Question

A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain conditions. The E-Glass fibers have a tensile modulus of 10x106 psi, and a tensile strength of 0.35x106 psi. The epoxy matrix has a tensile modulus of 0.45x106 psi, and a tensile strength of 9.0x103 psi. What is the strength of the composite material and what fraction of the load is carried by the E-Glass fibers?

Answer 1

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

$V_{f}=0.61$

$V_{m}=1-V_{f}$

     $=1-0.61$

     $=0.39$

$E_{f}=10 \ Mpa$

$\sigma_{f}=0.35 \ Mpa$

$E_{m}=0.45 \ Mpa$ , $\sigma_{m}=9\times 10^{-3} \ Mpa$

As we know,

⇒  $E_{e}=fE_{f}+mE_{m}$

On putting the estimated values, we get

⇒       $=0.61\times 10+0.39\times 0.95$

⇒       $=6.27 \ Mpa$

Now,

⇒  $\sigma_{c}=f\sigma_{f}+m\sigma_{m}$

On putting the estimated values, we get

⇒       $=0.61\times 0.35+0.39\times 0.009$

⇒       $=0.217 \ Mpa$

Therefore,

The load carried by fiber,

$=\frac{f\sigma_{f}}{\sigma_{c}}$

$=\frac{0.35\times 0.61}{0.217}$

$=0.98$ i.e., 98%