Chapter 12, Problem 12.17P

(a)

To determine

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

Answer to Problem 12.17P

The magnitude and direction of thermal stresses in matrix and fibers are $-171.31\times {10}^6\text{ Pa}$ and $57.10\times {10}^6\text{ Pa}$ respectively.

Explanation of Solution

Given:

Volume fraction of carbon fiberis $75\%$ .

Volume fraction of epoxy resin is $25\%$ .

Initial temperature is $\text{22}\text{°C}$ .

Final temperature is $260°\text{C}$ .

Elastic modulus of carbon fiber is $380\text{GPa}$ .

Ultimate tensile strength of carbon fiber is $2.4\text{}\text{GPa}$ .

Thermal expansion coefficient of carbon fiber is $-0.7\times {10}^{-6}{\text{K}}^{-1}$ .

Elastic modulus of Epoxy is $4\text{GPa}$ .

Ultimate tensile strength of epoxy resin is $0.07\text{ GPa}$

Thermal expansion coefficient of 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 coefficient of fiber, ${\alpha }_m$ is the thermal expansion coefficient of 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 ${\sigma }_C$ , $0.75$ for $V_f$ , $0.25$ for $V_m$ in equation (1).

  $\begin{array}{c}0=\left(0.75\right){\sigma }_f+\left(0.25\right){\sigma }_m\\ {\sigma }_m=-3{\sigma }_f\end{array}$

Substitute $-3{\sigma }_f$ for ${\sigma }_m$ in equation (2).

  ${\epsilon }_m=-\frac{3{\sigma }_f}{E_m}$

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

  $\left({\alpha }_f-{\alpha }_m\right)\left(T_f-T_i\right)+\left(\frac{{\sigma }_f}{E_f}+\frac{3{\sigma }_f}{E_m}\right)=0$

Substitute $-0.7\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_f$ , $60\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_m$ , $260°\text{C}$ for $T_f$ , $22°\text{C}$ for $T_i$ , $380\text{GPa}$ for $E_f$ and $4\text{ GPa}$ for $E_m$ in above expression.

  $\begin{array}{l}\left[\begin{array}{l}\left(-0.7\times {10}^{-6}°{\text{C}}^{-1}-60\times {10}^{-6}°{\text{C}}^{-1}\right)\left(260°\text{C}-22°\text{C}\right)\\ +{\sigma }_f\left(\frac{1}{380\text{}\text{GPa}}+\frac{1}{4\text{ GPa}}\right)\end{array}\right]=0\\ -0.014447+{\sigma }_f\left(0.253{\text{ GPa}}^{-1}\left(\frac{1\text{ GPa}}{{10}^9\text{ Pa}}\right)\right)=0\end{array}$

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

  $\begin{array}{l}{\sigma }_f=\frac{0.014447}{0.253\times {10}^{-9}{\text{ Pa}}^{-1}}\\ {\sigma }_f=57.10\times {10}^6\text{ Pa}\end{array}$

Calculate the value of stress in matrix.

  $\begin{array}{l}{\sigma }_m=-3{\sigma }_f\\ {\sigma }_m=-171.31\times {10}^6\text{ Pa}\end{array}$

Conclusion:

Thus, the magnitude and direction of thermal stresses in matrix and fibers are $-171.31\times {10}^6\text{ Pa}$ and $57.10\times {10}^6\text{ Pa}$ respectively.

(b)

To determine

Composite strain in cooling the composite.

Answer to Problem 12.17P

Composite strain in cooling the composite is $-0.02855$ .

Explanation of Solution

Given:

Initial temperature is $\text{22}\text{°C}$ .

Final temperature is $260°\text{C}$ .

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 $-171.31\times {10}^6\text{ Pa}$ for ${\sigma }_m$ and $4\text{ GPa}$ for $E_m$ in equation (2).

  $\begin{array}{c}{\epsilon }_m=\frac{-171.31\times {10}^6\text{ Pa}}{4\text{ GPa}\left(\frac{{10}^9\text{ Pa}}{1\text{ GPa}}\right)}\\ =-0.04283\end{array}$

Substitute $-0.04283$ for ${\epsilon }_m$ , $60\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_m$ , $260°\text{C}$ for $T_f$ and $22°\text{C}$ for $T_i$ in equation (4).

  $\begin{array}{c}{\epsilon }_C=\left(60\times {10}^{-6}°{\text{C}}^{-1}\right)\left(260°\text{C}-22°\text{C}\right)-0.04283\\ =-0.02855\end{array}$

Conclusion:

Thus, Composite strain in cooling the composite is $-0.02855$ .

(c)

To determine

Answer to Problem 12.17P

The component of design which can cause problem and justify your answer.

Explanation of Solution

The tensile stress in the fiber is less than the ultimate tensile strength of fiber and the stress in matrix is compressive in nature whereas the tensile strength of epoxy matrix is $\text{0}\text{.07 GPa}$ . For the material to prevent failure the ultimate compressive strength of matrix must be greater than the applied compressive stress on it.