Chapter 12, Problem 12.12P

(a)

To determine

The strain where the first change in elastic modulus occurs in composite material.

Answer to Problem 12.12P

Matrix will rupture before the fiber at the strain of $0.003$ .

Explanation of Solution

Given:

Elastic modulus of Boronis $400\text{GPa}$ .

Tensile strength of Boronis $2.8\text{GPa}$ .

Elastic modulus of Aluminumis $70\text{ GPa}$ .

Yield strength of Aluminumis $200\text{ MPa}$ .

Ultimate tensile strength of aluminum is $200\text{ MPa}$ .

Concept used:

Write the expression for failure strain in Fiber.

  ${\epsilon }_{fu}=\frac{{\sigma }_{fu}}{E_f}$ …… (1)

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

Write the expression for failure strain in Matrix.

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

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

Calculation:

Substitute $\text{2}\text{.8 GPa}$ for ${\sigma }_{fu}$ and $400\text{}\text{GPa}$ for $E_f$ in equation (1).

  $\begin{array}{c}{\epsilon }_{fu}=\frac{\text{2}\text{.8}\text{GPa}}{\text{400}\text{GPa}}\\ =0.007\end{array}$

Substitute $200\text{MPa}$ for ${\sigma }_{mu}$ and $70\text{}\text{GPa}$ for $E_m$ in equation (2).

  $\begin{array}{c}{\epsilon }_{mu}=\frac{200\text{MPa}}{\text{70 GPa}\left(\frac{{10}^3\text{ MPa}}{1\text{ GPa}}\right)}\\ =0.003\end{array}$

Conclusion:

Thus, matrix will rupture before the fiber at the strain of $0.003$ .

(b)

To determine

The composite stress where the change in elastic modulus occurs.

Answer to Problem 12.12P

The composite stress where the change in elastic modulus occurs is $800\text{MPa}$ .

Explanation of Solution

Given:

Volume percent of Boron fiber is $60\%$ .

Volume percent of Aluminumis $40\%$ .

Strain at fracture of Aluminumis $15\%$ .

Concept used:

Write the expression for stress carried by composite.

  ${\sigma }_c=V_f{\sigma }_f+V_m{\sigma }_m$ …… (3)

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 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.

Calculation:

Substitute $400\text{GPa}$ for $E_f$ and $0.003$ for $\epsilon$ in equation (4).

  $\begin{array}{c}{\sigma }_f=\left(0.003\right)\left(400\text{GPa}\right)\\ =1.2\text{GPa}\end{array}$

Substitute $0.6$ for $V_f$ , $0.4$ for $V_m$ , $1.2\text{GPa}$ for ${\sigma }_f$ and $0.20\text{ GPa}$ for ${\sigma }_m$ in equation (3).

  $\begin{array}{c}{\sigma }_c=0.6\left(1.2\right)+0.4\left(0.20\right)\\ =0.8\text{ GPa}\left(\frac{1000\text{ MPa}}{1\text{ GPa}}\right)\\ =800\text{ MPa}\end{array}$

Conclusion:

Thus, the composite stress where the change in elastic modulus occurs is $800\text{MPa}$ .

(c)

To determine

The elastic modulus at low strain and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fails.

Answer to Problem 12.12P

The elastic modulus at low strain is $268\text{ GPa}$ and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fail is $240\text{ GPa}$ .

Explanation of Solution

Concept used:

Write the expression for modulus of elasticity of composite.

  $E_c=V_f{E}_f+V_m{E}_m$ …… (5)

Here, $E_c$ is the modulus of elasticity in of composite, $E_f$ is the modulus of elasticity in fiber and $E_m$ is the modulus of elasticity of matrix.

Calculation:

Substitute $0.6$ for $V_f$ , $0.4$ for $V_m$ , $400\text{}\text{GPa}$ for $E_f$ and $70\text{ GPa}$ for $E_m$ in equation (5).

  $\begin{array}{c}{E}_c=\left(0.6\right)\left(400\right)+\left(0.4\right)\left(170\right)\\ =268\text{GPa}\end{array}$

After yielding of matrix the elastic modulus of composite is the elastic modulus of fiber in it. Therefore, the elastic modulus will be equal to 60% of $400\text{}\text{GPa}$ which will be equal to $240\text{GPa}$ .

Conclusion:

Thus, the elastic modulus at low strain is $268\text{ GPa}$ and the elastic modulus after Aluminum yields, if Aluminum yields before fiber fail is $240\text{ GPa}$ .

(d)

To determine

Fracture strength of composite.

Answer to Problem 12.12P

Fracture strength of composite is $1760\text{ MPa}$ .

Explanation of Solution

Concept used:

Write the expression for fracture strength.

  ${\sigma }_{ts}=V_f{\sigma }_f+V_m{\sigma }_m$ …… (6)

Here, ${\sigma }_{ts}$ is the fracture strength, ${\sigma }_f$ is the maximum strength of fiber and ${\sigma }_m$ is the maximum strength of matrix.

Calculation:

Substitute $0.6$ for $V_f$ , $0.4$ for $V_m$ , $2.8\text{ GPa}$ for ${\sigma }_f$ and $0.20\text{ GPa}$ for ${\sigma }_m$ in equation (6).

  $\begin{array}{c}{\sigma }_{ts}=\left(0.6\right)\left(2.8\right)+\left(0.4\right)\left(0.20\right)\\ =1.760\text{ GPa}\left(\frac{1000\text{ MPa}}{1\text{ GPa}}\right)\\ =1760\text{ MPa}\end{array}$

Conclusion:

Thus, the Fracture strength of composite is $1760\text{ MPa}$ .