The modulus of elasticity for titanium carbide (TiC) having 5 vol\% porosity is 310 GPa (45 \(\times\) \(\left.10^{6} \mathrm{psi}\right)\) (a) Compute the modulus of elasticity for the nonporous material. (b) At what volume percent porosity will the modulus of elasticity be 240 GPa ( \(35 \times 10^{6}\) psi)?

First, we need to rewrite the equation for \(E_{np}\). \(E_{np} = \frac{E}{(1-P)^n}\) Now, we can plug the given values into the equation: \( E = 310 \thinspace GPa = 310 \times 10^3 \thinspace MPa\) \(P = 5\% = 0.05\) Since \(n\) typically ranges from 1.5 to 2.5 for ceramic materials, we'll use the average value, \(n = 2\) (we can refine the value of \(n\) later if necessary). Now we plug these values into the equation and solve for \(E_{np}\): \(E_{np} = \frac{310 \times 10^3}{(1-0.05)^2}\) \(E_{np} \approx 343973 \thinspace MPa = 343.973 \thinspace GPa\) So the modulus of elasticity for the nonporous titanium carbide is approximately 343.973 GPa.

To find the volume percent porosity at a given modulus of elasticity, we'll need to solve our original equation for \(P\): \(P = 1 - \sqrt[n]{\frac{E}{E_{np}}}\) We're given the modulus of elasticity (\(E = 240 \thinspace GPa = 240 \times 10^3 \thinspace MPa\)). We'll use the \(E_{np}\) value that we calculated in part (a) and the same value of \(n = 2\). Now we plug these values and solve for \(P\): \(P = 1 - \sqrt[2]{\frac{240 \times 10^3}{343973}}\) \(P \approx 0.091\) To convert back to volume percent, simply multiply by 100: \(Volume\thinspace Percent\thinspace Porosity = P \times 100 = 0.091 \times 100 = 9.1\%\) So the volume percent porosity of the given modulus of elasticity (\(240 \thinspace GPa\)) will be approximately 9.1%.