Question 26.5: The strains at a point in a generally orthotropic ply are εx......

The strains at a point in a generally orthotropic ply are \varepsilon_{x}=0.005,\,\varepsilon_{y}=0.002,\,\mathrm{and}\gamma_{x y}=0.0002 referred to the xy reference axis system. If the ply angle is 45°, calculate the strains in the directions of the material axes.

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As in S.26.4, m²=0.5, n²=0.5, and mn=0.5. Eq. (26.25) then becomes

\left\{\begin{matrix} \varepsilon _{l} \\ \varepsilon _{t} \\ \gamma _{lt} \end{matrix} \right\} =\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\}             (26.25)

\left\{\begin{matrix} \varepsilon _{l} \\ \varepsilon _{t} \\ \gamma _{lt} \end{matrix} \right\} =\left[\begin{matrix} 0.5 & 0.5 & 0.5 \\ 0.5 & 0.5 & -0.5 \\ -1 & 1 & 0 \end{matrix} \right] \left\{\begin{matrix} 0.005 \\ 0.002 \\ 0.0002 \end{matrix} \right\}

Then

\begin{array}{l}{{\varepsilon_{l}=0.5\times0.005+0.5\times0.002+0.5\times0.0002=3.6\times10^{-3}}}\\ {{\varepsilon_{t}=0.5\times0.005+0.5\times0.002-0.5 \times 0.0002=3.4\times10^{-3}}}\\ {{\gamma_{\mathrm{lt}}=-1\times0.005+1\times0.002+0=-3.0\times10^{-3}}}\end{array}

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