Chapter 12, Problem 12.14P

(a)

To determine

The strain in composite when stress in matrix is $60\text{ MPa}$ .

Answer to Problem 12.14P

The strain in composite when stress in matrix reaches $60\text{ MPa}$ is $0.025$.

Explanation of Solution

Given:

Elastic modulus of polycarbonate matrix is $2.4\times {10}^9\text{ Pa}$ .

Maximum allowable stress in matrix is $60\text{ MPa}$ .

Concept used:

Write the expression for strain in composite.

  ${\epsilon }_C=\frac{{\sigma }_m}{E_m}$ …… (1)

Here, ${\epsilon }_C$ is the strain in composite, ${\sigma }_m$ is the stress in matrix and $E_m$ is the elastic modulus of matrix.

Calculation:

Substitute $60\text{ MPa}$ for ${\sigma }_m$ and $2.4\times {10}^9\text{ Pa}$ for $E_m$ in equation (1).

  $\begin{array}{c}{\epsilon }_C=\frac{60\text{ MPa}\left(\frac{{10}^6\text{Pa}}{1\text{}\text{MPa}}\right)}{2.4\times {10}^9\text{ Pa}}\\ =0.025\end{array}$

Conclusion:

Thus, the strain in composite when stress in matrix reaches $60\text{ MPa}$ is $0.025$.

(b)

To determine

The fiber stress for continuous fiber composite and identify if this is less than tensile strength of fiber.

Answer to Problem 12.14P

The fiber stress for continuous fiber composite is $3.25\text{ GPa}$ this is close to tensile strength of fiber.

Explanation of Solution

Given:

Elastic modulus of Fiber is $130\times {10}^9\text{ Pa}$ .

Concept used:

Write the expression for fiber stress.

  ${\sigma }_f={\epsilon }_C{E}_f$ …… (2)

Here, ${\sigma }_f$ is the fiber stress and $E_f$ is the modulus of elasticity of fiber.

Calculation:

Substitute $0.025$ for ${\epsilon }_C$ and $130\times {10}^9\text{ Pa}$ for $E_f$ in equation (2).

  $\begin{array}{c}{\sigma }_f=\left(0.025\right)\left(130\times {10}^9\text{ Pa}\right)\\ =3.25\times {10}^9\text{ Pa}\left(\frac{1\text{ GPa}}{{10}^9\text{ Pa}}\right)\\ =3.25\text{ GPa}\end{array}$

Conclusion:

Thus, the fiber stress for continuous fiber composite is $3.25\text{ GPa}$ this is close to tensile strength of fiber.

(c)

To determine

Critical fiber length

Answer to Problem 12.14P

Critical fiber length is $6.1\times {10}^{-4}\text{m}$ .

Explanation of Solution

Given:

Diameter of aramid fiber is $12\times {10}^{-6}\text{ m}$ .

Shear strength of Polycarbonate-aramid interface is $32\times {10}^6\text{ Pa}$ .

Concept used:

Write the expression for critical aspect ratio.

  $\frac{l_C}{d_f}=\frac{{\sigma }_f}{2{\tau }_{mf}^{*}}$ …… (3)

Here, $l_C$ is the critical fiber length, $d_f$ is the diameter of fiber and ${\tau }_{mf}^{*}$ is the maximum shear stress in interface.

Calculation:

Substitute $12\times {10}^{-6}\text{ m}$ for $d_f$ , $32\times {10}^6\text{ Pa}$ for ${\tau }_{fm}^{*}$ and $3.25\text{ GPa}$ for ${\sigma }_f$ in equation (3).

  $\begin{array}{c}\frac{l_C}{12\times {10}^{-6}\text{ m}}=\frac{3.25\text{ GPa}\left(\frac{{10}^9\text{ Pa}}{1\text{ GPa}}\right)}{2\left(32\times {10}^6\text{ Pa}\right)}\\ l_C=6.1\times {10}^{-4}\text{m}\end{array}$

Conclusion:

Thus, the Critical fiber length is $6.1\times {10}^{-4}\text{m}$ .

(d)

To determine

Average stress in chopped aramid fiber.

Answer to Problem 12.14P

Average stress in chopped aramid fiber is $1.6\text{ GPa}$ .

Explanation of Solution

Given:

Length of aramid fiber is $6\times {10}^{-4}\text{ m}$ .

Concept used:

Write the expression for average fiber stress.

  ${\overline{\sigma }}_f={\sigma }_f\left(1-\frac{l_C}{2l}\right)$ …… (4)

Here, $\overline{{\sigma }_f}$ is the average fiber stress and $l$ is the length of fiber.

Calculation:

Substitute $3.25\text{ GPa}$ for ${\sigma }_f$ , $6.1\times {10}^{-4}\text{m}$ for $l_C$ and $6\times {10}^{-4}\text{ m}$ for $l$ in equation (4).

  $\begin{array}{c}{\overline{\sigma }}_f=3.25\text{ GPa}\left(1-\frac{6.1\times {10}^{-4}\text{ m}}{2\left(6\times {10}^{-4}\text{ m}\right)}\right)\\ =1.6\text{ GPa}\end{array}$

Conclusion:

Thus, the average stress in chopped aramid fiber is $1.6\text{ GPa}$ .

(e)

To determine

Composite material stress.

Answer to Problem 12.14P

Composite material stress is $753\text{ MPa}$ .

Explanation of Solution

Given:

Volume percentage of aramid chopped fiber is $45\%$ .

Volume percent of polycarbonate matrix is $55\%$ .

Tensile strength of aramid fiber is $3.6\times {10}^9\text{ Pa}$ .

Tensile strength of Polycarbonate is $80\text{ MPa}$ .

Concept used:

Write the expression for composite material stress.

  ${\sigma }_C=V_f{\overline{\sigma }}_f+V_m{\sigma }_m$ …… (5)

Here, ${\sigma }_C$ is the composite material stress, $V_f$ is the volume fraction of fiber and $V_m$ is the volume fraction of matrix.

Calculation:

Substitute $0.45$ for $V_f$ , $0.55$ for $V_m$ , $1.6\text{ GPa}$ for ${\overline{\sigma }}_f$ and $60\text{ MPa}$ for ${\sigma }_m$ in equation (5).

  $\begin{array}{c}{\sigma }_C=\left(0.45\right)\left(1.6\text{ GPa}\right)\left(\frac{{10}^9\text{ Pa}}{1\text{ GPa}}\right)+\left(0.55\right)\left(60\text{ MPa}\right)\left(\frac{1{\text{0}}^6\text{ Pa}}{1\text{ MPa}}\right)\\ =720\times {10}^6\text{ Pa}+33\times {10}^6\text{ Pa}\\ \text{=753}\times {\text{10}}^6\text{ Pa}\left(\frac{1\text{ MPa}}{{10}^6\text{ Pa}}\right)\\ =753\text{ MPa}\end{array}$

Conclusion:

Thus, the composite material stress is $753\text{ MPa}$ .