Irwin Plastic Zone Size

The elastic solution for a crack says the stress at the tip is infinite. No material tolerates that, and none is asked to: it yields instead, over a small region ahead of the tip, and the stress is capped at something near the yield strength. Irwin's estimate of how big that region is comes from the crudest possible argument — find the distance at which the elastic stress would have fallen to the yield strength, and call that the zone.

Setting $\sigma_y = K/\sqrt{2\pi r}$ and solving gives $r = (1/2\pi)(K/\sigma_y)^2$, which is the plane-stress result on this page. (Irwin then noted that the load carried by the truncated part of the elastic field has to go somewhere, and redistributing it roughly doubles the zone. Different books quote the first estimate and the corrected one, so a factor of two between two sources is usually this and not an error.) The plane-strain case replaces the 2 with a 6, and that factor of three is the whole subject of constraint.

Why constraint changes the zone threefold. In a thin sheet the material at the crack tip is free to contract through the thickness as it stretches, so there is no stress in that direction and the state is plane stress. In a thick section it is not free — the bulk of the plate around it holds it — so a tensile stress develops through the thickness too. Yielding depends on the difference between the principal stresses, not on their absolute size, and adding a third tension raises the hydrostatic component without raising the shear. The material therefore needs far more load to yield at all. It yields less, so it does less plastic work, so it absorbs less energy, so the toughness is lower. That chain is the entire explanation of the thickness effect on $K_{IC}$: toughness falls as a plate gets thicker and then flattens once the interior is fully in plane strain, and the plane-strain value is the floor.

You can see both states on one fracture surface. The edges of a broken plate show shear lips at 45 degrees — plane stress, because a free surface cannot carry a through-thickness stress — and the centre is flat and brittle. The proportion of flat to shear is a direct visual reading of how close the section came to plane strain, and it is worth looking at before trusting any single toughness number in a report.

The real reason this page exists is as a validity check on every other page here. Linear elastic fracture mechanics assumes small-scale yielding: the plastic zone must be small compared with the crack length, the remaining ligament and the thickness, so that the elastic $K$ field still describes what surrounds it. ASTM E399 turns that into a hard rule — a valid plane-strain $K_{IC}$ requires thickness, crack length and ligament all to be at least $2.5(K_{IC}/\sigma_y)^2$. For a tough low-strength steel that demand can run to hundreds of millimetres of specimen, which is why plane-strain $K_{IC}$ values are scarce for exactly the materials structural engineers most want them for, and why elastic-plastic methods — the J-integral, crack-tip opening displacement, the failure-assessment diagram — exist at all. If your section is thinner than the rule requires, the number you measure on it is not $K_{IC}$, the plastic zone is not small, and this site's fracture pages are the wrong tool.

Two more things the simple expression does not tell you. The zone is not a circle: it is a butterfly or dogbone, lobed forward and to the sides, and the expression gives only its extent straight ahead. And a high yield strength is not automatically good news in fracture. For a fixed $K$ it shrinks the plastic zone, and a smaller zone consumes less energy. Very high strength alloys are strong and fragile at once, and that trade — strength against toughness, almost always pulling in opposite directions — is the central fact of alloy selection for anything that might be cracked.