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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
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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.

User Jason Higgins
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5 votes

Answer:

1.44 mm

Step-by-step 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* 393* 10^(9)* 0.9}{\pi* (16* 10^(6))^(2)= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

User Auntyellow
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