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

Step-by-step 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:

`(1)/(\rho_(ct)) = (w_f)/(w_cS_f)+ (w_m)/(w_cS_m)`

`(1)/(\rho_(ct)) = (3.0)/((4.0 * 2.4))+ (1.0)/((4.0* 1.3))`

`(1)/(\rho_(ct)) = (3.0)/(9.6)+ (1.0)/(5.2)`

`(1)/(\rho_(ct)) =0.505\\`

`\rho_(ct) =(1)/(0.505)`

`\mathbf{\rho_(ct) = 1.980 \ g/cm^3}`

The experimental density
`\rho _(ce)` is determined by using the equation:

`\rho _(ce) = (w_f + w_c)/((w_f )/(S_f) + (w_c )/(S_m) )`

`\rho _(ce) = (3.0 + 4.0)/((3.0 )/(2.4) + (4.0 )/(1.3) )`

`\rho _(ce) = (3.0 + 4.0)/(1.250 +3.077 )`

`\mathbf{\rho _(ce) = 1.620 \ g/cm^3}`

The void fraction is:
`= (\rho_(ct)-\rho_(ce))/(\rho_(ct))`

`= (1.980-1.620)/(1.980)`

= 0.1818