Answer
$a= 7.86 \times 10^{-4} m = 0.786 mm$
Work Step by Step
Required:
An MgO component must not fail when a tensile stress of 13.5 MPa (1960 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of MgO is $1.0 J/m^{2}$
Solution:
Using Equation 8.3 and using the modulus of elasticity of 225 GPa from Table 12.5, we find:
$σ_{c} = [\frac{2Eγ_{s}}{πa}]^{1/2}$
$a = \frac{2Eγ_{s}}{πσ_{c}^{2}} = \frac{2(225 \times 10^{9} N/m^{2})(1.0 N/m)}{π(13.5 \times 10^{6} N/m^{2})^{2}}\\ = 7.86 \times 10^{-4} m\\ = 0.786 mm$
