Answer
$a_{c} = 0.018 m = 18 mm$
Work Step by Step
Given:
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of $82.4 MPa \sqrt m (75.0 ksi \sqrt in.)$
Required:
If the plate is exposed to a tensile stress of 345 MPa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture assuming Y = 1.0
Solution:
Using Equation 8.7:
$a_{c} = \frac{1}{π} [\frac{K_{Ic}}{Yσ}]^{2} = \frac{1}{π} [\frac{82.4 MPa \sqrt m}{(1.0)(345 MPa)}]^{2} = 0.018 m = 18 mm$
