Choose a fibre and a matrix from the library and the page calculates the unidirectional ply, the stiffness and strength of a laminate, short-fibre efficiencies and a weight comparison with metals. With the defaults, T300 carbon fibre in 8552 epoxy at 60 % fibre volume, the rule of mixtures gives a longitudinal modulus of 139.9 GPa, close to the 140 GPa measured in Toray’s data sheet (60 %, a different epoxy). The transverse modulus is 7.96 GPa by the inverse rule of mixtures and 9.46 GPa by Halpin–Tsai. In the laminate tab, [0/±45/90]s of T300/5208 under Nx = 100 N/mm first fails in the 90° plies at 2.76 times the load (Tsai–Wu).
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Results
The micromechanics equations approximate straight, uniformly spaced fibres perfectly bonded to the matrix, with both phases linear elastic. Classical lamination theory treats a thin plate in plane stress (no interlaminar stresses or edge effects), and the failure criteria are empirical expressions applied to ply stresses. For design, use test data for your material and appropriate safety factors.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
Deriving the rule of mixtures, and how well it works
Pulled along the fibres, fibre and matrix stretch by the same amount like springs in parallel (isostrain). The stress is the volume-weighted average and E1 = VfEf1 + VmEm. Pulled across the fibres they can be taken to carry the same stress, like springs in series (isostress), giving 1/E2 = Vf/Ef2 + Vm/Em. These are the Voigt and Reuss estimates. Energy principles make the isostress formula a rigorous lower bound on the modulus; the isostrain formula is an upper bound only when fibre and matrix have the same Poisson’s ratio (when they differ, the true E1 is very slightly above the rule of mixtures; Hill 1964). Along the fibres the strain really is nearly uniform, so the E1 formula works well: T300 (230 GPa) at Vf 60 % in a 4.67 GPa epoxy gives 139.9 GPa, almost the 140 GPa of the unidirectional composite (Vf 60 %) in Toray’s data sheet. Across the fibres, stress concentrates in the matrix between fibres and the fibres take part of the load sideways, so the isostress formula is too low.
The Halpin–Tsai equations and the meaning of ξ
The Halpin–Tsai equations P = Pm(1 + ξηVf)/(1 − ηVf), η = (Pf/Pm − 1)/(Pf/Pm + ξ), are an interpolation built to fit exact elasticity solutions. ξ → 0 gives the inverse rule and ξ → ∞ the rule of mixtures, so ξ measures how effective the reinforcement is for the fibre shape and arrangement. For circular fibres in a square array, ξ = 2 is common for E2 and ξ = 1 for G12. Above about 60 % fibre the interaction between fibres raises the values, and ξ = 1 + 40Vf¹⁰ (Hewitt and de Malherbe) is sometimes used for G12. Carbon and aramid fibres are themselves anisotropic: their radial modulus is a few GPa to about 20 GPa, less than a tenth of the axial value. Using the axial Ef1 in the E2 calculation greatly overestimates E2.
Thermal expansion (Schapery) and residual stress
The longitudinal expansion is the stiffness-weighted average α1 = (Ef1αf1Vf + EmαmVm)/(Ef1Vf + EmVm). Carbon fibre contracts along its axis on heating (αf1 < 0) and is far stiffer than the matrix, so α1 is close to zero. Transversely, the matrix, held back along the fibres, also bulges sideways by the Poisson effect, so α2 = Vf(αf2 + νf12αf1) + Vm(1 + νm)αm − α1ν12, a large value close to that of the matrix. Because of this mismatch, when a laminate cools from the cure temperature (ΔT roughly −100 K to −160 K) the plies restrain each other and thermal residual stresses appear. In 90° plies they are transverse tension and bring first ply failure forward. Unsymmetric laminates warp on cooling for the same reason.
