From Eq. (26.14), the minor Poisson’s ratio is given by
\frac{\nu_{t l}}{E_t}=\frac{\nu_{l t}}{E_l} (26.14)
\nu_{ tl }=0.3 \times \frac{10000}{140000}=0.021
The reduced stiffnesses are then, from Eqs (26.17),
\left\{\begin{array}{l}\sigma_x \\\sigma_y \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}\frac{E_l}{1-\nu_{l t}\nu_{t l}} & \frac{\nu_{t l} E_l}{1-\nu_{lt} \nu_{t l}} & 0 \\\frac{\nu_{l t} E_t}{1-\nu_{l t} \nu_{t l}} &\frac{E_t}{1-\nu_{l t} \nu_{t l}} & 0 \\0 & 0 & G_{l t}\end{array}\right]\left\{\begin{array}{c}\varepsilon_l \\\varepsilon_t \\\gamma_{l t}\end{array}\right\} (26.17)\begin{aligned}& k_{11}=140000 /(1-0.3 \times 0.021)=140888 N / mm ^2 \\ & k_{22}=10000 /(1-0.3 \times 0.021)=10063 N / mm ^2 \\ & k_{33}=G_{1 t }=5000 N / mm ^2 \\ & k_{12}=0.021 \times 140000 /(1-0.3 \times 0.021)=2959 N / mm ^2 \end{aligned}
Also, k_{13}=0,\,k_{23}=0,\,k_{31}=0,\,k_{32}=0.
For plies 2 and 3, θ=45° so that m=n=1/√2. Then, from Eqs (26.38) and (26.31),
\begin{gathered}\left\{\begin{array}{c}\sigma_x \\\sigma_y \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}m^4k_{11}+m^2 n^2\left(2 k_{12}\right. & m^2n^2\left(k_{11}+k_{22}-4 k_{33}\right) & m^3n\left(k_{11}-k_{12}-2 k_{33}\right) \\\left.+4k_{33}\right)+n^4 k_{22} &+\left(m^4+n^4\right) k_{12} & +m n^3\left(k_{12}-k_{22}+2 k_{33}\right) \\m^2 n^2\left(k_{11}+k_{22}-4k_{33}\right) & n^4 k_{11}+m^2 n^2\left(2 k_{12}\right. & m n^3\left(k_{11}-k_{12}-2 k_{33}\right) \\+\left(m^4+n^4\right) k_{12} &\left.+4 k_{33}\right)+m^4 k_{22} & +m^3n\left(k_{12}-k_{22}+2 k_{33}\right) \\m^3 n\left(k_{11}-k_{12}-2 k_{33}\right) & m n^3\left(k_{11}-k_{12}-2 k_{33}\right) & m^2 n^2\left(k_{11}+k_{22}-2k_{12}\right. \\+mn^3\left(k_{12}+k_{22}+2 k_{33}\right) & +m^3n\left(k_{12}-k_{22}+2 k_{33}\right) & \left.-2k_{33}\right)+\left(m^4+n^4\right) k_{33}\end{array}\right]\left\{\begin{array}{c}\varepsilon_x \\\varepsilon_y \\\gamma_{x y}\end{array}\right\}\end{gathered} (26.31)\left\{\begin{array}{c}\sigma_x \\\sigma_y \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}\bar{k}_{11} & \bar{k}_{12} & \bar{k}_{13} \\\bar{k}_{12} & \bar{k}_{22} & \bar{k}_{23} \\\bar{k}_{13} & \bar{k}_{23} & \bar{k}_{33}\end{array}\right]\left\{\begin{array}{c}\varepsilon_x \\\varepsilon_y \\\gamma_{x y}\end{array}\right\} (26.38)
\bar{k}_{11}=(1 / \sqrt{2})^4 \times 140888+(1 /\sqrt{2})^2(1 / \sqrt{2})^2(2 \times 2959+4 \times 5000)+(1 / \sqrt{2})^4 \times 10063=44166 N / mm ^2
Similarly,
\begin{aligned}& \bar{k}_{22}=34300 N / mm ^2 \\& \bar{k}_{33}=36300 N / mm ^2 \\&\bar{k}_{12}=34300 N / mm ^2 \\&\bar{k}_{13}=32700 N / mm ^2 \\&\bar{k}_{21}=34300 N / mm ^2 \\&\bar{k}_{23}=32700 N / mm ^2\end{aligned}
For plies 1 and 4, θ=30° so that m=√3/2 and n=0.5. Again, from Eqs (26.38) and (26.31),
\begin{aligned}& \bar{k}_{11}=84800 N / mm ^2 \\& \bar{k}_{22}=19400 N / mm ^2 \\&\bar{k}_{33}=28400 N / mm ^2 \\&\bar{k}_{12}=26400 N / mm ^2 \\&\bar{k}_{13}=41900 N / mm ^2 \\&\bar{k}_{23}=14800 N / mm ^2\end{aligned}
