Question 26.3: The reduced stiffnesses in a unidirectional ply are k11 = 50......
The reduced stiffnesses in a unidirectional ply are k_{11}=50,000\;\mathrm{N/mm^{2}},k_{12}=k_{21}=4000\;\mathrm{N/mm^{2}}, k_{22}=15,000\ \mathrm{N/mm^{2},\,a n d}\ k_{33}=6000\ \mathrm{N/mm^{2}.} Calculate the elastic constants of the ply and also the reduced compliances.
From the development of Eqs (26.30),
\left\{\begin{matrix} \sigma _{x} \\ \sigma _{y} \\ \tau _{xy} \end{matrix} \right\} =\left[\begin{matrix} m^{2} & n^{2} & -2mn\\ n^{2} & m^{2} & 2mn \\ mn & -mn & m^{2}-n^{2} \end{matrix} \right] \left[\begin{matrix} k_{11} & k_{12} & 0 \\ k_{12} & k_{22} & 0 \\ 0 & 0 & k_{33} \end{matrix} \right] \times \left[\begin{matrix} m^{2} & n^{2} & mn \\ n^{2} & m^{2} & -mn \\ -2mn & 2mn & m^{2}-n^{2} \end{matrix} \right] \left\{\begin{matrix} \varepsilon _{x} \\ \varepsilon _{y} \\ \gamma _{xy} \end{matrix} \right\} (26.30)
k_{11}=E_{l}/(1-\nu_{\mathrm{lt}}\nu_{\mathrm{tl}}) (i)
k_{12}=\nu_{\mathrm{tl}}E_{l}/(1-\nu_{\mathrm{lt}}\nu_{\mathrm{tl}}) (ii)
Dividing Eq.(ii) by Eq. (i),
\nu_{tl}=k_{12}/k_{11}=4000/50000=0.08Also,
k_{21}=\nu_{\mathrm{lt}}E_{t}/(1-\nu_{\mathrm{lt}}\nu_{\mathrm{tl}}) (iii)
k_{22}=E_{t}/(1-\nu_{\mathrm{lt}}\nu_{\mathrm{tl}}) (iv)
Dividing Eq. (iii) by Eq. (iv)
\nu_{\mathrm{lt}}=k_{21}/k_{22}=4000/15000=0.27Then, from Eq. (i),
E_{l}=50000(1-0.27\times0.08)=48920\,\mathrm{N/mm^{2}}and from Eq. (iii),
E_{t}=4000(1-0.27\times0.08)/0.27=14495\,\mathrm{N/mm^{2}}Also
G=k_{33}=6000\,\mathrm{N/mm^{2}}The reduced compliances are then
\begin{array}{l}{{s_{11}=1/E_{l}=1/48\,920=2.04\times10^{-5}}}\\ {{s_{12}=-\nu_{tl}/E_{t}=-0.08/14495=-0.55\times10^{-5}}}\\ {{s_{21}=-\nu_{lt}/E_{l}=-0.27/48\,920=-0.55\times10^{-5}}}\\ {{s_{22}=1/E_{t}=6.90\times10^{-5}}}\\ {{s_{33}=1/G=1/6000=16.7\times10^{-5}}}\end{array}