Question

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

Answer 1

Answer:

critical stress required for the propagation is 27.396615 ×$10^{6}$ N/m²

Explanation:

given data

specific surface energy = 0.90 J/m²

modulus of elasticity E = 393 GPa = 393 ×$10^{9}$ 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 ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$    .....................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 ×$10^{-3}$ m

so now put value in equation 1 we get

( σc ) = $\sqrt{\frac{2E\gamma s}{\pi a}}$

( σc ) = $\sqrt{\frac{2*393*10^9*0.90}{\pi 0.3*10^{-3}}}$

( σc ) = 27.396615 ×$10^{6}$ N/m²

so critical stress required for the propagation is 27.396615 ×$10^{6}$ N/m²