Long-fibre composite

A long-fibre composite contains reinforcing fibres whose lengths are large enough that end regions do not dominate their load-carrying response. In a well-bonded continuous aligned specimen pulled along the fibres, the fibres and composite matrix undergo approximately the same axial strain. Their different Young's moduli then produce different axial stresses, and adding their forces gives the longitudinal rule of mixtures.

Load enters a fibre through shear stress at its interface with the composite matrix. Near a free fibre end the axial stress must rise from zero, so a short fibre can spend much of its length below its useful load. The critical fibre length expresses the associated scale in a simple strength-transfer model. Whether a fibre is mechanically long therefore depends on its diameter, the interface and the required stress, rather than on an absolute length in millimetres. Changing fibre direction still changes the response: long reinforcement does not remove anisotropy.

Fibre-matrix load transfer describes how interfacial shear stress transfers axial force between a reinforcing fibre and its surrounding composite matrix. For a cylindrical fibre of diameter $d$, axial force balance gives $d{\sigma }_f/dx=4\tau /d$ when $\tau$ is the appropriately signed interfacial shear stress.

The fibre stress is therefore small near a free end and builds over a transfer length. Short fibres may never reach their tensile strength before the interface slips or the matrix yields. The critical fibre length quantifies this competition in a simplified constant-shear model.

The critical fibre length is the minimum embedded fibre length needed to develop a specified tensile stress in a simple interfacial-load-transfer model. For diameter $d$, fibre strength ${\sigma }_f^{\ast }$ and constant limiting interfacial shear stress ${\tau }_i$,Each half of the fibre builds axial force from its end through interfacial shear; balancing that transferred force against the strength load gives the formula. It is a model-dependent criterion, not a universal length of the material. Nonuniform shear, imperfect bonding and fibre orientation modify actual composite behavior.

The fibre volume fraction is the fraction of a composite material's total volume occupied by reinforcing fibres:For a specimen with only fibres and a composite matrix, their fractions sum to one. Voids require a third fraction; they must not be silently counted as composite matrix. The fibre volume fraction is different from the fibre mass fraction because the constituents can have different mass densities.

For a void-free material, a measured fibre mass fraction $w_f$ giveswhere ${\rho }_f$ and ${\rho }_m$ are the constituent mass densities. For example, masses in the ratio $3:2$ and mass densities in the ratio $3:1$ give $V_f=1/3$, not $3/5$. Inserting $V_f$ into the longitudinal rule of mixtures estimates axial Young's modulus; orientation and fibre-matrix load transfer remain separate requirements.

A hybrid fibre composite contains two or more types of reinforcing fibre in a common composite matrix. Combining carbon fibres and glass fibres, for example, provides another design choice besides changing the amount or orientation of a single reinforcement. The word hybrid specifies the combination of reinforcements, not a guarantee that every mechanical property improves.

For continuous, parallel fibres, perfect bonding and elastic axial loading, equal strain gives the extension of the longitudinal rule of mixturesHere the $V_i$ are constituent fractions by volume and the $E_i$ are axial Young's moduli. This estimate describes the initial elastic response. It does not by itself predict tensile strength: when one fibre population breaks, its load is redistributed through the composite matrix, changing the stress carried by the surviving population.

The longitudinal rule of mixtures estimates the axial Young's modulus of continuous aligned fibres in a composite matrix:The model assumes perfect bonding, small elastic strain, no voids and loading parallel to the fibres. Equal extension over a common initial length gives equal axial strain $\epsilon$ in both phases. Adding their forces gives the average axial stress $\sigma =V_f{E}_f\epsilon +(1-V_f)E_m\epsilon$, which proves the formula. It is the axial Voigt elastic approximation.

For example, $V_f=0.6$, $E_f=70\mathrm{GPa}$ and $E_m=3\mathrm{GPa}$ give $E_L=43.2\mathrm{GPa}$. These are illustrative inputs, not universal constituent values. The formula makes reinforcement efficient because the stiffer constituent carries the larger stress at the common strain. Loading across the fibres changes the compatibility and traction conditions, so the same formula is not a general transverse-modulus law. Fibre waviness or poor fibre-matrix load transfer can also invalidate the ideal axial estimate.