Strain Energy Release Rate

Griffith worked in energy and Irwin worked in stress, and for a while it was not obvious they were describing the same thing. $G = K^2/E'$ is the proof that they are. The energy released per unit of new crack area is the square of the stress intensity divided by an effective modulus, and the two descriptions of a crack — how hard the tip is being pulled, and how much energy is available to pay for growth — are the same statement in different currencies.

$G$ is called the strain energy release rate, and the word "rate" is a trap: it is a rate with respect to crack AREA, not with respect to time. Its units are J/m², the same as surface energy, and for a perfectly brittle solid the critical value $G_c$ is exactly $2\gamma$ — the energy of the two faces the crack creates. For a metal $G_{IC}$ runs from about 10 to 100 kJ/m², four or five orders of magnitude above any surface energy, and every bit of that surplus is plastic work at the tip.

The effective modulus $E'$ is where plane stress and plane strain enter. In plane stress — a thin sheet, free to contract through its thickness — $E' = E$. In plane strain — a thick section that cannot contract, because the surrounding material holds it — $E' = E/(1-\nu^2)$. This page is written in the plane-strain form, $G = K^2(1-\nu^2)/E$, and you get plane stress by entering $\nu = 0$, which makes the bracket exactly 1. That is a small trick with a real virtue: it puts both cases on one page without a hidden switch, and the reader can see which one they asked for.

Be clear about how small that particular correction is. At $\nu = 0.3$ the bracket is 0.91, so plane strain gives about 9% less $G$ for the same $K$. That is the minor half of the plane stress versus plane strain story. The major half is that the measured TOUGHNESS differs enormously between the two states — a thin sheet can tolerate a $K$ two or three times the plane-strain $K_{IC}$ of the same alloy, because the material at the tip yields more freely and consumes far more plastic work. Confusing the 9% correction with the 200% toughness effect is a real and consequential mistake. Thickness is what decides which state you are in, and ASTM E399 sets the bar at a thickness of at least $2.5(K_{IC}/\sigma_y)^2$; below it, plane-strain toughness is a conservative floor rather than the answer.

Which currency to work in is largely a matter of trade. Metals work in $K$, because toughness is measured as $K_{IC}$ on a compact-tension specimen and the stress-intensity handbooks are all written in $K$. Polymers, adhesives and composites work in $G$, because their toughness is measured directly as energy in double-cantilever-beam and end-notched-flexure tests, and because for a delamination between two dissimilar plies there is no single modulus to divide by anyway. Both are right; this equation is the exchange rate.

One consequence worth carrying away. Because $K$ goes as $\sqrt{G}$, a material with ten times the fracture energy has only about three times the stress intensity toughness. But critical crack size goes as $K^2$, so that three-fold gain in $K$ restores the full ten-fold gain in tolerable crack length. The energy and the crack size scale together, one for one; $K$ is the square root that sits between them, and it is the reason toughness numbers feel less impressive than the improvements they represent.

Finally, $G$ is the ancestor of the J-integral. When yielding at the tip is no longer small, $K$ stops meaning anything but the energy argument survives — $J$ is what $G$ becomes when the material ahead of the crack is allowed to behave nonlinearly, and it is what elastic-plastic fracture assessment actually uses. In the fully elastic case $J$ equals $G$ exactly, which is a good sign that the generalisation was the right one.