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