Question

An aluminum oxide component must not fail when a tensile stress of 16 MPa is applied. Determine the maximum allowable surface crack length if the surface energy of aluminum oxide is 0.90 J/m2. The modulus of elasticity of this material is 393 GPa.

Answer 1

Answer:

0.88 mm

Explanation:

The maximum allowable surface crack, c is given by

$c=\frac {2E\gamma}{\pi \sigma_c^{2}}$ where c is cracking length, E is modulus of elasticity, $\gamma$ is surface energy and $\sigma_c$ is tensile stress.

Substituting $393\times 10^{9}$ for E, $16\times 10^{6}$ for $\sigma_c$ and 0.9 for $\gamma$ we obtain

$c=\frac {2\times 393\times 10^{9}\times 0.9}{\pi \times (16\times 10^{6})^{2}}=0.00088 m\approx 0.88mm$

Therefore, maximum allowable surface crack is 0.88 mm