Answer:
a) 23.39
b) 44977.08 N
c) 1922.92N
d) 454.31 MPa
e) 8.32 MPa
f) [tex] 3.47*10^-^3 [/tex]
Explanation:
a) fiber-matrix load ratio:
Let's use the formula :
[tex] \frac{F_f}{F_m} = \frac{V_f E_f}{V_m E_m}[/tex]
[tex] = \frac{0.3 * 131 GPa}{0.7 * 2.4 GPa} = 23.39 [/tex]
b & c)
Total load is given as:
Fc = Ff + Fm
46900 = Fm(23.39) + Fm
46900 = 24.39 Fm
Actual load carried by matrix=
[tex]F_m = \frac{46900}{24.39}[/tex]
= 1922.92N=> answer for option c
Actual load carried by fiber, Ff:
Ff = 46900 - 1922.92
Ff = 44977.08 N => answer option b
d)
Let's find area of fiber, A_f.
[tex] A_f = V_f * A_c[/tex]
Ac = Cross sectional area =300mm²
= 0.3 * 300 = 99 mm²
Area of matrix=
[tex] A_m = V_m * A_c[/tex]
= 0.7 * 300 = 231 mm²
Magnitude of the stress on the fiber phase:
[tex] \sigma _f= \frac{F_f}{A_f} [/tex]
[tex] \sigma _f= \frac{44977.08}{99} = 454.31 MPa [/tex]
e) Magnitude of the stress on the matrix phase.
[tex] \sigma _m = \frac{F_m}{A_m} [/tex]
[tex] \sigma _m = \frac{1922.92}{231} = 8.32 MPa[/tex]
f) Strain in fiber = [tex] \frac{\sigma _f}{E_f} [/tex]
[tex]= \frac{454.31*10^6}{131*10^9} = 3.47*10^-^3[/tex]
Strain in matrix = [tex] \frac{\sigma _m}{E_m} [/tex]
[tex]= \frac{8.32*10^6}{2.4*10^9} = 3.47*10^-^3[/tex]
Composite strain = [tex] (E_f *V_f) + (E_m * V_m) [/tex]
[tex] (3.47*10^-^3 * 0.3) + (3.47*10^-^3 * 0.7) = 3.47*10^-^3 [/tex]
