Question

A continuous and aligned fiber reinforced composite having a cross-sectional area of 1130 mm^2 (1.75 in.^2) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600 psi) and 2.75 MPa (400 psi), respectively; the force sustained by the fiber phase is 74,000 N (16,600 lbf); and the total longitudinal strain is 1.25 10^ -3, determine: (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the compos- ite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.

Answer 1

Answer:

a) $F_m=1803.013\ N$

b) $E=53665.84\ MPa$

c) $E_f=124800\ MPa$

   $E_m=2200\ MPa$

Explanation:

Given:

• cross-sectional area of reinforced composite, $A=1130\ mm^2$

• stress sustained by the fiber phase, $\sigma_f=156\ MPa$

• force sustained by the fiber phase, $F_f=74000\ N$

• Total strain on the composite, $\epsilon=1.25\times 10^{-3}$

• stress sustained in the matrix phase, $\sigma_m=2.75\ MPa$

Now, the area of fiber phase:

$A_f=\frac{F_f}{\sigma_f}$

$A_f=\frac{74000}{156}$

$A_f=474.359\ mm^2$

∴Area of matrix phase:

$A_m=A-A_f$

$A_m=1130-474.359$

$A_m=655.641\ mm^2$

(a)

Now the force sustained by the matrix phase:

$F_m=\sigma_m\times A_m$

$F_m=2.75\times 655.641$

$F_m=1803.013\ N$

(b)

Total stress on the composite:

$\sigma=\frac{(F_f+F_m)}{A}$

$\sigma=\frac{(74000+1803.013)}{1130}$

$\sigma=67.082\ MPa$

Now,Modulus of elasticity of the composite:

$E=\frac{\sigma}{\epsilon}$

$E=\frac{67.082}{1.25\times 10^{-3}}$

$E=53665.84\ MPa$

(c)

Since, strain will be same in this case throughout the material.

Now the modulus of elasticity of fiber phase:

$E_f=\frac{\sigma_f}{\epsilon}$

$E_f=\frac{156}{1.25\times 10^{-3}}$

$E_f=124800\ MPa$

Now the modulus of elasticity of matrix phase:

$E_m=\frac{\sigma_m}{\epsilon}$

$E_m=\frac{2.75}{1.25\times 10^{-3}}$

$E_m=2200\ MPa$