Chapter 17, Problem 17.29P

Interpretation Introduction

(a)

Interpretation:

Density of composite is to be calculated.

Concept introduction:

Rules of mixtures is:

  $\begin{array}{l}{\rho }_c={\displaystyle \sum \left(f_i{\rho }_i\right)}\text{for}\left\{\text{i}\text{=}\text{1}\mathrm{..........}\text{n}\right\}\\ {\rho }_c=f_1{\rho }_1+f_2{\rho }_2\mathrm{...................}{f}_n{\rho }_n\end{array}$

Where

  ${\rho }_c=$ density of composite

  $f_i=$ volume fraction of i component

  ${\rho }_i=$ density of i component

Volume fraction is defined as:

  $\begin{array}{l}f=\frac{V_1}{V_1+V_2}\\ f=\text{volume}\text{fraction}\\ V_1=\text{volume}\text{of}\text{first}\text{component}\\ V_2=\text{volume}\text{of}\text{second}\text{component}\\ \text{Where}{\text{V}}_{\text{1}}\text{and}{\text{V}}_{\text{2}}\\ {\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{weight}\text{of}\text{second}\text{component}}{\text{density}\text{of}\text{second}\text{component}}\\ \text{in}\text{terms}\text{of}\text{weight}\text{and}\text{density}\end{array}$

  $f=\frac{{\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}}{{\text{V}}_{\text{1}}\text{=}\frac{\text{weight}\text{of}\text{first}\text{component}}{\text{density}\text{of}\text{first}\text{component}}+V_2=\frac{\text{weight}\text{of}\text{second}\text{component}}{\text{density}\text{of}\text{second}\text{component}}}$

Answer to Problem 17.29P

The requiredvalue of volume fraction of composite = $2.558g/c{m}^3$

Explanation of Solution

Given information:

Weight of boron fiber in unidirectional orientation = $\text{5 kg}$

Weight of aluminum fiber in unidirectional orientation = $\text{8 kg}$

Based on given information:

Applying rule of mixing,

  ${\rho }_c=f_1{\rho }_1+f_2{\rho }_2$

  ${\rho }_c$ = density of composite

  $f_1$ = Volume fraction of boron

  $f_2$ = Volume fraction of aluminum

  ${\rho }_1$ = Density of boronis $2.36g/c{m}^3$

  ${\rho }_2$ = Density of aluminum is $2.7g/c{m}^3$

  ${\rho }_c=f_1{\rho }_1+f_2{\rho }_2$

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  $f_{boron}=\frac{V_{boron}}{V_{boron}+V_{Al}}$

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  $V_{boron}=\frac{m_B}{{\rho }_B}$

Conversion of mass in gram, therefore multiplying the units by 1000 grams.

  $\begin{array}{l}{V}_{boron}=\frac{5\times 1000}{2.36}\\ V_{boron}=2118.644g/c{m}^3\end{array}$

Calculation of volume of aluminum:

  $\begin{array}{l}{V}_{Al}=\frac{m_{Al}}{{\rho }_{Al}}\\ V_{Al}=\frac{8\times 1000}{2.7}\\ V_{Al}=2962.96g/c{m}^3\end{array}$

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum.

  $\begin{array}{l}{f}_B=\frac{V_{boron}}{V_{boron}+V_{Al}}\\ f_B=\frac{2118.644}{2118.644+2962.96}\\ f_B=0.41692\\ f_{Al}=\frac{V_{Al}}{V_{boron}+V_{Al}}\\ f_{Al}=\frac{2962.96}{2118.644+2962.96}\\ f_{Al}=0.5830\end{array}$

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Applying rule of mixing

  ${\rho }_c=f_1{\rho }_1+f_2{\rho }_2$

  $\begin{array}{l}{\rho }_c=\left(2.36\times 0.41692\right)+\left(2.7\times 0.5830\right)\\ {\rho }_c=0.9839+1.5741\\ {\rho }_c=2.558g/c{m}^3\end{array}$

Interpretation Introduction

(b)

Interpretation:

Modulus of elasticity parallel to fiber is to be calculated.

Concept introduction:

Modulus of elasticity is defined as the ratio of shear stress to shear strain.

Relation for modulus of elasticity is given as:

  $E_C=f_B{E}_B+f_{Al}{E}_{Al}$

  $f_B{E}_B$ are the volume fraction and modulus of elasticity for boron.

  $f_{Al}{E}_{Al}$ are the volume fraction and modulus of elasticity for aluminum.

