Answer :
Answer:
(a) 3290 N
(b) 52.9 Gpa
(c) 3.45 Gpa and 1378 Gpa for matrix and fibre phases respectively
Explanation:
(a)
The volume fraction of matrix is given by
[tex]v_m=1-v_f[/tex] where [tex]v_m[/tex] is volume fraction of matrix and [tex]v_f[/tex] is volume fraction of fibre
Moreover, the stress in matrix is given by
[tex]\sigma_m=\frac {F_m}{A_m}=\frac {F_m}{V_mA_c}[/tex] where [tex]F_m[/tex] is force on matrix and [tex]A_m[/tex] is the area of matrix, [tex]A_c[/tex] is cross-sectional area
Therefore, [tex]F_m=V_mA_c\sigma_m[/tex]
Substituting the figures in the question
[tex]F_m=0.631*(0.970*10^{-3})*(5.38*10^{6})=3290 N[/tex]
Therefore, force sustained by matrix is 3290 N
(b)
[tex]E_c=\frac {\sigma_c}{\epsilon}[/tex] and also
[tex]\sigma_c=\frac {F_m+F_f}{A_c}[/tex] therefore we combine these two equations and say
[tex]E_c=\frac {\sigma_c}{\epsilon}= \frac {F_m+F_f}{\epsilon A_c}[/tex]
Substituting the figures given
[tex]E_c=\frac {3290 N +76800 N}{(1.56*10^{-3}*(0.970*10^{-3})}=52.9*10^{9}=52.9 GPa[/tex]
(c)
The moduli of elasticity of matrix and fibre are given by
[tex]E_m=\frac {\sigma_m}{\epsilon_m}=\frac {\sigma_m}{\epsilon_c}[/tex]
Therefore,
[tex]E_m=\frac {5.38*10^{6}}{1.56*10^{-3}}=3.45*10^{9}=3.45 Gpa[/tex]
[tex]E_f=\frac {215*10^{6}}{1.56*10^{-3}}=1.378*10^{11}=1378 Gpa[/tex]
Therefore, moduli of elasticity for fibre and matrix phases are 1378 Gpa and 3.45 Gpa respectively