Determine the number of cycles - crack propagation:
The plate of above Example contained a crack of 50 mm length in the centre before it was subjected to cyclic stress. Determine the number of cycles the plate would undergo when final fracture occurs.smax = 100 MPa, smin = 30 MPa and law of crack propagation for material of plate is
da / dN = 10-8 (Δ K I )2.25 mm/cycle
The half crack length at which fracture will occur was calculated as ac = 57.17 mm.
∴ Δ K I = K I max - K I min
Using this value of Δ KI in law of fatigue crack propagation, i.e. the plate is likely to fracture under smax = 100 MPa,smin = 30 MPa if plate contains a central crack of 50 mm length, after 1048 cycles.
da / dN = 10-8 (124)2.25 (a)2.25 / 2
= 0.51´10- 3 (a)1.125