Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4MPa
`√(m)`(75.0ksi
`√(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 √(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 √(m) /(1.0)(345MPa)]^2`

simplifying we get

`=0.018m = 18mm`