Question 24.9: If the laminate in Example 24.8 comprises plies of the same ...

Initially we calculate the minor Poisson’s ratio ν_{t l} . From Eq. (24.14)

\frac{ν_{t 1}}{E_{t}}=\frac{ν_{l t}}{E_{l}}  (24.14)

Again, for a symmetric orthotropic ply the reduced stiffness terms, k_{13} and k_{23} are zero [Ref. 2]. Then, from Eq. (24.17)

\left\{\begin{array}{c}\sigma_{x} \\\sigma_{y} \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}\frac{E_{l}}{1-ν_{l t} ν_{t 1}} & \frac{ν_{t l} E_{l}}{1-ν_{l t} ν_{t l}} &0\\\frac{ν_{l t} E_{t}}{1-ν_{l t} ν_{t l}}&\frac{E_{t}}{1-ν_{l t} ν_{t l}} & 0 \\0 & 0 & G_{1 t}\end{array}\right]\left\{\begin{array}{c}\varepsilon_{l}\\\varepsilon_{t}\\\gamma_{l t}\end{array}\right\}  (24.17)

Similarly k_{12}=2958 N / mm ^{2}, k_{22}=10060 N / mm ^{2} \text { and } k_{33}=5000 N / mm ^{2}. The ply angle \theta=0^{\circ} so that in Eq. (24.31) m=\cos \theta=1 and \sin \theta=0 . Then

\left\{\begin{array}{c}\sigma_{x} \\\sigma_{y} \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{ccc}m^{4}k_{11}+m^{2}n^{2}\left(2k_{12}\right.&m^{2}n^{2}\left(k_{11}+k_{22}-4k_{33}\right)&m^{3}n\left(k_{11}-k_{12}-2k_{33}\right)\\\left.+4k_{33}\right)+n^{4}k_{22}&+\left(m^{4}+n^{4}\right)k_{12}&+mn^{3}\left(k_{12}-k_{22}+2k_{33}\right)\\m^{2}n^{2}\left(k_{11}+k_{22}-4k_{33}\right)&n^{4}k_{11}+m^{2}n^{2}\left(2k_{12}\right. & m n^{3}\left(k_{11}-k_{12}-2 k_{33}\right)\\+\left(m^{4}+n^{4}\right)k_{12}&\left.+4k_{33}\right)+m^{4} k_{22} & +m^{3} n\left(k_{12}-k_{22}+2 k_{33}\right)\\m^{3} n\left(k_{11}-k_{12}-2k_{33}\right)&mn^{3}\left(k_{11}-k_{12}-2k_{33}\right)&m^{2}n^{2}\left(k_{11}+k_{22}-2k_{12}\right.\\+mn^{3}\left(k_{12}+k_{22}+2k_{33}\right)& +m^{3} n\left(k_{12}-k_{22}+2k_{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\}  (24.31)

There are four identical plies in the laminate each 0.15 mm thick. Therefore from Eq. (24.45) etc.

A_{11}=\sum\limits_{ p =1}^{ N }\left(z_{ p }-z_{ p -1}\right) \bar{k}_{11}  (24.45)

The terms in the inverse of the [A] matrix are, then, from Eqs. (24.55)

a_{12}=-\left(A_{12}\right) /\left(A_{11} A_{22}-A{_{12}}^{2}\right)  (24.55)

Similarly a_{22}=16.65 \times 10^{-5}, a_{33}=33.33 \times 10^{-5}, a_{12}=-0.35 \times 10^{-5} From Eq. (24.56)

E_{x}=\frac{1}{t a_{11}}  (24.56)

From Eq. (24.59)

E_{y}=\frac{1}{t a_{22}}  (24.59)

From Eq. (24.57)

ν_{x y}=\frac{-\varepsilon_{y}}{\varepsilon_{x}}=\frac{-a_{12}}{a_{11}}  (24.57)

From Eq. (24.60)

ν_{y x}=\frac{-a_{12}}{a_{22}}  (24.60)

Also, from Eqs. (24.58) and (24.61) m_{x}=m_{y}=0.

\frac{\gamma_{x y}}{\varepsilon_{x}}=\frac{a_{13}}{a_{11}}=-m_{x}  (24.58)

m_{ y }=\frac{-a_{23}}{a_{22}}  (24.61)

Again, as in Example 24.8, the equivalent elastic constants for the laminate are the same as the elastic constants of the plies. This is to be expected since the four plies are identical and orientated in the same way.