Answer:
Critical Flaw Length=17.08 mm
The Critical flaw Length > 4mm, It means it is detectable.
Explanation:
Given Data:
Fracture Toughness=[tex]K_{tc}[/tex]=88MPa
Yield Strength=σ=710 MPa
Y=1.07
Solution:
Formula:
[tex]Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{Y*\sigma} )^2[/tex]
Since yield Strength is half, Critical Length will be:
[tex]Critical\ Length=\frac{1}{\pi } *(\frac{K_{tc}}{\frac{\sigma}{2} *Y} )^2\\Critical\ Length=\frac{1}{\pi } *(\frac{88MPa}{\frac{710MPa}{2} *1.07} )^2\\\\Critical\ Length=0.01708\ m[/tex]
Critical Flaw Length=17.08 mm
The Critical flaw Length > 4mm, It means it is detectable.
