A plate of steel with a central through-thickness flaw of length 16 mm is subjected to a stress of 350 MPa normal to the crack plane. If the yield strength of the material is 1400 MPa what is the plastic zone size and the effective stress intensity factor range at the crack tip? If a second plate of steel were heat treated to provide a yield strength of 385 MPa what much larger would the plastic zone size effective SIF be? 1 K ry = T 2 (Plane Stress) Keff = 1 - 2 20 Answer Part 1 Keff ry = 1 KI == \ y 2 == Part 2 1 - 1 -(+) 2 1/350(0.008) - 14002 350 (0.008) 1 - 1/350 2 1400 2 = 0.5 mm = 56.4 MPam ry Keff = 6.6 mm = 72.4 MPam