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A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa(50 ksi) 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? Why or why not? Assume that the parameter Y has a value of 1.0.

Respuesta :

Answer:

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

fracture will not be occure

Explanation:

given data

plane strain fracture toughness = 54.8 MPa

length of surface creak = 0.5 mm

Y = 1.0

solution

we apply here critical stress formula that is

critical stress [tex]\sigma c = \frac{K}{Y\sqrt{\pi * 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 = \frac{54.8 * 10^6}{1 \sqrt{\pi * 5 * 106{-4}}}[/tex]

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

here we can say that exposed stress 1030 MPa is less than critical stress 1382 MPa so that fracture will not be occure

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