Now, from Eqs (26.47),
A_{ ij }=\sum\limits_{ p =1}^{ N } t_{ p }\left(\bar{k}_{ ij }\right)_{ p } \quad( i =1,2,3, \ldots, j =1,2,3, \ldots) (26.47)
\begin{aligned}& A_{11}=2 \times 0.15(44166+84800)=3.9 \times 10^4 N / mm \\& A_{22}=2 \times 0.15(44166+19400)=1.9 \times 10^4 N / mm \\& A_{33}=2 \times 0.15(36300+28400)=1.9 \times 10^4 N /mm \\& A_{12}=2 \times 0.15(34300+26400)=1.8 \times 10^4 N / mm \\& A_{13}=2 \times 0.15(32700+41900)=2.2 \times 10^4 N / mm \\& A_{23}=2 \times 0.15(32700+14800)=1.4 \times 10^4 N / mm\end{aligned}
Then, from Eq. (26.53),
A A=A_{11} A_{22} A_{33}+2 A_{12} A_{23}A_{13}-A_{22} A_{13}{ }^2-A_{33} A_{12}{ }^2-A_{11} A_{23}{ }^2 (26.53)
A A=\left(3.9 \times 1.9 \times 1.9+2 \times 1.8 \times 1.4 \times 2.2-1.9 \times 2.2^2-1.9 \times 1.8^2-3.9 \times 1.4^2\right) \times 10^{12}
which gives
A A=2.2\times10^{12}
From Eqs (26.52),
\begin{aligned}& a_{11}=\left(A_{22} A_{33}-A_{23}{ }^2\right) / A A \\& a_{22}=\left(A_{11} A_{33}-A_{13}{ }^2\right) / A A \\& a_{33}=\left(A_{11} A_{22}-A_{12}{ }^2\right) / A A \\& a_{12}=\left(A_{13} A_{23}-A_{12} A_{33}\right) / A A \\& a_{13}=\left(A_{12}A_{23}-A_{22} A_{13}\right) / A A \\& a_{23}=\left(A_{12} A_{13}-A_{11} A_{23}\right) / A A\end{aligned} (26.52)
\begin{aligned}& a_{11}=\left(1.9 \times 1.9-1.4^2\right) \times 10^8 / 2.2 \times 10^{12}=0.75 \times 10^{-4} \\& a_{22}=\left(3.9 \times 1.9-2.2^2\right)\times 10^8 / 2.2 \times 10^{12}=1.17 \times 10^{-4} \\&a_{33}=\left(3.9 \times 1.9-1.8^2\right) \times 10^8 /2.2 \times 10^{12}=1.9 \times 10^{-4} \\& a_{12}=(2.2\times 1.4-1.8 \times 1.9) \times 10^8 / 2.2 \times 10^{12}=-0.15 \times 10^{-4} \\& a_{13}=(1.8 \times 1.4-1.9 \times 2.2) \times 10^8 / 2.2 \times 10^{12}=-0.75\times 10^{-4} \\& a_{23}=(1.8 \times 2.2-3.9 \times 1.4) \times 10^8 / 2.2 \times 10^{12}=-0.68\times 10^{-4}\end{aligned}
From Eq. (26.56),
E_{x}={\frac{1}{t a_{11}}} (26.56)
E_x=\frac{1}{4 \times 0.15 \times 0.75 \times 10^{-4}}=22222 N / mm ^2
From Eq. (26.59),
E_{y}={\frac{1}{t a_{22}}} (26.59)
E_y=\frac{1}{4 \times 0.15 \times 1.17 \times 10^{-4}}=14245 N / mm ^2
From Eq. (26.62),
\frac{\bar{\tau}_{x y}}{\gamma_{x y}}=\frac{1}{t a_{33}}=G_{x y} (26.62)
G_{x y}=\frac{1}{4 \times 0.15 \times 1.9 \times 10^{-4}}=8771 N / mm ^2
From Eq. (26.57),
\nu_{x y}=\frac{-\varepsilon_y}{\varepsilon_x}=\frac{-a_{12}}{a_{11}} (26.57)
\nu_{x y}=-\frac{\left(-0.15 \times 10^{-4}\right)}{0.75 \times 10^{-4}}=0.2
From Eq. (26.60),
\nu_{y x}=\frac{-a_{12}}{a_{22}} (26.60)
\nu_{y x}=-\frac{\left(-0.15 \times 10^{-4}\right)}{1.17 \times 10^{-4}}=0.13
From Eq. (26.58),
\frac{\gamma_{x y}}{\varepsilon_x}=\frac{a_{13}}{a_{11}}=-m_x (26.58)
m_x=-\frac{\left(-0.75 \times 10^{-4}\right)}{0.75 \times 10^{-4}}=1.0
From Eq. (26.61),
m_{\mathrm{y}}={\frac{-a_{23}}{a_{22}}} (26.61)
m_y=-\frac{\left(-0.68 \times 10^{-4}\right)}{1.17 \times 10^{-4}}=0.58