Question

Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction

Answer 1

Answer:

Explanation:

From the given information:

weight of fiber $w_f$ = 3.0 g

weight of composite specimen $w_c$ = 4.0 g

specimen composite weight in water $C_{wm}$ = 2.0 g

specific gravity of fiber $S_f$ = 2.4

specific gravity of matrix $S_m$ = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite $\rho_{ct}$ can be determined by using the formula:

$\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}$

$\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}$

$\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}$

$\dfrac{1}{\rho_{ct}} =0.505$

$\rho_{ct} =\dfrac{1}{0.505}$

$\mathbf{\rho_{ct} = 1.980 \ g/cm^3}$

The experimental density $\rho _{ce}$ is determined  by using the equation:

$\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }$

$\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }$

$\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }$

$\mathbf{\rho _{ce} = 1.620 \ g/cm^3}$

The void fraction is: $= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}$

$= \dfrac{1.980-1.620}{1.980}$

= 0.1818