Rule of mixtures

Consider a composite material under uniaxial tension ${\displaystyle \sigma _{\infty }}$. If the material is to stay intact, the strain of the fibers, ${\displaystyle \epsilon _{f}}$ must equal the strain of the matrix, ${\displaystyle \epsilon _{m}}$. Hooke's law for uniaxial tension hence gives

${\displaystyle {\frac {\sigma _{f}}{E_{f}}}=\epsilon _{f}=\epsilon _{m}={\frac {\sigma _{m}}{E_{m}}}}$

(1)

where ${\displaystyle \sigma _{f}}$, ${\displaystyle E_{f}}$, ${\displaystyle \sigma _{m}}$, ${\displaystyle E_{m}}$ are the stress and elastic modulus of the fibers and the matrix, respectively. Noting stress to be a force per unit area, a force balance gives that

${\displaystyle \sigma _{\infty }=f\sigma _{f}+\left(1-f\right)\sigma _{m}}$

(2)

where ${\displaystyle f}$ is the volume fraction of the fibers in the composite (and ${\displaystyle 1-f}$ is the volume fraction of the matrix).

If it is assumed that the composite material behaves as a linear-elastic material, i.e., abiding Hooke's law ${\displaystyle \sigma _{\infty }=E_{c}\epsilon _{c}}$ for some elastic modulus of the composite ${\displaystyle E_{c}}$ and some strain of the composite ${\displaystyle \epsilon _{c}}$, then equations 1 and 2 can be combined to give

${\displaystyle E_{c}\epsilon _{c}=fE_{f}\epsilon _{f}+\left(1-f\right)E_{m}\epsilon _{m}.}$

Finally, since ${\displaystyle \epsilon _{c}=\epsilon _{f}=\epsilon _{m}}$, the overall elastic modulus of the composite can be expressed as[7]

${\displaystyle E_{c}=fE_{f}+\left(1-f\right)E_{m}.}$

Now let the composite material be loaded perpendicular to the fibers, assuming that ${\displaystyle \sigma _{\infty }=\sigma _{f}=\sigma _{m}}$. The overall strain in the composite is distributed between the materials such that

${\displaystyle \epsilon _{c}=f\epsilon _{f}+\left(1-f\right)\epsilon _{m}.}$

The overall modulus in the material is then given by

${\displaystyle E_{c}={\frac {\sigma _{\infty }}{\epsilon _{c}}}={\frac {\sigma _{f}}{f\epsilon _{f}+\left(1-f\right)\epsilon _{m}}}=\left({\frac {f}{E_{f}}}+{\frac {1-f}{E_{m}}}\right)^{-1}}$

since ${\displaystyle \sigma _{f}=E\epsilon _{f}}$, ${\displaystyle \sigma _{m}=E\epsilon _{m}}$.[7]