Chapter 12, Problem 12.16P

(a)

To determine

The magnitude and direction of thermal stresses in matrix and fibers.

Answer to Problem 12.16P

The magnitude and direction of thermal stresses in matrix and fibers are $26.1\times {10}^6\text{}\text{Pa}$ and $-26.1\times {10}^6\text{}\text{Pa}$ respectively.

Explanation of Solution

Given:

Volume fraction of uniaxial glass fiber is $50\%$ .

Volume fraction of epoxy resin is $50\%$ .

Initial temperature is $\text{150}\text{°C}$ .

Final temperature is $25°\text{C}$ .

Elastic modulus of glass fiber is $76\text{}\text{GPa}$ .

Thermal expansion coefficientof glass fiberis $5\times {10}^{-6}°{\text{C}}^{-1}$ .

Elastic modulus of Epoxyis $4\text{GPa}$ .

Thermal expansion coefficientof Epoxy is $60\times {10}^{-6}°{\text{C}}^{-1}$ .

Concept used:

Write the expression for stress carried by composite.

  ${\sigma }_c=V_f{\sigma }_f+V_m{\sigma }_m$ …… (1)

Here, ${\sigma }_c$ is the stress in of composite, ${\sigma }_f$ is the stress in fiber, $V_f$ is the volume fraction of fiber, ${\sigma }_m$ is the stress in matrix and $V_m$ is the volume fraction of matrix.

Write the expression for strain in Fiber.

  ${\epsilon }_f=\frac{{\sigma }_f}{E_f}$

Here, ${\epsilon }_f$ is the strain in Fiber, ${\sigma }_f$ is the tensile strength of Fiber and $E_f$ is the modulus of elasticity of Fiber.

Write the expression for strain in Matrix.

  ${\epsilon }_m=\frac{{\sigma }_m}{E_m}$ …… (2)

Here, ${\epsilon }_m$ is the strain in Matrix, ${\sigma }_m$ is the tensile strength of Matrix and $E_m$ is the modulus of elasticity of Matrix.

Write the expression for the total strain in the composite.

  $\left({\alpha }_f-{\alpha }_m\right)\left(T_f-T_i\right)+\left({\epsilon }_f-{\epsilon }_m\right)=0$ …… (3)

Here, ${\alpha }_f$ is the thermal expansion coefficientof fiber, ${\alpha }_m$ is the thermal expansion coefficientof matrix, $T_i$ is the initial in temperature of composite and , $T_f$ is the final in temperature of composite.

Calculation:

The composite stress is zero because there is no application of stress to the composite material in the axial direction.

Substitute $0\text{ MPa}$ for ${\sigma }_C$ , $0.5$ for $V_f$ , $0.5$ for $V_m$ in equation (1).

  $\begin{array}{c}0=\left(0.5\right){\sigma }_f+\left(0.5\right){\sigma }_m\\ {\sigma }_m=-{\sigma }_f\end{array}$

Substitute $-{\sigma }_f$ for ${\sigma }_m$ in equation (2).

  ${\epsilon }_m=-\frac{{\sigma }_f}{E_m}$

Substitute $-\frac{{\sigma }_f}{E_m}$ for ${\epsilon }_m$ and $\frac{{\sigma }_f}{E_f}$ for ${\epsilon }_f$ in equation (3).

  $\left({\alpha }_f-{\alpha }_m\right)\left(T_f-T_i\right)+\left(\frac{{\sigma }_f}{E_f}+\frac{{\sigma }_f}{E_m}\right)=0$

Substitute $5\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_f$ , $60\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_m$ , $25°\text{C}$ for $T_f$ , $150°\text{C}$ for $T_i$ , $76\text{}\text{GPa}$ for $E_f$ and $4\text{ GPa}$ for $E_m$ in above expression.

  $\begin{array}{l}\left[\begin{array}{l}\left(5\times {10}^{-6}°{\text{C}}^{-1}-60\times {10}^{-6}°{\text{C}}^{-1}\right)\left(25°\text{C}-150°\text{C}\right)\\ +{\sigma }_f\left(\frac{1}{76\text{}\text{GPa}}+\frac{1}{4\text{ GPa}}\right)\end{array}\right]=0\\ 6.875\times {10}^{-3}+{\sigma }_f\left(0.263{\text{ GPa}}^{-1}\left(\frac{1\text{ GPa}}{{10}^9\text{ Pa}}\right)\right)=0\end{array}$

Simplify above expression for ${\sigma }_f$ .

  $\begin{array}{l}{\sigma }_f=-\frac{6.875\times {10}^{-3}}{0.263\times {10}^{-9}{\text{ Pa}}^{-1}}\\ {\sigma }_f=-26.14\times {10}^6\text{ Pa}\end{array}$

Calculate the value of stress in matrix.

  $\begin{array}{l}{\sigma }_m=-{\sigma }_f\\ {\sigma }_m=26.14\times {10}^6\text{ Pa}\end{array}$

Conclusion:

Thus, the magnitude and direction of thermal stresses in matrix and fibers are $26.1\times {10}^6\text{}\text{Pa}$ and $-26.1\times {10}^6\text{}\text{Pa}$ respectively.

(b)

To determine

Composite strain in cooling the composite.

Answer to Problem 12.16P

Composite strain in cooling the composite is $-0.00097$ .

Explanation of Solution

Given:

Initial temperature is $\text{150}\text{°C}$ .

Final temperature is $25°\text{C}$ .

Concept used:

Write the expression for stress in composite.

  ${\epsilon }_C={\alpha }_m\left(T_f-T_i\right)+{\epsilon }_m$ …… (4)

Here, ${\epsilon }_C$ is the stress in composite

Calculation:

Substitute $26.1\times {10}^6\text{}\text{Pa}$ for ${\sigma }_m$ and $4\text{ GPa}$ for $E_m$ in equation (2).

  $\begin{array}{c}{\epsilon }_m=\frac{26.1\times {10}^6\text{ Pa}}{4\text{ GPa}\left(\frac{{10}^9\text{ Pa}}{1\text{ GPa}}\right)}\\ =0.00653\end{array}$

Substitute $0.00653$ for ${\epsilon }_m$ , $60\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_m$ , $25°\text{C}$ for $T_f$ and $150°\text{C}$ for $T_i$ in equation (4).

  $\begin{array}{c}{\epsilon }_C=\left(60\times {10}^{-6}°{\text{C}}^{-1}\right)\left(25°\text{C}-150°\text{C}\right)+0.00653\\ =-0.00097\end{array}$

Conclusion:

Thus, Composite strain in cooling the composite is $-0.00097$ .