Question

A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 31%. The remainder is vinyl ester. The density of the vinyl ester is 0.79 g/cm3, and its modulus of elasticity is 4.04 GPa. The density of E-glass is 3.011 g/cm3, and its modulus of elasticity is 80 GPa. A section of composite 1.00 cm by 25.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the modulus of elasticity of the composite in GPa.

Answer 1

Full Question

Calculate the modulus of elasticity of the composite

I. In the longitudinal direction of the glass fibre

2. In the perpendicular direction of glass fibre

Answer:

1. 27.5876 GPa

2. 5.725 GPa

Explanation:

Given

Dimensions of the fibreglass is given by: 1.00 cm x 25.00 cm x 200.00 cm

Volume = 1 * 25 * 200 = 5000cm³

P1 = The volume fraction of the E-glass is 31%

P2 = The remainder 69% is Vinyl Ester

Volume of Vinyl Ester= 69% * 5000cm³ = 3450cm³

I. In the longitudinal direction of the glass fibre

Where

Modulus of elasticity of glass, E1 = 80 Gpa

Modulus of elasticity of vinyl ester, E2 = 4.04 Gpa

Modulus of elasticity = E1P1 + E2P2

Modulus of elasticity of vinyl esters = 31% * 80 + 69% * 4.04

Modulus = 24.8 + 2.7876

Modulus = 27.5876 GPa

ii. In the perpendicular direction of glass fibre

This is solved by

E1.E2/(P1.E2 + P2.E1)

= (80 * 4.04)/(31% * 4.04 + 69% * 80)

= 323.2/(1.2524 + 55.2)

= 5.725177317527687

= 5.725 GPa