Question

A magnesium oxide component must not fail when a tensile stress of 14 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.

Answer 1

Answer:

$x=0.730*10^{-3}m$

Explanation:

From the question we are told that:

Tensile stress $\sigma_c = 14 MPa=>14*10^6$

Modulus of elasticity $E=225GPa=>225*10^9$

Surface energy of MgO $\gamma=1N/m$

Generally the equation for maximum allowable surface crack length  is mathematically given by

$x=\frac{2E \gamma}{\pi\sigma_c^2}$

$x=\frac{2(225*10^9)(1)}{3.142*(14*10^6)^2}$

$x=0.730*10^{-3}m$