Chapter 17, Problem 17.33P

Interpretation Introduction

(a)

Interpretation:

The weight of part for the given condition is to be calculated if the part is made up of aluminum reinforced with boron fiber.

Concept introduction:

The rule of mixtures for particulate composites 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

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

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

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.

Answer to Problem 17.33P

The requiredvalue of weight of part= $267141.76\text{gram}$

Explanation of Solution

Given information:

Modulus of elasticity for titanium = $16\times {10}^6\text{psi}$

Part of 1000 pound is made

Considering modulus of elasticity parallel to fiber.

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 boron is $2.36g/c{m}^3$

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

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

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

  $V_{part}=\frac{m_{part}}{{\rho }_{part}}$

Mass of part = $454\text{kg}$

Density of part = $4.507{\text{kg/m}}^{\text{3}}$

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

  $\begin{array}{l}{V}_{part}=\frac{454\times 1000}{4.507}\\ V_{part}=100732.19{\text{cm}}^3\end{array}$

The volume of part is $100732.19{\text{cm}}^3$

Using the relation of modulus of elasticity

  $E_C=f_B{E}_B+f_{Al}{E}_{Al}$

Using the relation of volume fraction that is the sum of sum of all volume fraction is equal to one which is given by relation,

  $\begin{array}{l}{f}_B+f_{Al}=1\\ f_{Al}=1-f_B\end{array}$

Substituting the values in modulus of elasticity formula:

  $\begin{array}{l}{E}_c=110\times {10}^3\text{MPa}\\ E_B=379\times {10}^3\text{MPa}\\ E_{Al}=69\times {10}^3\text{MPa}\\ E_C=f_B{E}_B+f_{Al}{E}_{Al}\\ 110\times {10}^3=\left(f_B\right)\left(379\times {10}^3\right)+\left(1-f_B\right)\left(69\times {10}^3\right)\\ f_B=0.1322\end{array}$

Calculation of volume fraction of aluminum:

  $\begin{array}{l}{f}_B=0.1322\\ f_{Al}=1-0.1322\\ f_{Al}=0.8678\end{array}$

Applying rule of mixing using the values of calculated volume fraction and density of both the components.

  $\begin{array}{l}{\rho }_c=f_1{\rho }_1+f_2{\rho }_2\\ {\rho }_c=2.36\times 0.1322+2.7\times 0.867\\ {\rho }_c=2.652g/c{m}^3\end{array}$

Calculation of weight of part using the density of part which is defined as the ratio of mass per unit volume, given by relation:

  $\begin{array}{l}{\rho }_c=\frac{m_{part}}{V_{part}}\\ m_{part}={\rho }_c\times V_{part}\\ m_{part}=2.652\times 100732.19\\ m_{part}=267141.76\text{gram}\end{array}$

Thus, the required value of mass of part is $267141.76\text{gram}$

Interpretation Introduction

(b)

Interpretation:

The weight of part for polyester is to be calculated if the part is reinforced with carbon fiber.

Concept introduction:

The rule of mixtures for particulate composites:

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

Where

  ${\rho }_c=$ density of composite

  $f_i=$ volume fraction of i component

  ${\rho }_i=$ density of i component

Volume fraction is defined as

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

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

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

Relation for modulus of elasticity is given as:

  $E_C=f_C{E}_C+f_{polyester}{E}_{polyester}$

  $f_C{E}_C$ are the volume fraction and modulus of elasticity for carbon.

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

Answer to Problem 17.33P

The required value of weight of part for polyester is $147737.85\text{gram}$

Explanation of Solution

Given information:

Modulus of polyester = 650000 psi

Calculation of volume fraction of carbon and polyesters using the relation based on rule of mixing, which states that the sum of volume fraction is equal to one given by relation:

  $\begin{array}{l}{f}_C+f_{polyester}=1\\ f_{polyester}=1-f_C\end{array}$

Putting the values of volume fraction of polyester in terms of volume fraction of carbon using the formula of modulus of elasticity:

  $\begin{array}{l}{E}_C=f_C{E}_C+\left(1-f_C\right)E_{polyester}\\ E_C=110\times {10}^3\text{MPa}\\ E_C=531\times {10}^3\text{MPa}\\ E_{polyester}=4482\text{MPa}\end{array}$

Substituting the values in the formula of modulus of elasticity:

  $\begin{array}{l}110\times {10}^3=f_C\times \left(531\times {10}^3\right)+\left(1-f_C\right)\times \left(4482\right)\\ f_C=0.201\end{array}$

Calculation of volume fraction of polyester using the relation:

