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damaskus [11]
3 years ago
12

A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

crack length if the surface energy of magnesium oxide is 1.0 J/m2. The modulus of elasticity of this material is 225 GPa.
Physics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, \sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa

Surface energy of magnesium oxide, SE = 1.0\ J/m^{2}

Modulus of elasticity of the material, E = 225 GPa = 225\times 10^{9}\ Pa

Now,

To calculate the maximum allowable surface crack length:

L = \frac{2E\times SE}{\sigma^{2}\pi }

L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm

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