Question

Boron coated with SiC (or Borsic) reinforced aluminum containing aligned 20 vol% fibers is an important high-temperature, lightweight composite material. Estimate the modulus of elasticity perpendicular to the fibers. The modulus of elasticity of the fibers is 55 X 10^6 psi and the modulus of elasticity of aluminum is 10 X 10^6 psi.
a. 19 X 10^6 psi
b. 55 X 10^6 psi
c. 12 X 10^6 psi
d. 10 X 10^6 psi

Answer 1

Answer:

Option C is correct.

Modulus of elasticity of the composite perpendicular to the fibers = (12 × 10⁶) psi

Explanation:

For combination of materials, the properties (especially physical properties) of the resulting composite is a sum of the fractional contribution of each material thay makes up the composite.

In this composite,

The fibres = 20 vol%

Aluminium = 80 vol%

Modulus of elasticity of the composite

= [0.2 × E(fibres)] + [0.8 × E(Al)]

Modulus of elasticity of the fibers = E(fibres) = (55 × 10⁶) psi. =

Modulus of elasticity of aluminum = E(Al) = (10 × 10⁶) psi.

But modulus of elasticity of the composite perpendicular to the fibers is given in the expression.

[1 ÷ E(perpendicular)]

= [0.2 ÷ E(fibres)] + [0.8 ÷ E(Al)]

[1 ÷ E(perpendicular)]

= [0.2 ÷ (55 × 10⁶)] + [0.8 ÷ (10 × 10⁶)]

= (3.636 × 10⁻⁹) + (8.00 × 10⁻⁸)

= (8.3636 × 10⁻⁸)

E(perpendicular) = 1 ÷ (8.3636 × 10⁻⁸)

= 11,961,722.5 psi = (11.96 × 10⁶) psi

= (12 × 10⁶) psi

Hope this Helps!!!