key takeaways
- Thickness is read at a stated pressure; compressibility = (T₁ − T₂) ÷ T₁ × 100. Flexural rigidity G = W c³ with bending length c = overhang ÷ 2 (cantilever at 41.5°) and W in N/m². Drape coefficient F = (A_d − A_s) ÷ (A_D − A_s) × 100, and a low F means more drapeable.
- Shear rigidity governs conformity to curved surfaces. Crease recovery angle measures resilience after folding. Strip strength of woven fabric ≈ yarns across the strip × yarn force × utilisation below 1.
- Tear tests load only a few yarns at a time, so tear force is far below strip strength, and yarn mobility raises it. Bursting is biaxial, with membrane tension N = P R ÷ 2 and R = (a² + h²) ÷ (2h).
- Pills form from fuzz and persist when fibres are strong; Martindale and ICI box test pilling, Martindale and Taber test abrasion, and results from different methods are not interchangeable.
- KES-F measures tension (LT, WT, RT), shear (G, 2HG), bending (B, 2HB), compression (LC, WC, RC) and surface (MIU, MMD, SMD) at low stress; FAST is a simpler set. Hysteresis terms show energy not recovered, and hand-value equations are fabric-class specific.
- Air permeability = Q ÷ A at a stated pressure drop. Wetting needs θ < 90°; breakthrough pressure Δp = 2 γ |cos θ| ÷ r. Lucas–Washburn gives h² = γ r cos θ t ÷ (2 η). WVT = Δm ÷ (A t); R_ct = (T_s − T_a) ÷ q, with 1 clo = 0.155 m²·K/W and R ≈ t ÷ k_eff.
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.
A fabric has a thickness of 3.20 mm at a light pressure of 1 kPa and 2.40 mm at a heavy pressure of 50 kPa. Taking compressibility as (T₁ − T₂) ÷ T₁ × 100 with T₁ the thickness at the light pressure, what is it?
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Answer: B — 25.0%
Compressibility = (3.20 − 2.40) ÷ 3.20 × 100 = 25.0%. Dividing the lost thickness by the compressed thickness gives 33.3%, which is the usual slip, and 75.0% is the ratio of the two thicknesses.Two fabrics X and Y have the same bending length in a cantilever test, but fabric X has twice the mass per unit area of fabric Y. Using G = W c³, what is the ratio of the flexural rigidity of X to that of Y?
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Answer: B — 2
With c equal, G is proportional to W, so doubling W doubles G. The cube applies to c, not W, so 8 is the slip of cubing the weight ratio. An equal rigidity of 1 would need X to have a bending length smaller by the cube root of 2.Which fabric property chiefly determines whether a woven cloth can be shaped over a double-curved surface, such as a hemisphere, without wrinkling?
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Answer: C — A low in-plane shear rigidity, which lets the yarn grid trellis
To cover double curvature the cloth must change the angle between warp and weft, which is in-plane shear, so low shear rigidity is the enabling property. A high tensile modulus makes the cloth resist the extension that would otherwise help, and air permeability and crease recovery are unrelated to conformity.A fabric is folded flat (taken as 0°) and, after the test, the two arms spring back to an angle of 126°, where 180° would be complete recovery. What is the percentage recovery?
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Answer: C — 70%
Percentage recovery = 126 ÷ 180 × 100 = 70%. The figure 54 is the unrecovered angle in degrees, and quoting the angle itself as a percentage gives the impossible value 126%.A woven fabric gives a trouser-tear force that is far lower than the force in a strip tensile test on the same cloth. What is the main reason?
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Answer: A — In tearing, only a few yarns at the tip of the tear carry the load at any moment, whereas a strip test loads all the yarns across the width
The load is concentrated on the few yarns being broken at the tear tip, each carrying a large share, while the strip test shares the load across the whole width. The rate difference is not the cause, only the slit is cut and not the yarns being tested, and the Elmendorf test also reports an average force.A polyester knitted fabric keeps unsightly pills on its surface after wear, whereas a similar cotton fabric loses its pills quickly. What is the usual explanation?
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Answer: B — The strong, abrasion-resistant polyester fibres that anchor the pill do not break, so the pill stays attached
A pill is held by anchor fibres, and strong polyester anchors resist breaking, so pills persist, while weaker cotton anchors break and the pills drop off. Cotton can pill. The water-uptake and melting explanations are wrong, since polyester absorbs little water and cotton does not melt.In a hydrostatic head test, water does not wet the pores of a fabric (contact angle above 90°, the same liquid and surface throughout). If the effective pore radius is halved, how does the water pressure needed for breakthrough, Δp = 2 γ |cos θ| ÷ r, change?
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Answer: D — It doubles
Δp varies as 1/r, so halving the radius doubles the pressure the fabric withstands. A finer pore holds water out better. Quadrupling would need the radius to fall to a quarter, and halving would be the wrong direction.Air flows through a circular fabric specimen of area 20.0 cm² at a volume flow of 0.45 L/s under the stated pressure drop. What is the air permeability, in mm/s?
