Poisson’s ratio derivation
The Poisson’s ratio νij describes the contraction in the j-direction due to a stress in the i-direction.
\[{\nu _{ij}} = \frac{{ - {\varepsilon _j}}}{{{\varepsilon _i}}}\]
For a transversely isotropic material such as a composite the following relationship exists:
\[\frac{{{\nu _{12}}}}{{{E_1}}} = \frac{{{\nu _{21}}}}{{{E_2}}}\]
For a composite under axial tension the Poisson strains for the two constituents are:
\[{\varepsilon _{2f}} = - {\nu _f}{\varepsilon _{1f}} = - {\nu _f}\frac{{{\sigma _{1f}}}}{{{E_f}}}\]
\[{\varepsilon _{2m}} = - {\nu _m}{\varepsilon _{1m}} = - {\nu _m}\frac{{{\sigma _{1m}}}}{{{E_m}}}\]
The total transverse strain is simply the weighted sum of the two Poisson strains:
\[{\varepsilon _2} = - \left[ {f{\nu _f}\frac{{{\sigma _{1f}}}}{{{E_f}}} + \frac{{(1 - f){\nu _m}{\sigma _{1m}}}}{{{E_m}}}} \right]\]
Under axial loading the equal axial strain condition applies.
\[∴\;{\varepsilon _2} = - \left[ {f{\nu _f}{\varepsilon _1} + (1 - f){\nu _m}{\varepsilon _1}} \right]\]
\[{\nu _{12}} = - \frac{{{\varepsilon _2}}}{{{\varepsilon _1}}} = {f^{}}{\nu _f} + {(1 - f)^{}}{\nu _m} \;\;\rm{and}\;\;{\nu _{21}} = - \frac{{{\varepsilon _1}}}{{{\varepsilon _2}}} = \left[ {{f^{}}{\nu _f} + {{(1 - f)}^{}}{\nu _m}} \right]\frac{{{E_2}}}{{{E_1}}}\]