  $\begin{array}{l}f=\frac{V_1}{V_1+V_2}\\ f=\text{volume}\text{fraction}\\ V_1=\text{volume}\text{of}\text{first}\text{component}\\ V_2=\text{volume}\text{of}\text{second}\text{component}\end{array}$

Answer to Problem 17.29P

The required value of modulus of elasticity parallel to fibers is $E_C=198239.68MPa$

Explanation of Solution

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  $f_{boron}=\frac{V_{boron}}{V_{boron}+V_{Al}}$

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  $V_{boron}=\frac{m_B}{{\rho }_B}$

Conversion of mass in gram, therefore multiplying the units by 1000 grams:

  $\begin{array}{l}{V}_{boron}=\frac{5\times 1000}{2.36}\\ V_{boron}=2118.644g/c{m}^3\end{array}$

Calculation of volume of aluminum:

  $\begin{array}{l}{V}_{Al}=\frac{m_{Al}}{{\rho }_{Al}}\\ V_{Al}=\frac{8\times 1000}{2.7}\\ V_{Al}=2962.96g/c{m}^3\end{array}$

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum:

  $\begin{array}{l}{f}_B=\frac{V_{boron}}{V_{boron}+V_{Al}}\\ f_B=\frac{2118.644}{2118.644+2962.96}\\ f_B=0.41692\\ f_{Al}=\frac{V_{Al}}{V_{boron}+V_{Al}}\\ f_{Al}=\frac{2962.96}{2118.644+2962.96}\\ f_{Al}=0.5830\end{array}$

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Substituting the following values in the formula of modulus of elasticity:

  $\begin{array}{l}{E}_C=f_B{E}_B+f_{Al}{E}_{Al}\\ E_B=379\times {10}^{3}MPa\\ E_{Al}=69\times {10}^{3}MPa\\ f_B=0.41692\end{array}$

  $f_{Al}=0.5830$

  $\begin{array}{l}{E}_C=0.41692\times 379\times {10}^3+0.5830\times 69\times {10}^3\\ E_C=198239.68MPa\end{array}$

The required value of modulus of elasticity is 198239.68 MPa.

Interpretation Introduction

(c)

Interpretation:

Modulus of elasticity perpendicular to fiber is to be calculated.

Concept introduction:

Modulus of elasticity is defined as the ratio of shear stress to shear strain

Relation for modulus of elasticity perpendicular to fiber is given as:

  $\frac{1}{E_C}=\frac{f_B}{E_B}+\frac{f_{Al}}{E_{Al}}$

  $f_B{E}_B$ are the volume fraction and modulus of elasticity for boron

  $f_{Al}{E}_{Al}$ are the volume fraction and modulus of elasticity for aluminum

  $\begin{array}{l}f=\frac{V_1}{V_1+V_2}\\ f=\text{volume}\text{fraction}\\ V_1=\text{volume}\text{of}\text{first}\text{component}\\ V_2=\text{volume}\text{of}\text{second}\text{component}\end{array}$

Answer to Problem 17.29P

The required value of modulus of elasticity perpendicular to fibers is $E_C=104719.40MPa$

Explanation of Solution

Calculation of volume fraction of boron is defined as the ratio of volume of boron to total volume:

  $f_{boron}=\frac{V_{boron}}{V_{boron}+V_{Al}}$

Calculation of volume of boron, expressed as the ratio of mass of boron to density of boron:

  $V_{boron}=\frac{m_B}{{\rho }_B}$

Conversion of mass in gram, therefore multiplying the units by 1000 grams:

  $\begin{array}{l}{V}_{boron}=\frac{5\times 1000}{2.36}\\ V_{boron}=2118.644g/c{m}^3\end{array}$

Calculation of volume of aluminum:

  $\begin{array}{l}{V}_{Al}=\frac{m_{Al}}{{\rho }_{Al}}\\ V_{Al}=\frac{8\times 1000}{2.7}\\ V_{Al}=2962.96g/c{m}^3\end{array}$

Calculation of volume fraction of boron and aluminum on substituting the value of volume for boron and aluminum:

  $\begin{array}{l}{f}_B=\frac{V_{boron}}{V_{boron}+V_{Al}}\\ f_B=\frac{2118.644}{2118.644+2962.96}\\ f_B=0.41692\\ f_{Al}=\frac{V_{Al}}{V_{boron}+V_{Al}}\\ f_{Al}=\frac{2962.96}{2118.644+2962.96}\\ f_{Al}=0.5830\end{array}$

Volume fraction of boron = 0.41692

Volume fraction of aluminum = 0.5830

Substituting the following values in the formula of modulus of elasticity

  $\begin{array}{l}{E}_B=379\times {10}^{3}MPa\\ E_{Al}=69\times {10}^{3}MPa\\ f_B=0.41692\end{array}$

  $f_{Al}=0.5830$

  $\begin{array}{l}\left(\frac{1}{E_C}\right)=\left(\frac{0.41692}{379\times {10}^3}\right)+\left(\frac{0.5830}{69\times {10}^3}\right)\\ E_C=104719.40MPa\end{array}$

The required value of modulus of elasticity is 104719.40 MPa.