Answer
σc = 33.55MPa
Work Step by Step
Required:
If the specific surface energy for aluminum oxide is $0.90 J/m^{2}$, then using data in Table 12.5, compute the critical stress required for the propagation of an internal crack of length 0.40 mm.
Solution:
Using Equation 8.3 and using the modulus of elasticity of 393 GPa for aluminum oxide from Table 12.5:
$σ_{c} = [\frac{2Eγ_{s}}{π{a}/2}]^{1/2} = [\frac{2(393 \times 10^{9} N/m^{2})(0.90 J/m^{2})}{π(0.40 \times 10^{-3} m/2)}]^{1/2} \\= 33.55 \times 10^{6} N/m^{2} = 33.55 MPa$
