Stress Intensity Factor

Before 1920 the strength of a part was its stress against its strength, and that was the whole story. It failed to explain why glass breaks at a thousandth of the strength its atomic bonds promise, why the Liberty ships split in half in cold harbours, and why the Comet airliners came apart at the corners of their windows. The answer in every case was a crack, and the equation that describes what a crack does is $K = Y\sigma\sqrt{\pi a}$.

The remarkable thing George Irwin showed in 1957 is that the entire elastic stress field around a crack tip has the same shape in every cracked body. Approach the tip and the stress rises as $1/\sqrt{r}$ — towards infinity, which is why a sharp crack has no stress concentration factor in the ordinary sense. Only the magnitude of that field changes from one situation to another, and one number captures it. That number is $K$. Two entirely different components with the same $K$ have identical conditions at the crack tip, which is why a small laboratory specimen can certify a large structure at all.

Its units are the strangest on this site: MPa·$\sqrt{\text{m}}$, a stress times the square root of a length. It is not a stress and not an energy; it is the coefficient of the singularity, and there is no everyday quantity to compare it with. Get used to the numbers instead. A structural steel runs 50 to 200 MPa·$\sqrt{\text{m}}$, an aluminium alloy 20 to 45, a titanium alloy 40 to 100, a hardened tool steel perhaps 20, and a ceramic 1 to 5. Ordinary window glass is under 1.

The geometry factor $Y$ is where all the difficulty lives. It is exactly 1 for one case only: a through-thickness crack in the middle of an infinite plate under uniform remote tension. Every real component departs from that, and $Y$ is how the departure is accounted for. A single edge crack in a wide plate gives 1.12 — the 12% is the free surface allowing the crack faces to open more. An embedded circular flaw gives $2/\pi = 0.637$. A semicircular surface flaw gives about 0.73 at its deepest point. As a crack grows across a finite plate, $Y$ climbs steeply and runs to infinity as the remaining ligament vanishes. Handbook compendia — Tada, Paris and Irwin; Rooke and Cartwright; the annexes of BS 7910 and API 579 — collect perhaps a hundred standard configurations, and a real weld toe, nozzle corner or fillet root often matches none of them well enough to be worth arguing about. Getting $Y$ from a finite-element model is ordinary practice, not a luxury, and a fracture assessment whose $Y$ is a guess is a fracture assessment whose answer is a guess.

Half or whole: the arithmetic mistake everyone makes once. For a CENTRE crack, $a$ is the HALF length — the crack measures $2a$ across the plate and $a$ runs from the centre to one tip. For an EDGE crack, $a$ is the WHOLE depth from the surface. Feed the full length of a centre crack in as $a$ and $K$ comes out $\sqrt{2}$ too large, 41% high, and the assessment is needlessly conservative. Make the error the other way — halving an edge crack — and $K$ is 29% low, which is unsafe. Every page in this set says which $a$ it wants, and it is worth reading the label every time.

Now the hard truth about toughness. $K_{IC}$ is quoted in tables as though it were a property like density, and it is not. It falls with section thickness, steeply, until full plane-strain constraint is reached and only then goes flat — the same alloy in thin sheet can carry two or three times the plane-strain number. It falls through the ductile-to-brittle transition as temperature drops, and for a ferritic steel that fall can be a factor of five over forty degrees, which is precisely what happened to the Liberty ships in the North Atlantic. It falls with loading rate. And it falls with time in service, through neutron irradiation in reactor vessels, temper embrittlement in alloy steels held hot, and hydrogen from cathodic protection or from sour service. A handbook plane-strain value is conservative for a thin warm section and can be dangerously optimistic for one that is colder, faster loaded, or older than the specimen ever was.

And toughness enters the answer squared. Rearranged for crack size, $a_c = (K_{IC}/Y\sigma)^2/\pi$. Every uncertainty in toughness is doubled in the crack you would tolerate. This is why the conversion factor between ksi·$\sqrt{\text{in}}$ and MPa·$\sqrt{\text{m}}$ deserves its four figures — 1.0988434, not 1.1: nine percent on $K$ is nineteen percent on $a_c$, and North American aerospace and pressure-vessel work moves between the two systems constantly.

Finally, know where linear elastic fracture mechanics stops. It assumes the plastic zone at the tip is small compared with the crack, the ligament and the thickness. When it is not — a tough steel at room temperature, a thin section, a high applied stress — the elastic solution is not describing the tip at all, and the honest tools are the J-integral or crack-tip opening displacement, which is what the failure-assessment diagrams in BS 7910 and API 579 are built on. A critical crack size that comes out larger than the wall thickness is the calculation telling you politely that the part will leak or yield before it fractures, and that is usually good news.