Question

An aluminum oxide component must not fail when a tensile stress of 12.5 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:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

$C=\frac {2E\gamma}{\pi \sigma_c^{2}}$ where E is the modulus  of elasticity, $\gamma$ is surface energy and $\sigma_c$ is tensile stress

Substituting the given values

$C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm$

The maximum allowable surface crack is 1.44 mm