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

0.88 mm

Step-by-step 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* 10^(9)` for E,
`16* 10^(6)` for
`\sigma_c` and 0.9 for
`\gamma` we obtain

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

Therefore, maximum allowable surface crack is 0.88 mm