  $\begin{array}{l}{f}_C+f_{polyester}=1\\ f_{polyester}=1-f_C\end{array}$

  $\begin{array}{l}{f}_{polyester}=1-f_C\\ f_{polyester}=1-0.201\\ f_{polyester}=0.799\end{array}$

Calculation of density of part using rule of mixing and substituting the value,

  $\begin{array}{l}{\rho }_c=f_1{\rho }_1+f_2{\rho }_2\\ {\rho }_1=1.9g/c{m}^3\\ {\rho }_2=1.36g/c{m}^3\\ f_{polyester}=0.799\\ f_C=0.201\\ {\rho }_c=0.201\times 1.9+0.799\times 1.36\\ {\rho }_c=1.466g/c{m}^3\end{array}$

Calculation of weight of part if polyester is reinforced with carbon fiber based on its relation with mass and volume.

Defining the relation as density is defined as the ratio of mass per unit volume stated as:

  $\begin{array}{l}{\rho }_c=\frac{m_{part}}{V_{part}}\\ m_{part}={\rho }_c\times V_{part}\end{array}$

Substituting the value of V calculated from part (a) as $100732.19$

  ${\text{cm}}^3$ and density as $1.466$

  $\text{gram}{\text{/cm}}^3$

  $\begin{array}{l}{m}_{part}=1.466\times 100732.19\\ m_{part}=147737.85\text{gram}\end{array}$

Thus, the calculated value of weight of part is $147737.85\text{gram}$

Interpretation Introduction

(c)

Interpretation:

The Specific Modulus fortitanium, boron and carbon is to be calculated.

Concept introduction:

Specific modulus is defined as the ratio of modulus of elasticity to density, given by relation

  $\begin{array}{l}\text{Specific Modulus =}\frac{\text{moldulus}\text{of}\text{elasticity}\left(\sigma \right)}{\text{density}\left(\rho \right)}\\ \end{array}$

Answer to Problem 17.33P

The required value of specific modulus for titanium= $2.44\times {10}^8\text{cm}$

The required value of specific modulus for boron = $4.14\times {10}^8\text{cm}$

The required value of specific modulus for carbon = $7.67\times {10}^8\text{cm}$

Explanation of Solution

Calculation of specific modulus for titanium, using the values from part (a):

  $E_c=110\times {10}^3$ MPa

Conversion of mega pascal to $kg/c{m}^2$ :

  $\begin{array}{l}{\text{110×10}}^{\text{3}}\text{MPa = 110}{\text{×10}}^{\text{3}}{\text{×10}}^{\text{6}}\text{pascal}\\ {\text{110×10}}^{\text{9}}\text{pascal}{\text{= 110×10}}^{\text{9}}{\text{N/m}}^{\text{2}}\\ {\text{110×10}}^{\text{9}}{\text{N/m}}^{\text{2}}\text{=}{\text{110×10}}^{\text{5}}{\text{N/cm}}^{\text{2}}=110\times {10}^6{\text{kg/cm}}^{\text{2}}\end{array}$

Calculation of specific modulus for titanium alloy:

  $\begin{array}{l}\text{Specific Modulus =}\frac{{\text{110×10}}^{\text{6}}{\text{kg/cm}}^{\text{2}}\text{×}\frac{\text{1000}\text{gram}}{\text{1}\text{kg}}}{\text{4}\text{.507}{\text{g/cm}}^{\text{3}}}\\ \text{Specific Modulus =}2.44\times {10}^8\text{cm}\end{array}$

Similarly calculating specific composition of boron

Modulus of elasticity of boron = $110\times {10}^6\times 100\text{g}{\text{/cm}}^{\text{2}}$

Density of boron = $2.652{\text{g/cm}}^{\text{3}}$

  $\begin{array}{l}\text{Specific Modulus =}\frac{{\text{110×10}}^{\text{6}}{\text{kg/cm}}^{\text{2}}\text{×}\frac{\text{1000}\text{gram}}{\text{1}\text{kg}}}{2.652{\text{g/cm}}^{\text{3}}}\\ \text{Specific Modulus =}4.14\times {10}^8\text{cm}\end{array}$

Calculation of specific modulus for carbon

  $\begin{array}{l}\text{Specific Modulus =}\frac{{\text{110×10}}^{\text{6}}{\text{kg/cm}}^{\text{2}}\text{×}\frac{\text{1000}\text{gram}}{\text{1}\text{kg}}}{1.468{\text{g/cm}}^{\text{3}}}\\ \text{Specific Modulus =}7.67\times {10}^8\text{cm}\end{array}$

Thus the required values of specific modulus for titanium, boron and carbon are $2.44\times {10}^8\text{cm}$, $4.14\times {10}^8\text{cm}$ and $7.67\times {10}^8\text{cm}$.