Chapter 17, Problem 17.53P

Interpretation Introduction

Interpretation:

The modulus of elasticity of micro laminate for the parallel and perpendicular condition to the kevlar fiber should be determined. Also, the advantages of arall material and its comparison with unreinforced aluminum is to be explained.

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

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

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

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

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

$E_C=f_1{E}_1+f_2{E}_2$

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

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

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:

$\frac{1}{E_C}=\frac{f_1}{E_1}+\frac{f_2}{E_2}$

$f_1{E}_1$ are the volume fraction and modulus of elasticity for first component.

$f_2{E}_2$ are the volume fraction and modulus of elasticity for second component.

Answer to Problem 17.53P

The required value of modulus of elasticity parallel to fiber is $E_C=698896\times {10}^5\text{pascal}$.

The required value of modulus of elasticity perpendicular to fiber is $E_C=698896\times {10}^5\text{pascal}$.

Explanation of Solution

Given information:

Micro laminate, Arall is produced using aluminum and epoxy reinforced with Kevlar fiber. The composite is made using 5 sheets of aluminum and 4 sheets of epoxy.

Thickness of five sheets of aluminum = $0.4\text{mm}$.

Thickness of four sheets of epoxy reinforced with Kevlar fiber = $0.2\text{mm}$.

Volume fraction of Kevlar in sheets = $55\%$.

Modulus of elasticity of epoxy = $0.5\times {10}^6\text{psi}$

From given information,

Calculation of volume fraction,

$f_{Al}=\frac{V_{Al}}{V_{total}}$

Where,

$f_{Al}$ is the volume fraction of aluminum.

$V_{Al}$ is the volume of aluminum.

$V_{total}$ is the total volume of all components.

Calculation of total volume, based on aluminum, Kevlar, and epoxy,

${\text{V}}_{\text{Al}}=\text{number}\text{of}\text{sheet}\text{×}\text{thickness}$

Number of sheets of aluminum = 5

Thickness of aluminum sheet = $0.4\text{mm}$

Putting the values in formula,

$\begin{array}{l}{\text{V}}_{\text{Al}}=\text{number}\text{of}\text{sheet}\text{×}\text{thickness}\\ {\text{V}}_{\text{Al}}=5\times 0.4\\ {\text{V}}_{\text{Al}}=2{\text{mm}}^3\end{array}$

Calculation of volume of epoxy, as per given condition:

Volume fraction of Kevlar = $55\%$

Volume of Kevlar = $0.55$

Using rule of mixing for calculation of volume fraction

$\begin{array}{l}{f}_{epoxy}+f_{kevlar}=1\\ f_{epoxy}=1-f_{kevlar}\\ f_{kevlar}=0.55\\ f_{epoxy}=1-0.55\\ f_{epoxy}=0.45\end{array}$

Calculating the volume of kevlar reinforced with epoxy,

$V_{kevlar}=\text{volume}\text{fraction}\text{×}\text{number}\text{of}\text{sheet}\text{×}\text{thickness}$

Thickness of four sheets of epoxy reinforced with Kevlar fiber = $0.2\text{mm}$.

Volume fraction of Kevlar in sheets = $55\%$.

$\begin{array}{l}{V}_{kevlar}=0.55\times 4\times 0.2\\ V_{kevlar}=0.44{\text{mm}}^3\end{array}$

Calculation of volume of epoxy in kevlar fiber,

$V_{epoxy}=\text{volume}\text{fraction}\left(f_{epoxy}\right)\times \text{number}\text{of}\text{sheet}\text{×}\text{thickness}$

$f_{epoxy}=0.45$

Thickness of four sheets of epoxy reinforced with Kevlar fiber = $0.2\text{mm}$.

