. Calculate the following:

a. the modulus of elasticity,
b. the tensile strength,
c. the fraction of the load carried by the fiber,

for the following composite material stressed under isostrain conditions.
The composite consists of a continuous glass‐fiber‐reinforced‐epoxy resin produced by
Using 60 percent by volume of E-glass fibers having a modulus of elasticity of
Ef =7.24× 104 MPa and a tensile strength of 2.4 GPa and a hardened epoxy resin with a
Modulus of Em = 3.1 × 103 MPa and a tensile strength of 0.06 GPa.

1 answer

To calculate the modulus of elasticity, tensile strength, and the fraction of the load carried by the fiber for the given composite material under isostrain conditions, we can use the following formulas:

a) Modulus of Elasticity (Ec):
Ec = Ef * Vf + Em * Vm

Where:
Ef = Modulus of elasticity of the fiber = 7.24 × 10^4 MPa
Em = Modulus of elasticity of the matrix = 3.1 × 10^3 MPa
Vf = Volume fraction of the fiber = 60% = 0.60
Vm = Volume fraction of the matrix = 100% - 60% = 40% = 0.40

Ec = (7.24 × 10^4 MPa * 0.60) + (3.1 × 10^3 MPa * 0.40)
Ec = 43440 MPa + 1240 MPa
Ec = 44680 MPa

b) Tensile Strength (σc):
σc = σf * Vf + σm * Vm

Where:
σf = Tensile strength of the fiber = 2.4 GPa = 2400 MPa
σm = Tensile strength of the matrix = 0.06 GPa = 60 MPa

σc = (2400 MPa * 0.60) + (60 MPa * 0.40)
σc = 1440 MPa + 24 MPa
σc = 1464 MPa

c) Fraction of the load carried by the fiber (η_fiber):
η_fiber = Ef * Vf / Ec

η_fiber = (7.24 × 10^4 MPa * 0.60) / 44680 MPa
η_fiber = (43440 MPa) / 44680 MPa
η_fiber = 0.972

The fraction of the load carried by the fiber is approximately 0.972, or 97.2%.

In summary:
a) Modulus of Elasticity (Ec) = 44680 MPa
b) Tensile Strength (σc) = 1464 MPa
c) Fraction of the load carried by the fiber (η_fiber) = 0.972, or 97.2%