Answer:
maximum allowable surface crack length is 1.08 mm
Explanation:
given data
stress σ = 250 MPa
length of a surface crack = 1.6 mm
plane-strain fracture toughness = 62 MPa [tex]\sqrt{m}[/tex]
solution
we will calculate here design parameter by design stress equation that is
[tex]\sigm = \frac{K}{Y\sqrt{\pi * a} }[/tex] ...................1
here σ is stress given 250 and a is length of a surface crack and K is plane strain fracture toughness so Y will be here
250 = [tex]\frac{250}{Y*\sqrt{\pi * 1.6*10^{-3}} }[/tex]
Y = 3.49798
so
now we find maximum allowable surface crack length for another alloy using relation
[tex]\sigm = \frac{K}{Y\sqrt{\pi * a} }[/tex] ................2
so a will be here
a = [tex]\frac{1}{\pi} ( \frac{K}{\sigma * Y})^2[/tex]
put here value
a = [tex]\frac{1}{\pi} ( \frac{51*\sqrt{m}}{250 * 3.49798})^2[/tex]
a = 0.001082 m
a = 1.08 mm
so maximum allowable surface crack length is 1.08 mm