# Another way of expressing the energies

What we have been doing so far is to calculate how U varies with c, and then find the equilibrium value of c by differentiating U with respect to c. Looking at the terms they fall into two types:

Those that come from the loading system, U_{E} and U_{F}: let’s add these together and call the sum U_{M}

- These give the driving force for cracking

Those that are associated with the material, U_{S}

- This gives the resistance to cracking

Equilibrium will occur when

$${{{\rm{d}}U{\rm{(}}c{\rm{)}}} \over {{\rm{d}}c}} = 0$$

Breaking our energies into the two different types, gives

$$ - {{{\rm{d}}{U_{\rm{M}}}} \over {{\rm{d}}c}} = {{{\rm{d}}{U_{\rm{S}}}} \over {{\rm{d}}c}}$$

From our expression for cracking in tension

$${{{\rm{d}}{U_{\rm{S}}}} \over {{\rm{d}}c}} = 2R$$

and

$${{{\rm{d}}{U_{\rm{M}}}} \over {{\rm{d}}c}} = - {{2\pi {\sigma ^2}} \over E}{\rm{ }}c = 2G$$

where G is the energy per unit area of crack and so is often called the strain energy release rate, or the crack driving force. Now the turning point occurs when

G = R

This is our condition for cracking, that the crack driving force, G, equals the fracture resistance of the material, R.