A unidirectional carbon fibre epoxy laminate has fibre volume fraction 0.66 and matrix volume fraction 0.34, with fibre modulus 340 GPa, matrix modulus 3.05 GPa, fibre strength 3648 MPa and matrix strength 60 MPa. Using the rule of mixtures, find the longitudinal stiffness and tensile strength when the load runs parallel to the fibres.

Step-by-step solution

1. 1

   Understand why a simple weighted average is legitimate here. Loading parallel to the fibres puts fibre and matrix under the same strain (the iso-strain, or Voigt, assumption). Each phase then contributes force in proportion to its own stiffness and to the fraction of the cross-section it occupies, which is exactly what the rule of mixtures encodes:

   $E_{11}=V_fE_f+V_mE_m$

   Load the laminate transversely instead and this formula fails — you would need the inverse rule of mixtures.

2. 2

   Check the volume fractions add to 1. $V_f+V_m=0.66+0.34=1.00$. If they did not, one of them would have to be recomputed from the weight fractions and densities before going any further.

3. 3

   Compute the longitudinal stiffness.

   $E_{11}=(0.66)(340)+(0.34)(3.05)=224.4+1.037=225.437\ \text{GPa}$

   Rounded to one decimal place, $E_{11}=225.4\ \text{GPa}$. Note that the resin contributes about $1$ GPa out of $225$ — under $0.5\%$. In a fibre-dominated direction the matrix is essentially just glue.

4. 4

   Compute the longitudinal strength the same way. The same iso-strain reasoning gives

   $\sigma_{11}=V_f\sigma_f+V_m\sigma_m=(0.66)(3648)+(0.34)(60)=2407.68+20.4=2428.08\ \text{MPa}$

   so $\sigma_{11}=2428.1\ \text{MPa}$.

5. 5

   Sanity-check the magnitudes and the units. Both answers must lie between the matrix value and the fibre value, and both do ($3.05<225.4<340$ GPa; $60<2428.1<3648$ MPa), sitting close to the fibre end because $V_f$ is high. Keep GPa with GPa and MPa with MPa — mixing the two by a factor of $1000$ is the classic error in this calculation.