Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 89 MPa (81.00 ksi). If the plate is exposed to a tensile stress of 336 MPa (48730 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 0.92 for Y.

Answer 1

Step-by-step explanation:

The given data is as follows.

`K_(k)` = 89 MPa,
`\sigma` = 336 MPa

Y = 0.92

Now, we will calculate the length of critical interior flaw as follows.

`a_(c) = (1)/(\pi)((K_(k))/(\sigma Y))^(2)`

=
`(1)/(\pi)((89)/(336 * 0.92))^(2)`

=
`(656.38)/(3.14)`

= 209.04 mm

Thus, we can conclude that minimum length of a surface crack that will lead to fracture is 209.04 mm.