Engineering Materials In an aligned and continuous glass fib

(Engineering Materials)

In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 92 % of a load applied in the longitudinal direction.

(a) Using the data provided, determine the volume fraction of fibers that will be required.

(b) What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi).

76 (11,000)

(a):........

(b):......... MPa

Solution

Let Vf , Vm, represent volume fraction of Fiber and Matrix respectively.

Let Fc be the uni axial load applied on the composite in the longitudinal direction of fiber.

And Ff and Fm are loads taken by fiber and matrix respectively.

Implies Fc = Ff + Fm

Let sc , sf, sm be the stresses of composite, fibre and matrix respectively.

And Ac, Af, Am be the area of composite, fibre and matrix respectively.

sc * Ac = sf * Af + sm * Am.

Let ec, ef, em be the strains in composite, fibre and matrix respectively.

Let Ec, Ef, Em be the young’s modulus of the composite, fibre and matrix respectively.

Therefore

Ec * ec* Ac = Em * em * Am + Ef * ef * Af.

The strains in the composite , fibre and matrix are equal, implies :

Ec = ((Em * Am) /Ac) + ((Ef *Af)/Ac)

By definition of volume fraction

Ec = Em * Vm + Ef * Vf.     This is called rule of mixtures.

Ec = 3 *(1-Vf) + 72.5 * Vf Þ Ec = 3 + 69.5 Vf.

The load shared by the fibre is given by:

( Ff /Fc ) = (Ef / Ec) * Vf.

Load taken by fibre is given by 92 % of composite. Ef = 72.5 GPa, And Em = 3 GPa.

0.92 = (72.5/(3+69.5Vf)) *Vf

72.5 Vf = 63.94 Vf + 2.76    Þ Vf = 0.3224.

Volume fraction of fibre is 0.3224.

Volume Fraction of matrix is Vm = 1 – Vf = 1 – 0.3224 = 0.6776.

Tensile strength of fibre at failure is 3400 MPa.

Tensile strength of matrix at fibre failure is 30Mpa.

Therefore Tensile Strength of composite at fiber failure of matrix stress of 30 MPa is

sc = sf * Vf + sm *Vm = 3400 * 0.3224 + 30 * 0.6776 = 1116.488 MPa.