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A specimen of a 4340-steel alloy with a plane strain fracture toughness of 54.8 MPa sqrt(m) (50 ksi sqrt(in.)) is exposed to a stress of 1030 MPa (150,000 psi). Will this specimen experience fracture if the largest surface crack is 0.5 mm (0.02 in.) long

Respuesta :

Answer:

critical stress [tex]\sigma _c[/tex] = 1382.67 MPa

Explanation:

given data

plane strain fracture toughness = 54.8 MP

length of surface creak = 0.5 mm

we take here

parameter Y = 1.0

solution

we apply critical stress formula that is

critical stress [tex]\sigma _c[/tex] = [tex]\frac{K}{Y\sqrt{\pi \times a} }[/tex]   .............................1

here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1

critical stress [tex]\sigma _c[/tex]  =  [tex]\frac{54.8 \times 10^6}{1 \sqrt{\pi \times 5 \times 106{-4}}}[/tex]    

solve it we get

critical stress  = 1382.67 MPa

As exposed stress 1030 MPa is less than critical stress 1382 MPa

so that fracture will not be occur here

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