Question 26.11: If the laminate of P.26.10 is subjected to a membrane load i......
If the laminate of P.26.10 is subjected to a membrane load intensity N_{x}=100\,\mathrm{N/mm}, investigate the strength of the laminate given that X_{T}\!=\!1200\;\mathrm{N/m}m^{2},\;X_{\mathrm{C}}\!=\!1000^{\mathrm{2}}\mathrm{N/m}m^{2},\;Y_{\mathrm{T}}\!=\!75\;\mathrm{N/m}m^{2}, Y_{\mathrm{C}}\!=\!200\ \mathrm{N/m}m^{2},\ \mathrm{and}\ S=70\ \mathrm{N/mm^{2}}.
Therefore a direct stress failure of the plies occurs in the transverse direction together with a failure in shear.
The values of \bar{k}_{11} etc. and a_{11} etc. have been calculated in Ex. 26.10. The ply stresses are then given, in matrix form by (see Eq. (iv) of Ex. 26.12)
Simplifying,
which gives
\begin{aligned}& \sigma_x=196.9 N / mm ^2 \\& \sigma_y=-13.1 N / mm ^2 \\& \tau_{x y}=-1.5 N / mm ^2\end{aligned}The ply angle is 45° so that m=1/√2=n. Then, from Eq. (26.24),
which gives
\begin{aligned}& \sigma_{ l }=90.4 N / mm ^2 \\& \sigma_{ t }=93.4 N / mm ^2 \\& \tau_{ lt }=-105.0 N /mm ^2\end{aligned}Then
\begin{aligned}& \sigma_l / X_{ T }=90.4 /1200=0.08 \\& \sigma_{ t } / Y_{ T }=93.4 / 75=1.25 \\& \tau_{ lt } / S=105 / 70=1.5\end{aligned}Therefore the plies fail in the transverse direction and also in shear.