Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is exposed to a tensile stress of 345 MPa (50040 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.04 for Y.

Answer 1

minimum length of a surface crack is 15.043 mm

Step-by-step explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α =
`(1)/(\pi) ((K)/(\sigma Y))^2` ....................1

put here value we get

α =
`(1)/(\pi) ((78*√(10^3) )/(345*1.04))^2`

α = 15.043 mm

so minimum length of a surface crack is 15.043 mm