Longitudinal tensile strength is set by failure strains
Pulled along the fibres, fibre and matrix have the same strain. Common resins stretch further than carbon fibre can (1.5 % for T300), so the fibres break first and the strength is σ1u = Vfσf + Vmσm′, where σm′ is the matrix stress when the fibres break. With few fibres the matrix alone carries Vmσm after they break, so reinforcement only pays off above the critical volume fraction Vcrit. When the matrix fails first (for example 4.8 % strain E-glass in a brittle epoxy), the strength is the larger of two expressions: the fibres alone after cracking, or failure of the whole at matrix cracking. Real strengths are lower than this because fibre strength scatters (fibres break at weak points and load concentrates on neighbours) and alignment is imperfect. Toray’s measured 1820 MPa for unidirectional T300 is about 85 % of this formula. Compressive strength is controlled by microbuckling of fibres (kink bands) and cannot be obtained from it.
Converting volume fraction, mass fraction and voids
Micromechanics uses volume fractions. Prepregs and resin formulations are specified by mass, so convert with the densities: Vf = (Wf/ρf)/(Wf/ρf + Wm/ρm). Carbon fibre (1.77 g/cm³) with PEEK (1.30 g/cm³) at 42 % resin by mass (Wf 58 %) gives Vf of about 50 %. In a real laminate the fibre content is measured by burning off or digesting the resin and weighing the fibres (ASTM D3171), and voids are found from the density (ASTM D2734). Voids do more than lower density and E1 as a zero-stiffness phase: acting as stress raisers, they are reported to cut transverse modulus and flexural properties by about 1.5 % per 1 % voids, and interlaminar shear strength by much more. The packing limit for equal circular fibres is π/4 = 78.5 % for a square array and π/(2√3) = 90.7 % for a hexagonal one.
Processing route and actual volume fraction
Volume fraction is set by the process, not the material. Hand lay-up tends to give low volume fractions and many voids; vacuum infusion processes such as VARTM give higher volume fractions than hand lay-up. Autoclave-cured prepreg achieves high, consistent volume fractions, and the unidirectional tests in Toray’s fibre data sheets are normalised to Vf 60 %. Injection-moulded short-fibre compounds have short fibres whose orientation follows the flow and changes through the thickness, so they need a different approach (the short-fibre tab). The same fibre and resin give different volume fraction, voids and fibre waviness by different routes, and so different stiffness and strength; keep this in mind when using library values.
Assumptions of classical lamination theory and the ABD matrix
Classical lamination theory assumes that straight lines normal to a thin plate stay straight and normal after bending (Kirchhoff), that each ply is in plane stress and linear elastic, and that the plies are perfectly bonded. Then the strain is ε = ε⁰ + zκ (mid-plane strain ε⁰, curvature κ) and the ply stress is σ = Q̄(ε⁰ + zκ). Integrating through the thickness relates the force resultants N and moments M by {N; M} = [A B; B D] {ε⁰; κ}. A = ΣQ̄t is the in-plane stiffness, D = ΣQ̄(difference of z³)/3 the bending stiffness, and B = ΣQ̄(difference of z²)/2 the stretching–bending coupling. Effective in-plane moduli come from the inverse of ABD, as Ex = 1/(h a*11). Interlaminar stresses (the cause of delamination), free-edge effects and transverse shear in thick plates are not part of the theory.
Symmetric, balanced and quasi-isotropic laminates
In a symmetric laminate, plies mirror each other about the mid-plane, so B = 0: it does not bend when pulled and does not warp when cooled. In a balanced laminate there are equal numbers of +θ and −θ plies, so A16 = A26 = 0 and tension produces no shear strain. Even balanced laminates such as [±45]s have D16 and D26 ≠ 0, because the +45° and −45° plies sit at different heights, so they twist when bent. Equal numbers of plies at angles spaced by 180°/n (n ≥ 3), as in [0/±45/90]s, make A isotropic in the plane: Ex is the same in every direction (quasi-isotropic). D is still not isotropic, so the bending stiffness varies with direction. Quasi-isotropic laminates are common in parts whose load direction is uncertain and in parts with holes, because they avoid weak directions without fibres. The price is much of the specific stiffness of the unidirectional ply (for T300/5208, E1 is 181 GPa but the quasi-isotropic Ex is about 70 GPa).
Failure criteria, first ply and last ply
The maximum stress criterion says a ply fails when any of σ1, σ2 or τ12 in ply axes reaches its strength (X, X′, Y, Y′, S). It has no interaction between components but tells you the mode (fibre tension, matrix tension, shear and so on). Tsai–Hill extends the von Mises idea to anisotropy as a quadratic, so combinations of stresses matter; for materials with different tensile and compressive strengths X and Y are chosen by the sign of the stress. Tsai–Wu is a tensor polynomial with linear and quadratic terms that handles tension–compression differences in one expression. F12 needs biaxial tests, and F12* = −1/2 (equivalent to von Mises) is widely used. In the World-Wide Failure Exercise (WWFE) no criterion predicted all the test cases. In a laminate the remaining plies can carry load after the first ply fails (usually matrix cracking in 90° plies), so first ply failure (FPF) and last ply failure (LPF) are considered separately.
Short fibres: critical length and efficiencies
Load enters a short fibre gradually from its ends by interfacial shear stress τ. The force balance πd²σ/4 = πd(l/2)τ gives the shortest length at which the stress at the middle of the fibre reaches its strength σf, the critical length lc = σfd/(2τ). Fibres shorter than lc pull out without breaking, and even longer fibres are under-used over lc/2 at each end. Kelly–Tyson, assuming the interface slides at constant τ, estimates the strength of aligned short fibres for l ≥ lc as Vfσf(1 − lc/2l) + Vmσm′. For the modulus, Cox’s shear-lag theory, from the shear coupling between a fibre and the surrounding matrix, gives the length efficiency ηl = 1 − tanh(βl/2)/(βl/2). Orientation is handled by Krenchel’s efficiency η0 = Σ an cos⁴θn (3/8 for random in plane, 1/5 for random in 3D), which multiplies the fibre term of the modulus. τ depends strongly on the fibre surface treatment and sizing, so use values measured by fragmentation or pull-out tests.
Comparing at equal stiffness and weight
How to compare materials depends on what limits the design. If a tie must have a given stiffness, mass scales with ρ/E; if a panel must have a given bending stiffness, with ρ/∛E; if it must carry a given load, with ρ/σ (Ashby’s material indices). Because the bending stiffness of a panel grows with the cube of its thickness, low density pays off strongly. With this tool’s default values and steel (S355) set to 1, the mass of a panel with the same bending stiffness is about 0.50 for aluminium alloy 6061-T6, about 0.28 for [0/±45/90]s CFRP (NTNU CF A, Vf 55 %) and about 0.50 for the same layup in GFRP (E-glass). In tension, quasi-isotropic CFRP has Ex of only about 50 GPa, below aluminium alloy (69 GPa), so the mass gap shrinks to about 0.85 against 1.05. GFRP is about 2.7 times heavier than steel for the same tensile stiffness, and even as a panel in bending it only matches aluminium alloy (about 0.50). On strength, judged by first-ply failure (cracking of the 90° plies), GFRP is about as heavy as steel (about 1.1); judged by final failure it does better. The compare tab lines these calculations up under the same conditions and links to the index lines of the material selection chart.
When these calculations do not apply
Fibre waviness and crimp (woven or stitched fabrics), misalignment and fibre contact at high volume fractions take the material away from the micromechanics formulas. Matrix non-linearity (the curved shear stress–strain response), softening by temperature and moisture, and viscoelasticity (creep) are not covered. Classical lamination theory leaves out interlaminar stresses, three-dimensional stresses near free edges and holes, and transverse shear in thick plates. The failure criteria look only at ply stresses; delamination, impact damage, fatigue and compressive buckling need separate treatment. Library values are typical values from manufacturers, textbooks and lecture notes, not design allowables. Dashed fields hold assumed values with no source.
