Answer:
It is not possible to produce continued and oriented fiber.
Explanation:
To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:
[tex]E = (1-V_f)E_m+V_fE_f[/tex]
Where,
E is the longitudinal modulus of Elasticity
[tex]V_f[/tex] is the fiber volume ratio
[tex]E_m[/tex] is the elastic modulus of the matrix
[tex]E_f[/tex] is the elastic modulus of the fibers
We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is
[tex]E_f = 131Gpa[/tex]
Moreover from the statement,
[tex]E = 35Gpa[/tex]
[tex]E_m = 3.4Gpa[/tex]
Replacing in the previous equation,
[tex]35 = 3.4 (1-V_f)+131V_f[/tex]
[tex]V_f = 0.25 \rightarrow longitudinal[/tex]
To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,
[tex]E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}[/tex]
Our values are given in this case as:
[tex]E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa[/tex]
Replacing,
[tex]5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}[/tex]
[tex]V_f = 0.351 \rightarrow transversal[/tex]
From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.
