Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 M P a m. If the plate is exposed to a tensile stress of 345 MPa during service use, determine the minimum length of a surface crack that will lead to fracture. Assume a Y = 1.0.

Answer 1

The minimum length of a surface crack that will lead to fracture is 18.16 mm

Step-by-step explanation:

The critical flaw size is

`a_(c) =(1)/(\pi ) ((K_(c) )/(oY) )^(2)`

where

Kc = plane strain fracture toughness = 82.4 MPa√m

o = imposed stress = 345 MPa

Y = dimensionless parameter = 1

Replacing

`a_(c) =(1)/(\pi ) ((82.4)/(1*345) )^(2) =18.16x10^(-3) m=18.16mm`

Answer 2

Answer for the question is given in the attachment.

Step-by-step explanation: