# What about tension?

It is not as easy to estimate U_{E} for tension. Griffith used Inglis’ analysis for the overall stress and strain fields in a cracked body, under a constant applied load. This gives U_{E} as

$${U_{\rm{E}}} = {{\pi {c^2}{\sigma ^2}} \over E}$$

In other words there is an **increase** in the elastic strain energy as the crack grows.

However when the crack grows, work is done by the applied force, F, and is equal in magnitude to twice the change in elastic strain energy. (Think of the work done by the applied force and the elastic energy changes on a uniform bar loaded in tension.) As it is work done **on** the system, the sign is opposite to that of U_{E}, so that

$$ - {U_{\rm{F}}} + {U_{\rm{E}}} = - {{\pi {c^2}{\sigma ^2}} \over E}$$

where U_{F} is the work done by the applied force, U_{E} is the elastic energy change on cracking and U_{S} is the work required to create two new surfaces.

The work associated with creating new crack faces, U_{S}.

U_{S} = 2 c R

where R is the fracture energy.

Combining the terms of this energy expression we obtain:

$$U = - {{\pi {\sigma ^2}{c^2}} \over E} + 2cR$$

The energy function is plotted in the following animation:

## How does cracking occur now?

The energy terms above vary with c as shown. Although there is an equilibrium, it is unstable in tension whereas it was stable in wedging.

Adding the energies and differentiating gives the equilibrium crack length c_{e} as

$${c_{\rm{e}}} = {{E{\rm{ }}R} \over {\pi {\sigma ^2}}}$$

Since this is an unstable equilibrium, once it is surpassed, fracture will occur, but will it be catastrophic?