Answer
$E_{0} = 341.69 GPa$
Work Step by Step
Given:
The modulus of elasticity for titanium carbide (TiC) having 5 vol% porosity is 310 GPa.
Required:
Modulus of elasticity for the nonporous material
Solution:
Using Equation 12.9;
$E_{0} =\frac{E}{(1 -1.9P + 0.9P^{2})} = \frac{(310 GPa)}{(1 -1.9(0.05) + 0.9(0.05)^{2})} = 341.69 GPa$