Question

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 88 () and a yield strength of 710 MPa (51490 psi). The flaw size resolution limit of the flaw detection apparatus is 4 mm (0.1575 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.07, what is the critical flaw length

Answer 1

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.

Step-by-step explanation:

Given Data:

Fracture Toughness=
`K_(tc)`=88MPa

Yield Strength=σ=710 MPa

Y=1.07

Solution:

Formula:

`Critical\ Length=(1)/(\pi ) *((K_(tc))/(Y*\sigma) )^2`

Since yield Strength is half, Critical Length will be:

`Critical\ Length=(1)/(\pi ) *((K_(tc))/((\sigma)/(2) *Y) )^2\\Critical\ Length=(1)/(\pi ) *((88MPa)/((710MPa)/(2) *1.07) )^2\\\\Critical\ Length=0.01708\ m`

Critical Flaw Length=17.08 mm

The Critical flaw Length > 4mm, It means it is detectable.