The Griffiths equation describes the relationship between applied nominal stress and crack length at fracture, i.e. when it becomes energetically favourable for a crack to grow. Griffith was concerned with the energetics of fracture, and considered the energy changes associated with incremental crack extension.

For a loaded brittle body undergoing incremental crack extension,
the only contributors to energy changes are the energy of the new fracture surfaces (two
surfaces per crack tip) and the change in potential energy in the body. The surface
energy term (*S*) represents energy absorbed in crack growth, while the some stored
strain energy (*U*) is released as the crack extends (due to unloading of regions
adjacent to the new fracture surfaces). Surface energy has a constant value per unit
area (or unit length for a unit thickness of body) and is therefore a linear function of
(crack length), while the stored strain energy released in crack growth is a function of
(crack length)^{2}, and is hence parabolic. These changes are indicated in
the figure below:

The next step in the development of Griffith's
argument was consideration of the rates of energy change with crack extension, because the
critical condition corresponds to the maximum point in the total energy curve, i.e. *dW/da*
= 0, where *a* = *a**. For crack lengths greater than this value
(under a given applied stress), the body is going to a lower energy state, which is
favourable, and hence fast fracture occurs. *dW/da* = 0 occurs when *dS/da*
= *dU/da*. The sketch below shows these energy rates, or differentials with
respect to *a*.

*R* is the resistance to crack growth (= *dS/da*)
and *G* is the strain energy release rate (= *dU/da*).

When fracture occurs, *R* = *G* and we
can define *G _{crit} as the critical value of strain energy release*, and
equate this to

where, to get the fracture stress in MPa (the standard SI
engineering unit), the critical strain energy release rate is in N/m, E is in N/m^{2},
and a is in m. This provides an answer in N/m^{2} (Pa), which needs to be
divided by 10^{6} to get the standard engineering unit of MPa. In plane
strain: