# If the surface energy of a magnesium oxide – nickel oxide (MgO-NiO) solid solution is 1.23 J/m2 and its elastic modulus is 208

If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.23 J/m2 and its elastic modulus is 208 GPa, what is the maximum length of an internal elliptical flaw, in mm, if the ceramic part must survive an applied tensile stress of 97 MPa normal to this elliptical flaw

## This Post Has 3 Comments

1. aidanfbussiness says:

0.0173 mm

Explanation:

The maximum length of eliptical flow will be given by

$l=\frac {2Er}{\pi\sigma^{2}}$ where l is the maximum length of elliptical flow, E is the modulus of elasticity, r is the surface energy of magnesium oxide and nickel, $\sigma$ is the tensile stress.

Substituting 208 Gpa for E, 1.23 for r and 97 Mpa for tensile stress then

$l=\frac {2*208*10^{9}*1.23}{\pi\times (97*10^{6})^{2}}=0.0000173 m= 0.0173 mm$

2. Expert says:

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a true.

3. Expert says:

you forgot the rest of the question sorry