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.s_{max} = 100 MPa, s_{min} = 30 MPa and law of crack propagation for material of plate is
da / dN = 10^{-8} (Δ K_{ I} )2.25 mm/cycle
Solution
The half crack length at which fracture will occur was calculated as a_{c} = 57.17 mm.
∴ Δ K I = K _{I} _{max} - K _{I} _{min}
Using this value of Δ K_{I} in law of fatigue crack propagation, i.e. the plate is likely to fracture under s_{max} = 100 MPa,s_{min} = 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}