If the specific surface energy for aluminum oxide is 0.90 j/m2 and its modulus of elasticity is (393 gpa), compute the critical stress required for the propagation of an internal crack of length 0.6 mm. mpa

critical stress required for the propagation is 27.396615 × N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 × N/m²

internal crack length = 0.6 mm

to find out

critical stress required for the propagation

solution

we will apply here critical stress formula for propagation of internal crack

( σc ) =    .....................1

here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm  = 0.3 × m

so now put value in equation 1 we get

( σc ) =

( σc ) =

( σc ) = 27.396615 × N/m²

so critical stress required for the propagation is 27.396615 × N/m²

answer:

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