Question

A unidirectional composite of a resin is prepared with continuous fibers, with volume fraction of fiber = 0.7. Moduli: resin = 9 GPa, fiber = 90 GPa. Sample breadth = 4 mm, thickness = 1 mm. Tensile strain = 0.5%. Find the tensile force applied on the sample. (Round off to nearest integer)

Hint:

Always use the rule of mixtures for composite modulus: \(E_c = V_f E_f + V_m E_m\). Then use stress–strain relation \(\sigma = E_c \varepsilon\), and finally multiply with area for force.

Answer 1

Correct Answer: 1300

Step 1: Effective modulus of composite (rule of mixtures).
\[ E_c = V_f E_f + V_m E_m \] \[ E_c = (0.7)(90) + (0.3)(9) = 63 + 2.7 = 65.7 \; \text{GPa} \]

Step 2: Stress in composite.
\[ \sigma = E_c \times \varepsilon = 65.7 \times 10^9 \times 0.005 = 3.285 \times 10^8 \; \text{Pa} \]

Step 3: Cross-sectional area of sample.
\[ A = \text{breadth} \times \text{thickness} = (4 \times 10^{-3})(1 \times 10^{-3}) = 4 \times 10^{-6} \; \text{m}^2 \]

Step 4: Force.
\[ F = \sigma \times A = 3.285 \times 10^8 \times 4 \times 10^{-6} = 1314 \; \text{N} \] Check again!
Wait → strain given is 0.5% = 0.005. Already used. Good. But calculation: \[ 65.7 \times 10^9 \times 0.005 = 3.285 \times 10^8 \, \text{Pa} \] Now: \[ F = 3.285 \times 10^8 \times 4 \times 10^{-6} = 1314 \; \text{N} \] Correct nearest integer = 1314 N. Final Answer: \[ \boxed{1314 \; \text{N}} \]