# Starting out question

Hi

This is my first post so bear with me.

I have a question and an answer but have little idea how they got the answer.  The question relates to a sample with load being applied at both ends.

Question

The load and displacement at which cracking commences are X1 and u1 when the crack
area is A1. Later the load is X2 and the displacement is u2 when the crack area is A2.
Assuming displacement reversibility, and that the ‘R’ locus is a straight line joining X-
1u1= and X2u2, show that the fracture toughness R is given by
R = (X1u2 –X2u1)/2 (A2 – A1)

Stored energy when cracks area is A1 is

½ X1u1 and

when the crack area is A2, the stored energy is

½ X2u2

The work done by the external forces is

X2 (u2-u1) + ½ (X1- X2)(u2 – u1)

So ½ X1u1 - ½ X2u2 + X2 (u2-u1) + ½ (X1- X2)(u2 – u1) = R(A2 – A1)

Simplifying gives the result required

Am I right to think that X2 (u2-u1) is the surface energy and ½ X1u1 - ½ X2u2 is the -(change in Potential energy)? What is ½ (X1- X2)(u2 – u1) and why would you minus the change in potential energy from the work of external forces to get the work to fracture. I just can't figure it out.

Edward 