# Chapter 12 - Structures and Properties of Ceramics - Questions and Problems - Page 508: 12.49a

$E_{0} = 264.54 GPa$

#### Work Step by Step

Given: The modulus of elasticity for spinel $(MgAl_{2}O_{4})$ having 5 vol% porosity is 240 GPa ($35 \times 10 ^{6} psi$). Required: modulus of elasticity for the nonporous material Solution: Rearranging Equation 12.9 to solve $E_{0}$ and considering that $P = 0.05$: $E_{0} =\frac{E}{1 -1.9P + 0.9P^{2}} = \frac{(240 GPa)}{1 -1.9(0.05) + 0.9(0.05)^{2}} = 264.54 GPa$

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