# 9.7: 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, UE and UF: let’s add these together and call the sum UM

• These give the driving force for cracking

Those that are associated with the material, US

• This gives the resistance to cracking

Equilibrium will occur when

$\frac{\mathrm{d} U(c)}{\mathrm{d} c}=0$

Breaking our energies into the two different types, gives

$-\frac{\mathrm{d} U_{\mathrm{M}}}{\mathrm{d} c}=\frac{\mathrm{d} U_{\mathrm{s}}}{\mathrm{d} c}$

From our expression for cracking in tension

$\frac{\mathrm{d} U_{\mathrm{s}}}{\mathrm{d} c}=2 R$

and

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

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.