Question

Continuous and aligned fiber-reinforced composite with cross-sectional area of 300 mm2 (0.47 in.2) is subjected to a longitudinal load of 46900 N (10500 lbf). Assume Vf = 0.3, Vm = 0.7, Ef = 131 GPa and Em = 2.4 GPa. (a) Calculate the fiber-matrix load ratio. (b) Calculate the actual load carried by fiber phase. (c) Calculate the actual load carried by matrix phase. (d) Compute the magnitude of the stress on the fiber phase. (e) Compute the magnitude of the stress on the matrix phase. (f) What strain is expected by the composite?

Answer 1

Answer:

a) 23.39

b) 44977.08 N

c) 1922.92N

d) 454.31 MPa

e) 8.32 MPa

f) $3.47*10^-^3$

Explanation:

a) fiber-matrix load ratio:

Let's use the formula :

$\frac{F_f}{F_m} = \frac{V_f E_f}{V_m E_m}$

$= \frac{0.3 * 131 GPa}{0.7 * 2.4 GPa} = 23.39$

b & c)

Total load is given as:

Fc = Ff + Fm

46900 = Fm(23.39) + Fm

46900 = 24.39 Fm

Actual load carried by matrix=

$F_m = \frac{46900}{24.39}$

= 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.

$A_f = V_f * A_c$

Ac = Cross sectional area =300mm²

= 0.3 * 300 = 99 mm²

Area of matrix=

$A_m = V_m * A_c$

= 0.7 * 300 = 231 mm²

Magnitude of the stress on the fiber phase:

$\sigma _f= \frac{F_f}{A_f}$

$\sigma _f= \frac{44977.08}{99} = 454.31 MPa$

e) Magnitude of the stress on the matrix phase.

$\sigma _m = \frac{F_m}{A_m}$

$\sigma _m = \frac{1922.92}{231} = 8.32 MPa$

f) Strain in fiber = $\frac{\sigma _f}{E_f}$

$= \frac{454.31*10^6}{131*10^9} = 3.47*10^-^3$

Strain in matrix = $\frac{\sigma _m}{E_m}$

$= \frac{8.32*10^6}{2.4*10^9} = 3.47*10^-^3$

Composite strain = $(E_f *V_f) + (E_m * V_m)$

$(3.47*10^-^3 * 0.3) + (3.47*10^-^3 * 0.7) = 3.47*10^-^3$