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Answer: B — 225
Q = 0.45 L/s = 4.5 × 10⁻⁴ m³/s and A = 20 cm² = 2.0 × 10⁻³ m², so Q ÷ A = 0.225 m/s = 225 mm/s. The value 0.225 is the same quantity in m/s but the wrong unit, and the other values come from slipping a factor of ten.In a cup test, the cup mouth has an area of 30.0 cm² and the cup gains 0.45 g of water vapour through the fabric in 6.0 h. What is the water-vapour transmission rate in g/m²/day?
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Answer: C — 600
0.45 g ÷ 0.0030 m² = 150 g/m² in 6 h, and 6 h is a quarter of a day, so the rate is 150 × 4 = 600 g/m²/day. The value 150 is the 6-hour figure, and 25 is that figure per hour.Which of the following statements about the Kawabata (KES-F) low-stress parameters are correct?
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Answer: A — B is the bending rigidity per unit width; B — MIU is the mean coefficient of friction of the surface
B (bending) and MIU (surface friction) are defined correctly. RC is the compressional resilience, the tensile resilience being RT, and shear stiffness G is measured together with tension on the first instrument, not on the compression one.Which of the following statements about fabric strength tests are correct?
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Answer: A — A strip tensile result is quoted per width of the specimen, as fabric thickness is ill-defined; B — The strip strength of a woven fabric is about the number of yarns across the strip times the single-yarn force times a utilisation factor below 1
Strip results are reported per width, and the yarn-count-times-force estimate is the usual strength model. Bursting is a biaxial test, and in a tear only the yarns at the tip carry the load.Which of the following statements about comfort-related fabric tests are correct?
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Answer: A — In the Lucas–Washburn relation the square of the wicking height grows in proportion to time; B — A higher dry thermal resistance R_ct means less heat loss through the fabric for the same temperature difference; C — One clo equals 0.155 m²·K/W
The statements on Lucas–Washburn, thermal resistance and the clo are standard: h² ∝ t when gravity is ignored, heat flux = ΔT ÷ R_ct, and 1 clo = 0.155 m²·K/W. Flow through a fabric is proportional to the pressure drop only at low flows, because inertial losses add a non-linear part at higher flows.In a cantilever stiffness test a fabric of mass per unit area 180 g/m² has an overhang length of 6.0 cm when its tip reaches the 41.5° inclined plane. Take the bending length c = overhang ÷ 2, g = 9.81 m/s² and G = W c³ with W the weight per unit area in N/m². The flexural rigidity per unit width, in µN·m to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 47.7
c = 6.0 ÷ 2 = 3.0 cm = 0.030 m. W = 0.180 kg/m² × 9.81 = 1.766 N/m². G = 1.766 × 0.030³ = 1.766 × 2.7 × 10⁻⁵ = 4.77 × 10⁻⁵ N·m = 47.7 µN·m. Cubing the overhang (6.0 cm) instead of the bending length gives eight times this value.In a Cusick drape test a circular specimen of diameter 30 cm is draped over a supporting disc of diameter 18 cm, and the area of the shadow of the draped specimen is 400 cm². Using F = (A_d − A_s) ÷ (A_D − A_s) × 100, where A_D is the area of the specimen and A_s the area of the disc, the drape coefficient in per cent, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 32.2
A_D = π × 15² = 706.9 cm² and A_s = π × 9² = 254.5 cm². F = (400 − 254.5) ÷ (706.9 − 254.5) × 100 = 145.5 ÷ 452.4 × 100 = 32.2%. Using the diameters instead of the radii in the areas gives different, wrong values.A fabric is burst by a pneumatic diaphragm. The clamped circular area has radius a = 15.0 mm, the fabric dome has a height h = 10.0 mm at bursting, and the pressure is P = 400 kPa. Treating the dome as a spherical cap, the membrane tension per unit width N = P R ÷ 2, with R = (a² + h²) ÷ (2h), in N/mm is ____. (give two decimal places)
Numerical answer — type the value.
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Answer: 3.25
R = (15.0² + 10.0²) ÷ (2 × 10.0) = 325 ÷ 20 = 16.25 mm = 0.01625 m. N = 400 × 10³ × 0.01625 ÷ 2 = 3250 N/m = 3.25 N/mm. Using the clamp radius 15.0 mm as R would give 3.00 N/mm.Water of surface tension 0.072 N/m and viscosity 1.0 mPa·s rises by capillary action in a fabric whose pores act as capillaries of radius 15 μm with contact angle cos θ = 0.7. Ignoring gravity and using h² = γ r cos θ t ÷ (2 η), the height reached in 16 s, in mm to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 77.8
h² = 0.072 × 15 × 10⁻⁶ × 0.7 × 16 ÷ (2 × 1.0 × 10⁻³) = 1.2096 × 10⁻⁵ ÷ 2 × 10⁻³ = 6.048 × 10⁻³ m². h = 0.0778 m = 77.8 mm. Taking h² as the height itself, or leaving the factor 2 out, would give a wrong value.A fabric layer of thickness 3.0 mm has an effective thermal conductivity of 0.040 W/m·K, and its thermal resistance is taken as R = t ÷ k_eff. With 1 clo = 0.155 m²·K/W, the thermal insulation of the layer in clo, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.48
R = 0.0030 m ÷ 0.040 W/m·K = 0.075 m²·K/W. In clo: 0.075 ÷ 0.155 = 0.484, so 0.48. Using the thickness in mm without converting to metres gives a value one thousand times too large.