Volume is mass divided by density. In 100 g of composite, the fiber volume is 70/1.80 ≈ 38.9 cm³ and the resin volume is 30/1.20 = 25.0 cm³, so the fiber volume fraction is Vf = 38.9/(38.9 + 25.0) ≈ 0.61, about 61%. Because the fiber is denser than the resin, the volume fraction is smaller than the mass fraction. About 70% takes the mass fraction directly as the volume fraction; rule-of-mixtures calculations need the volume fraction. About 78% is the value calculated with the fiber and resin densities swapped. Real moldings may contain voids, so Vf and void content are checked by density measurement or the burn-off method.
ALoading parallel to the fibers is approximated by the isostrain form and loading perpendicular to them by the isostress form
BLoading parallel to the fibers is approximated by the isostress form and loading perpendicular to them by the isostrain form
CThe isostrain form holds in both directions, and Young's modulus does not depend on the fiber orientation
Show answer and explanation
Answer: A. Loading parallel to the fibers is approximated by the isostrain form and loading perpendicular to them by the isostress form
When loaded parallel to the fibers, fiber and matrix stretch by the same amount like springs connected in parallel, so the isostrain form E₁ = VfEf + VmEm applies. When loaded perpendicular to the fibers, fiber and matrix carry the same stress like springs in series, so the isostress form 1/E₂ = Vf/Ef + Vm/Em is used as an approximation. At the same volume fraction, the isostrain form gives an upper bound on Young's modulus and the isostress form a lower bound (the Voigt and Reuss bounds). The choice applying the isostress form to parallel loading and the isostrain form to perpendicular loading swaps parallel and perpendicular. The modulus of unidirectional material differs greatly between the fiber and transverse directions, so saying it does not depend on direction is also wrong. The actual transverse modulus is often somewhat higher than the isostress form predicts, and corrections such as the Halpin–Tsai equations are used.
Perpendicular to the fibers, fiber and resin are treated as lying in series and carrying the same stress, so 1/E₂ = Vf/Ef + Vm/Em is used. 1/E₂ = 0.50/72 + 0.50/3.0 = 0.00694 + 0.1667 = 0.1736 GPa⁻¹, so E₂ = 1/0.1736 ≈ 5.76 GPa, about 5.8 GPa. About 37.5 GPa is the modulus in the fiber direction from the isostrain form 0.50 × 72 + 0.50 × 3.0. About 0.17 GPa forgets to take the reciprocal. The transverse modulus stays close to that of the compliant resin, only about 15% (about 1/6.5) of the modulus in the fiber direction. This is why unidirectional material is not used in parts loaded perpendicular to the fibers.
The specific modulus is 140/1.6 = 87.5 for CFRP and 210/7.9 ≈ 26.6 for steel (both in GPa/(g/cm³)), so the ratio is 87.5/26.6 ≈ 3.3. CFRP has the lower Young's modulus itself, but its density is about one fifth, so at equal weight it is about 3.3 times as stiff. About 0.67 compares only the moduli (140/210), and about 4.9 looks only at the density ratio (7.9/1.6). Note that this value applies only in the fiber direction; perpendicular to the fibers, the specific modulus is far lower than that of steel. Real structures compensate for direction by laminating, so the specific-stiffness advantage is smaller than this value.
ANear zero at the ends, rising away from them through interfacial shear, and constant in the middle if the fiber is long enough
BUniform along the whole length because the interfacial shear stress is constant, and independent of fiber length
CMaximum at both ends, because the end faces take the load directly, and decreasing toward the middle of the fiber
Show answer and explanation
Answer: A. Near zero at the ends, rising away from them through interfacial shear, and constant in the middle if the fiber is long enough
Almost no load is transferred through the end faces of the fiber, so the stress in the fiber is nearly zero at its ends. Moving away from the ends, load enters little by little through the interfacial shear stress τ, and the stress rises roughly in proportion to the distance from the end. If the fiber is long enough, beyond a certain distance the strains of fiber and matrix match and the stress reaches a constant maximum. The length needed for this maximum to reach the tensile strength of the fiber is the critical fiber length lc, and the strength is reached at a distance lc/2 from the end. If the fiber is much longer than lc, the low-stress end portions are a small fraction, and the reinforcing effect approaches that of continuous fibers. A constant interfacial shear stress means that the fiber stress rises at a constant rate, not that the stress itself is uniform. The explanation that load enters through the end faces and the stress is greatest at the ends is the reverse.
