A unidirectional prepreg carbon fibre has Ef = 242 GPa, Em = 7.82 GPa, Vf = 0.606, fibre Poisson ratio 0.22 and matrix Poisson ratio 0.387. Use the rule of mixtures to find the longitudinal stiffness E11 to the nearest GPa and the Poisson ratio to three decimal places.

Step-by-step solution

1. 1

   Use the parallel form for the longitudinal direction. Loaded along the fibres, both phases stretch by the same amount, so their stiffness contributions add in proportion to volume:

   $E_{11}=V_f E_f+V_m E_m.$

   This is the opposite of the transverse case, where compliances add instead.

2. 2

   Find the matrix volume fraction.

   $V_m=1-V_f=1-0.606=0.394.$

3. 3

   Compute the two contributions to E11.

   $V_f E_f=0.606\times 242=146.652\ \text{GPa},\qquad V_m E_m=0.394\times 7.82=3.081\ \text{GPa}.$

   The fibres carry almost all the load — $98\%$ of the total — which is the whole point of an aligned composite.

4. 4

   Add and round.

   $E_{11}=146.652+3.081=149.733\ \text{GPa}\approx 150\ \text{GPa}.$

5. 5

   Apply the same volume-weighted average to Poisson's ratio. The major Poisson's ratio follows the identical parallel rule:

   $\nu_{12}=V_f\nu_f+V_m\nu_m=0.606(0.22)+0.394(0.387).$

6. 6

   Evaluate and check.

   $0.13332+0.152478=0.285798\approx 0.286.$

   As a sanity check, both results must lie between the two constituent values: $7.82<149.7<242$ ✓ and $0.22<0.286<0.387$ ✓. Note that $\nu_{12}$ leans toward the matrix value even though the fibres dominate the stiffness — the volume fractions here are nearly equal, and Poisson's ratio is not weighted by modulus.