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AlexFokin [52]
3 years ago
13

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

of elasticity of 35 GPa (5×106psi) and 5.17 GPa (7.5×105psi), respectively? Why or why not?Assume that the elastic modulus of the epoxy is 3.4 GPa (4.93×105psi).Also, the value ofEfor aramid fibers is 131GPa.
Physics
1 answer:
Masja [62]3 years ago
5 0

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

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b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

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