Answer:
1.5510 mm
Explanation:
Plane strain fracture toughness = 45 MPa√m
failing stress ( б ) = 300 MPa
maximum length of surface crack ( a )= 0.95 mm
Determine maximum allowable surface crack length ( in mm )
we will make use of this relationship for Design stress equation to determine the value of Y
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 1 )
k = 45 MPa√m
б = 300 MPa
a = 0.95 mm
y = ?
From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )
hence ; y = 2.7457
Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m
we will apply the equation
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 2 )
K = 57.5 MPa√m
б = 300 MPa
y = 2.7457
a ( maximum allowable surface crack ) = ?
from equation make a subject of the equation
a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]
a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex] = 1.5510 mm
