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.

Respuesta :

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

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

Substituting the given values

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

The maximum allowable surface crack is 1.44 mm