Critical Fibre Length

A continuous fibre running the whole length of a part is gripped at both ends by whatever the part is bolted to, and it carries its share of the load directly. A discontinuous fibre — a chopped strand in an injection moulding, a whisker, a piece of a fibre that broke — has free ends, and a free end carries no tension at all. Load can only get into it through shear along its curved surface, building up from zero at each tip. The question this equation answers is how much length that takes: $l_c = \sigma_f d / (2\tau_i)$.

The derivation is a two-line force balance and it is worth doing, because it shows where the 2 comes from. Consider half a fibre. The shear delivered over that half of the surface is $\tau_i \cdot \pi d \cdot (l/2)$. The tension it has to build up to is $\sigma_f \cdot \pi d^2/4$. Set them equal, cancel $\pi d$, and $l = \sigma_f d / (2\tau_i)$. The factor of two is there because the fibre is loaded from both ends and each end only has to deliver half.

What the answer means is a hard split between two kinds of composite. A fibre shorter than $l_c$ can never break, no matter how hard the part is pulled: the shear runs out before the tension reaches the fibre's strength, and the fibre slides out of the resin instead. That is pull-out, and a moulding full of sub-critical fibres fails at a fraction of the load its fibre content suggests. A fibre much longer than $l_c$ spends most of its length at full stress and behaves nearly like a continuous one. In between, the average stress in the fibre is reduced roughly by a factor $(1 - l_c/2l)$.

Put numbers to it. A 7 µm carbon filament with a 3500 MPa strength and a 40 MPa interfacial bond gives $l_c = 0.31$ mm — an aspect ratio $l_c/d$ of about 44. That is why injection-moulded short-fibre compounds fall so far short of laminate properties: the pellets may start at 3 mm, but every pass through a screw, a check valve and a gate breaks them further, and what ends up in the part is frequently near or below the critical length. It also explains why the moulding process matters as much as the compound, and why long-fibre thermoplastic and direct-compounding processes exist at all — they are attempts to keep the fibres above $l_c$.

Read backwards, the equation answers the more practical question: given the fibre length actually present in the part, what is the most stress those fibres can ever reach? If that comes out below the filament's own strength, the fibres are pulling out and the strength you paid for is not being used. The fix is almost always longer fibres or a better interface, not a stronger fibre.

Two honesty points about the inputs. $\sigma_f$ is not a single number. Brittle filaments follow Weibull statistics: strength is set by the worst flaw present, a longer fibre contains more flaws, and so a long fibre is genuinely weaker than a short one. The critical length therefore depends slightly on the length you are asking about, and the equation cannot represent that. And $\tau_i$ is the least reliable quantity on the page. It is a property of a bond a few nanometres thick, it moves with the sizing chemistry, the resin, the cure cycle and the compressive grip the resin keeps on the fibre as it shrinks, and it is measured by single-fibre pull-out or fragmentation tests that do not always agree with one another. The fragmentation test is itself this equation run backwards: pull a single embedded filament until the fragments are all too short to break again, and the settled fragment lengths fall between $l_c/2$ and $l_c$.

A last thought that cuts against the obvious. A stronger interface is not always better. Pull-out dissipates energy, and it is one of the reasons a composite can be tougher than either of its constituents; a ceramic-matrix composite is deliberately given a weak interface so that fibres pull out rather than the crack running straight through. The interface strength that maximises strength and the one that maximises toughness are not the same number.