Chapter 12, Problem 12.8P

(a)

To determine

The elastic modulus of composite in axial direction.

Answer to Problem 12.8P

The elastic modulus of composite in axial direction is $99.5\text{}\text{GPa}$ .

Explanation of Solution

Given:

Volume percent of Kevlar fiber is $75\%$ .

Volume percent of epoxy is $25\%$ .

Elastic modulus of Kevlar is $131\times {10}^9\text{ Pa}$ .

Elastic modulus of epoxy is $5\times {10}^9\text{Pa}$ .

Concept used:

Write the expression for elastic modulus of composite.

  $E_{\text{C}}=E_m{V}_m+E_f{V}_f$ …… (1)

Here, $E_{\text{C}}$ is the modulus of elasticity of composite, $E_m$ is the modulus of elasticity of matrix, $V_m$ is the volume fraction of matrix, $E_f$ is the modulus of elasticity fiber and $V_f$ is the volume fraction of fiber.

Calculation:

Substitute $131\text{GPa}$ for $E_f$ , $0.75$ for $V_f$ , $5\text{}\text{GPa}$ for $E_m$ and $0.25$ for $V_m$ in equation (1).

  $\begin{array}{c}{E}_{\text{C}}=\left(0.25\right)\left(5\text{ GPa}\right)+\left(0.70\right)5\left(131\text{GPa}\right)\\ =99.5\text{ GPa}\end{array}$

Conclusion:

Thus, the elastic modulus of composite in axial direction is $99.5\text{}\text{GPa}$ .

(b)

To determine

The elastic modulus of composite in transverse direction.

Answer to Problem 12.8P

The elastic modulus of composite in transverse direction is $18\text{ GPa}$ .

Explanation of Solution

Concept used:

Write the expression for elastic modulus of E-glass composite in the direction perpendicular to fiber axis.

  ${E^{\prime }}_C=\frac{E_f{E}_m}{V_f{E}_m+V_m{E}_f}$ …… (2)

Here, ${E^{\prime }}_C$ is the transverse elastic modulus of material.

Calculation:

Substitute $131\text{ GPa}$ for $E_f$ , $0.75$ for $V_f$ , $5\text{ GPa}$ for $E_m$ and $0.25$ for $V_m$ in equation (2).

  $\begin{array}{c}{E^{\prime }}_C=\frac{\left(131\right)\left(5\right)}{\left(0.75\right)\left(5\right)+\left(0.25\right)\left(131\right)}\\ =18\text{ GPa}\end{array}$

Conclusion:

Thus, the elastic modulus of composite in transverse direction is $18\text{ GPa}$ .

(c)

To determine

The axial strainin composite due to axial stress.

Answer to Problem 12.8P

The axial strain in composite due to axial stress is $0.005$ .

Explanation of Solution

Given:

Axial stress is $0.5\times {10}^9\text{}\text{Pa}$ .

Concept used:

Write the expression for axial strain in composite.

  $\epsilon =\frac{\sigma }{E_{\text{C}}}$ …… (3)

Here, $\epsilon$ is the axial strain in composite and $\sigma$ is the average composite stress.

Calculation:

Substitute $0.5\times {10}^9{\text{N/m}}^2$ for $\sigma$ and $99.5\text{ GPa}$ for $E_C$ in equation (3).

  $\begin{array}{c}\epsilon =\frac{0.5\times {10}^9{\text{N/m}}^2}{99.5\text{ GPa}\left(\frac{{10}^9{\text{N/m}}^2}{1\text{GPa}}\right)}\\ =0.005\end{array}$

Conclusion:

Thus, the axial strain in composite due to axial stress is $0.005$ .

(d)

To determine

The stresses in the fiber and in matrix of composite due to axial stress.

Answer to Problem 12.8P

The stresses in the fiber and in matrix of composite due to axial stress is $0.665\text{GPa}$ and $0.025\text{ GPa}$ respectively.

Explanation of Solution

Given:

Transverse stress is $4\times {10}^6\text{ Pa}$ .

Tensile strength of Kevlar is $4.0\times {10}^9\text{Pa}$ .

Concept used:

Write the expression for stress in fiber by considering the iso-strain model for axial strain.

  ${\sigma }_f=\epsilon E_f$ …… (4)

Here, ${\sigma }_f$ is the stress in fiber.

Write the expression for stress in matrix by considering the iso-strain model for axial strain.

  ${\sigma }_m=\epsilon E_m$ …… (5)

Here, ${\sigma }_m$ is the stress in fiber.

Calculation:

Substitute $0.005$ for $\epsilon$ and $131\text{ GPa}$ for $E_f$ in equation (4).

  $\begin{array}{c}{\sigma }_f=\left(0.005\right)\left(131\right)\\ =0.655\text{ GPa}\end{array}$

Substitute $0.005$ for $\epsilon$ and $5\text{ GPa}$ for $E_m$ in equation (5).

  $\begin{array}{c}{\sigma }_m=\left(0.005\right)\left(5\right)\\ =0.025\text{ GPa}\end{array}$

Conclusion:

Thus, the stresses in the fiber and in matrix of composite due to axial stress is $0.665\text{GPa}$ and $0.025\text{ GPa}$ respectively.

(e)

To determine

Whether each of the component can withstand the axial stress or not, if not what is the fracture stress.

Explanation of Solution

The calculated stress for fiber is very less than the minimum tensile strength of Kevlar. This indicates that the fiber is sufficiently string to withstand the stress.

Matrix can also hold the tensile stress as the minimum tensile strength of epoxy resin is larger than the calculate stress in it.

(f)

To determine

The transverse fracture stress.

Answer to Problem 12.8P

The transverse fracture stress is $1.1\times {10}^6\text{ Pa}$ .

Explanation of Solution

Given:

Tensile strength of epoxy is $50\times {10}^6\text{Pa}$ .

Concept used:

Write the expression for transverse tensile strength of composite.

  ${\sigma }_{2u}={\sigma }_{mu}\left(1-2{\left(\frac{v_f}{\pi }\right)}^{0.5}\right)$ …… (1)

Here, ${\sigma }_{2u}$ is the transverse tensile strength of composite, ${\sigma }_{mu}$ is the tensile strength of composite and $v_f$ is the volume fraction of fiber.

Calculation:

Substitute $50\times {10}^6\text{ Pa}$ for ${\sigma }_{mu}$ and $0.75$ for $v_f$ in equation (1).

  $\begin{array}{c}{\sigma }_{2u}=\left(50\times {10}^6\text{ Pa}\right)\left(1-2{\left(\frac{0.75}{\pi }\right)}^{0.5}\right)\\ =\left(50\times {10}^6\text{ Pa}\right)\left(1-0.98\right)\\ =1.1\times {10}^6\text{ Pa}\end{array}$

Conclusion:

Thus, the transverse fracture stress is $1.1\times {10}^6\text{ Pa}$ .