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Rus_ich [418]
3 years ago
15

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.
Engineering
1 answer:
aivan3 [116]3 years ago
4 0

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

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