(a) 23.4
The fiber-to-matrix load ratio is given by
[tex]\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}[/tex]
where
[tex]E_f = 131 GPa[/tex] is the fiber elasticity module
[tex]E_m = 2.4 GPa[/tex] is the matrix elasticity module
[tex]V_f=0.3[/tex] is the fraction of volume of the fiber
[tex]V_m=0.7[/tex] is the fraction of volume of the matrix
Substituting,
[tex]\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4[/tex] (1)
(b) 44,594 N
The longitudinal load is
F = 46500 N
And it is sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
We can rewrite (1) as
[tex]F_m = \frac{F_f}{23.4}[/tex]
And inserting this into (2):
[tex]F=F_f + \frac{F_f}{23.4}[/tex]
Solving the equation, we find the actual load carried by the fiber phase:
[tex]F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N[/tex]
(c) 1,906 N
Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
Using
F = 46500 N
[tex]F_f = 44594 N[/tex]
We can immediately find the actual load carried by the matrix phase:
[tex]F_m = F-F_f = 46,500 N - 44,594 N=1,906 N[/tex]
(d) 437 MPa
The cross-sectional area of the fiber phase is
[tex]A_f = A V_f[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_f=0.3[/tex], we have
[tex]A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the fiber phase is
[tex]\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa[/tex]
(e) 8.0 MPa
The cross-sectional area of the matrix phase is
[tex]A_m = A V_m[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_m=0.7[/tex], we have
[tex]A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the matrix phase is
[tex]\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa[/tex]
(f) [tex]3.34\cdot 10^{-3}[/tex]
The longitudinal modulus of elasticity is
[tex]E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa[/tex]
While the total stress experienced by the composite is
[tex]\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa[/tex]
So, the strain experienced by the composite is
[tex]\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}[/tex]
