Answer:
The total load carried by the fiber will be "98%".
Explanation:
The given values are:
[tex]V_{f}=0.61[/tex]
[tex]V_{m}=1-V_{f}[/tex]
[tex]=1-0.61[/tex]
[tex]=0.39[/tex]
[tex]E_{f}=10 \ Mpa[/tex]
[tex]\sigma_{f}=0.35 \ Mpa[/tex]
[tex]E_{m}=0.45 \ Mpa[/tex] , [tex]\sigma_{m}=9\times 10^{-3} \ Mpa[/tex]
As we know,
⇒ [tex]E_{e}=fE_{f}+mE_{m}[/tex]
On putting the estimated values, we get
⇒ [tex]=0.61\times 10+0.39\times 0.95[/tex]
⇒ [tex]=6.27 \ Mpa[/tex]
Now,
⇒ [tex]\sigma_{c}=f\sigma_{f}+m\sigma_{m}[/tex]
On putting the estimated values, we get
⇒ [tex]=0.61\times 0.35+0.39\times 0.009[/tex]
⇒ [tex]=0.217 \ Mpa[/tex]
Therefore,
The load carried by fiber,
[tex]=\frac{f\sigma_{f}}{\sigma_{c}}[/tex]
[tex]=\frac{0.35\times 0.61}{0.217}[/tex]
[tex]=0.98[/tex] i.e., 98%
