Chapter 12.15, Problem 85AAP

To determine

The tensile modulus of the composite in the direction of the fibers.

The elastic strain in the composite.

Answer to Problem 85AAP

The tensile modulus of the composite in the direction of the fibers is $\text{347}\text{GPa}$.

The elastic strain in the composite is $\text{4}\text{.19}\times {10}^{-3}$.

Explanation of Solution

Write the equation for the rule of mixture for the ceramic-matrix composites.

  $E_{\text{comp}}=f_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}{E}_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}+f_{\text{SiC}}{E}_{\text{SiC}}$                                                                                   (I)

Here, modulus of elasticity of composite is $E_{\text{comp}}$, modulus of elasticity of silicon carbide fiber is $E_{\text{SiC}}$, volume fraction of silicon carbide fiber is $f_{\text{SiC}}$, modulus of elasticity of aluminum oxide matrix is $E_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}$ and volume fraction of aluminum oxide matrix is $f_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}$.

Write the equation to calculate the elastic strain in the composite.

  $\begin{array}{c}{\epsilon }_{\text{comp}}=\frac{{\sigma }_{\text{comp}}}{E_{\text{comp}}}\\ =\frac{\left(P_{\text{comp}}/A_{\text{comp}}\right)}{E_{\text{comp}}}\end{array}$                                                                                              (II)

Here, elastic strain in the composite is ${\epsilon }_{\text{comp}}$, tensile strength of composite is ${\sigma }_{\text{comp}}$, load applied on the composite is $P_{\text{comp}}$ and surface area of the composite is $A_{\text{comp}}$.

Conclusion:

Substitute $\text{340}\text{GPa}$ for $E_{\text{SiC}}$, $\text{350}\text{GPa}$ for $E_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}$, $0.70$ for $f_{{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}}$ and $0.30$ for $f_{\text{SiC}}$ in Equation (I).

 $\begin{array}{c}{E}_{\text{comp}}=\left(0.70\right)\left(\text{350}\text{GPa}\right)+\left(0.30\right)\left(\text{340}\text{GPa}\right)\\ =347\text{GPa}\end{array}$

Thus, the tensile modulus of the composite in the direction of the fibers is $\text{347}\text{GPa}$.

Substitute $\text{8}\text{MN}$ for $P_{\text{comp}}$, $\text{55}{\text{cm}}^2$ for $A_{\text{comp}}$ and $347\text{GPa}$ for $E_{\text{comp}}$ in Equation (II).

 $\begin{array}{c}{\epsilon }_{\text{comp}}=\frac{\left(\text{8}\text{MN}/\text{55}{\text{cm}}^2\right)}{347\text{GPa}}\\ =\frac{\left(\frac{\text{8}\text{MN}}{\text{55}{\text{cm}}^2\times \frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^{\text{2}}}}\right)}{347\times {10}^3\text{MPa}}\\ =\text{4}\text{.19}\times {10}^{-3}\end{array}$

Thus, the elastic strain in the composite is $\text{4}\text{.19}\times {10}^{-3}$.