$\begin{array}{l}{V}_{epoxy}=0.45\times 4\times 0.2\\ V_{epoxy}=0.36{\text{mm}}^3\end{array}$

Calculation of total volume of arall,

$\begin{array}{l}{V}_{tatal}={\text{V}}_{\text{Al}}+{\text{V}}_{\text{epoxy}}+{\text{V}}_{\text{kevlar}}\\ {\text{V}}_{\text{Al}}=2{\text{mm}}^3\\ {\text{V}}_{\text{epoxy}}=0.36{\text{mm}}^3\\ {\text{V}}_{\text{kevlar}}=0.44{\text{mm}}^3\\ {\text{V}}_{\text{total}}=2+0.36+0.44\\ {\text{V}}_{\text{total}}=2.8{\text{mm}}^3\end{array}$

Calculation of volume fraction of aluminum,

$f_{Al}=\frac{V_{Al}}{V_{total}}$

${\text{V}}_{\text{Al}}=2{\text{mm}}^3$

${\text{V}}_{\text{total}}=2.8{\text{mm}}^3$

$\begin{array}{l}{f}_{Al}=\frac{2}{2.8}\\ f_{Al}=0.714\end{array}$

Calculation of volume fraction of Kevlar,

$f_{kevlar}=\frac{V_{kevlar}}{V_{total}}$

${\text{V}}_{\text{kevlar}}=0.44{\text{mm}}^3$

${\text{V}}_{\text{total}}=2.8{\text{mm}}^3$

$\begin{array}{l}{f}_{kevlar}=\frac{0.44}{2.8}\\ f_{kevlar}=0.157\end{array}$

Calculation of volume fraction of epoxy,

$f_{epoxy}=\frac{V_{epoxy}}{V_{total}}$

${\text{V}}_{\text{epoxy}}=0.36{\text{mm}}^3$

${\text{V}}_{\text{total}}=2.8{\text{mm}}^3$

$\begin{array}{l}{f}_{epoxy}=\frac{0.36}{2.8}\\ f_{epoxy}=0.128\end{array}$

Calculation of modulus of elasticity parallel to fiber,

$E_C=f_{Al}{E}_1+f_{epoxy}{E}_2+f_{kevlar}{E}_3$

$E_1$ be the modulus of elasticity for aluminum = $70\times {10}^9\text{pascal}$.

$E_2$ be the modulus of elasticity for epoxy = $3450\times {10}^6\text{pascal}$.

$E_3$ be the modulus of elasticity for Kevlar = $124\times {10}^9\text{pascal}$.

$f_{Al}=0.714$

$f_{kevlar}=0.157$

$f_{epoxy}=0.128$

Putting the values,

$\begin{array}{l}{E}_C=70\times {10}^9\times 0.714+3450\times {10}^6\times 0.128+124\times {10}^9\times 0.157\\ E_C=698896\times {10}^5\text{pascal}\end{array}$

The required value of modulus of elasticity parallel to fiber is $E_C=698896\times {10}^5\text{pascal}$.

Calculation of modulus of elasticity perpendicular to fiber,

$\frac{1}{E_C}=\frac{f_1}{E_1}+\frac{f_2}{E_2}+\frac{f_3}{E_3}$

$E_1$ be the modulus of elasticity for aluminum = $70\times {10}^9\text{pascal}$.

$E_2$ be the modulus of elasticity for epoxy = $3450\times {10}^6\text{pascal}$.

$E_3$ be the modulus of elasticity for Kevlar = $124\times {10}^9\text{pascal}$.

$f_{Al}=0.714$

$f_{kevlar}=0.157$

$f_{epoxy}=0.128$

Putting the values,

$\begin{array}{l}\frac{1}{E_C}=\frac{0.714}{70\times {10}^9}+\frac{0.157}{3450\times {10}^6}+\frac{0.128}{124\times {10}^9}\\ E_C=199096\times {10}^5\text{pascal}\end{array}$

The required value of modulus of elasticity perpendicular to fiber is $E_C=698896\times {10}^5\text{pascal}$.

The advantages of arall over unreinforced aluminum is as follows:

1. Arall possess high fatigue strength as compared with unreinforced aluminum.
2. Arall possess the properties of high corrosion resistance and wear resistance.
3. The mechanical properties of arall does not get affected by humidity.

Conclusion

$E_C=698896\times {10}^5\text{pascal}$ is the required value of modulus of elasticity parallel to fiber.

$E_C=698896\times {10}^5\text{pascal}$ is the required value of modulus of elasticity perpendicular to fiber.