Respuesta :
Answer:
a) [tex]F_m=1803.013\ N[/tex]
b) [tex]E=53665.84\ MPa[/tex]
c) [tex]E_f=124800\ MPa[/tex]
[tex]E_m=2200\ MPa[/tex]
Explanation:
Given:
- cross-sectional area of reinforced composite, [tex]A=1130\ mm^2[/tex]
- stress sustained by the fiber phase, [tex]\sigma_f=156\ MPa[/tex]
- force sustained by the fiber phase, [tex]F_f=74000\ N[/tex]
- Total strain on the composite, [tex]\epsilon=1.25\times 10^{-3}[/tex]
- stress sustained in the matrix phase, [tex]\sigma_m=2.75\ MPa[/tex]
Now, the area of fiber phase:
[tex]A_f=\frac{F_f}{\sigma_f}[/tex]
[tex]A_f=\frac{74000}{156}[/tex]
[tex]A_f=474.359\ mm^2[/tex]
∴Area of matrix phase:
[tex]A_m=A-A_f[/tex]
[tex]A_m=1130-474.359[/tex]
[tex]A_m=655.641\ mm^2[/tex]
(a)
Now the force sustained by the matrix phase:
[tex]F_m=\sigma_m\times A_m[/tex]
[tex]F_m=2.75\times 655.641[/tex]
[tex]F_m=1803.013\ N[/tex]
(b)
Total stress on the composite:
[tex]\sigma=\frac{(F_f+F_m)}{A}[/tex]
[tex]\sigma=\frac{(74000+1803.013)}{1130}[/tex]
[tex]\sigma=67.082\ MPa[/tex]
Now,Modulus of elasticity of the composite:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]E=\frac{67.082}{1.25\times 10^{-3}}[/tex]
[tex]E=53665.84\ MPa[/tex]
(c)
Since, strain will be same in this case throughout the material.
Now the modulus of elasticity of fiber phase:
[tex]E_f=\frac{\sigma_f}{\epsilon}[/tex]
[tex]E_f=\frac{156}{1.25\times 10^{-3}}[/tex]
[tex]E_f=124800\ MPa[/tex]
Now the modulus of elasticity of matrix phase:
[tex]E_m=\frac{\sigma_m}{\epsilon}[/tex]
[tex]E_m=\frac{2.75}{1.25\times 10^{-3}}[/tex]
[tex]E_m=2200\ MPa[/tex]
