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

Answer:

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

Explanation:

The critical flaw size is

$a_{c} =\frac{1}{\pi } (\frac{K_{c} }{oY} )^{2}$

where

Kc = plane strain fracture toughness = 82.4 MPa√m

o = imposed stress = 345 MPa

Y = dimensionless parameter = 1

Replacing

$a_{c} =\frac{1}{\pi } (\frac{82.4}{1*345} )^{2} =18.16x10^{-3} m=18.16mm$

Answer 2

Answer:

Answer for the question is given in the attachment.

Explanation: