Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4MPa$\sqrt{m}$(75.0ksi$\sqrt{in}$). 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. Assume a value of 1.0 for Y.

Answer 1

Given:

plane strain fracture toughness= $82.4 MPa \sqrt{m}$

tensile stress = $345 Mpa$

$Y= 1.0$

we know the formula:

$sigma=K/Y\sqrt{\pi *lambda$

Rearranging the formula

$lambda=1/\pi [K/Y*sigma]^2$

inserting values in formula

$=1/3.14[82.4 MPa \sqrt{m} /(1.0)(345MPa)]^2$

simplifying we get

$=0.018m = 18mm$