In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94 % of a load applied in the longitudinal direction. Modulus of Elasticity [GPa(psi)] Tensile Strength [MPa(psi)] Glass fiber 72.5 (10.5 ×106) 3400 (4.9 ×105) Nylon 6,6 3.0 (4.35 ×105) 76 (11,000) (a) Using the data provided, determine the volume fraction of fibers that will be required.

volume fraction of fibers that will be required;V_f = 0.397

Explanation:

We are given;

Modulus of Elasticity of glass fibre; E_f = 72.5 GPa

Modulus of Elasticity of nylon 6,6; E_m = 3 GPa

%of load fibres are able to carry; F_f = 94% = 0.94

Now, the volume fraction can be determined from the formula;

F_f/F_m = (E_f•V_f)/(E_m(1 - V_f))

Where F_m is given by;

F_m = (1 - F_f). Thus, we now have;

F_f/(1 - F_f) = (E_f•V_f)/(E_m(1 - V_f))

Where;

V_f is volume fraction.

The other terms are stated above.

Thus, plugging in the relevant values to obtain;

0.94/(1 - 0.94) = (72.5•V_f)/(3(1 - V_f))

15.667 = (72.5•V_f)/(3 - 3V_f))

This give;

15.667(3 - 3V_f)) = (72.5•V_f)

47 - 46V_f = (72.5•V_f)

72.5V_f + 46V_f = 47

118.5V_f = 47

V_f = 47/118.5

V_f = 0.397

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