A generally orthotropic ply is subjected to direct stresses of 120 N/mm² and 60 N/mm² parallel to the x and y reference axes, respectively, together with a shear stress of 80 N/mm². If the ply angle is 45°, determine the direct and shear stresses referred to the material axes.
Since the ply angle is 45°, the trigonometric terms in Eqs (26.24) are
\left\{\begin{matrix} \sigma _{l} \\ \sigma _{t} \\ \tau _{lt} \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} \sigma _{x} \\ \sigma _{y} \\ \tau _{xy} \end{matrix} \right\} (26.24)
m^{2}=0.5,\;\;n^{2}=0.5,\;\;m n=0.5Then, substituting in Eqs (26.24),
\left\{\begin{matrix} \sigma _{l} \\ \sigma _{t} \\ \tau _{lt} \end{matrix} \right\} =\left[\begin{matrix} 0.5 & 0.5 & 1 \\ 0.5 & 0.5 & -1 \\ -0.5 & 0.5 & 0 \end{matrix} \right] \left\{\begin{matrix} 120 \\ 60 \\ 80 \end{matrix} \right\}so that
\begin{array}{l}{{\sigma_{1}=0.5\times120+0.5\times60+1\times80=170\mathrm{N/mm^{2}}}}\\ {{\sigma_{\mathrm{t}}=0.5\times120+0.5\times60-1\times80=10\mathrm{N/mm^{2}}}}\\ {{\tau_{\mathrm{lt}}=-0.5\times120+0.5\times60+0=-30\mathrm{N/mm^{2}}}}\end{array}