AThe Young's moduli in the 0° and 90° directions are equal, and at 45° both plies contribute half each, so the laminate is fully isotropic in plane
BOnly the 0° direction has a high modulus; in the 90° direction no fibers carry load, so the modulus is that of the resin alone
CThe Young's moduli in the 0° and 90° directions are nearly equal, but in the 45° direction in-plane shear dominates and the modulus is lower
Show answer and explanation
Answer: C. The Young's moduli in the 0° and 90° directions are nearly equal, but in the 45° direction in-plane shear dominates and the modulus is lower
In [0/90]ₛ, loading in the 0° direction is carried along the fibers of the 0° plies and loading in the 90° direction along the fibers of the 90° plies, so the moduli in the two directions are nearly equal. Loaded at 45°, however, the fibers of every ply are 45° off the load direction, and the deformation becomes mainly in-plane shear supported by the resin, so the modulus drops considerably. To approach in-plane isotropy, ±45° plies are added to make a quasi-isotropic layup such as [0/±45/90]ₛ. Saying the 45° direction is the same overlooks this point. In the 90° direction the fibers of the 90° plies carry the load, so the modulus is not that of the resin alone. Note that in any layup the through-thickness direction is supported by the resin, so susceptibility to delamination remains.
FAQ
Should I use the inverse rule of mixtures or Halpin–Tsai for E2?
The inverse rule is a lower bound, so the real E2 is normally higher. For unidirectional composites with circular fibres, Halpin–Tsai (ξ = 2) is closer to measurements. For carbon and aramid fibres the radial modulus of the fibre is low, so it matters to enter the radial value, not the axial one, as Ef2.
How does the ply E(θ) differ from the laminate Ex?
The ply E(θ) is the modulus of a unidirectional ply pulled at θ to its fibres; it falls quickly even for small angles. The laminate Ex is the in-plane modulus of plies in several directions combined, from the inverse of the ABD matrix. For a quasi-isotropic laminate Ex is the same in every direction.
Is a laminate with B ≠ 0 always bad?
With B ≠ 0 the laminate bends when pulled and warps on cooling after cure, so parts that must keep their shape are usually symmetric. Some designs use coupling deliberately (bend–twist coupled blades, for example), but they still need the warping to be calculated.
Why can a laminate still carry load after the first ply fails?
The first failure is usually matrix cracking in plies transverse to the load, while their fibres are intact. The cracked plies lose transverse and shear stiffness and shed load to the others. Load can be carried until the fibres in the 0° plies break, so last ply failure can be well above first ply failure. Cracks can start delamination, leakage and fatigue, though, so designs often use first ply failure as the limit.
What value should I use for the Tsai–Wu F12*?
Strictly it comes from biaxial tests, but these are difficult, so F12* = −1/2 (F12 = −½√(F11F22)), recommended by Tsai and Hahn, is widely used. F12* must lie between −1 and 1 for the failure surface to be closed, and its value changes the predicted strength under biaxial loads.
Should I enter the mass fraction or the volume fraction?
Either. Resin content of prepregs and formulations is often given by mass; choose mass fraction and the densities convert it to a volume fraction. The formulas use volume fraction; for carbon fibre and epoxy the volume fraction is about 5–10 points lower than the mass fraction.
Are the Voigt and Reuss formulas really upper and lower bounds?
The lower bound is rigorous: a uniform stress satisfies equilibrium, so the complementary energy calculated with it is at least the true value (principle of minimum complementary energy), and the isostress modulus is at most the true one (Paul 1960). The upper bound reduces to VfEf + VmEm only when fibre and matrix have the same Poisson’s ratio. When they differ, the energy bound is a slightly larger expression, and the true longitudinal E1 is also very slightly above the rule of mixtures (Hill 1964). In practice E1 = VfEf1 + VmEm is accurate enough.