Question 24.8: Figure 24.13 shows the cross-section in the xz plane of a sy...

Since the laminate is symmetric and isotropic the elastic constants for each ply are

Also, for an isotropic laminate (see Eq. (1.50))

\gamma=\frac{2(1+ν)}{E} \tau  (1.50)

For an isotropic ply, which is a special case of an orthotropic ply, the reduced stiffness terms k_{13} and k_{23} in Eq. (24.31) are zero. Also, from Eq. (24.17)

\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}-2 k_{33}\right)&m^{2}n^{2}\left(k_{11}+k_{22}-2k_{12}\right. \\+m n^{3}\left(k_{12}+k_{22}+2 k_{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)

\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)

For an isotropic ply there is no ply angle so that in effect, \theta=0^{\circ} and in Eq. (24.38)

\left\{\begin{array}{c}\sigma_{x} \\\sigma_{y} \\\tau_{x y}\end{array}\right\}=\left[\begin{array}{lll}\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_{xy}\end{array}\right\}  (24.38)

From Eq. (24.45) etc. and since there are four identical plies each 0.15 mm thick

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

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

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

Similarly  a_{33}=61.9 \times 10^{-6}

From Eq. (24.56) the equivalent Young’s modulus is given by

E_{x}=\frac{1}{t a_{11}}  (24.56)

and from Eq. (24.57) Poisson’s ratio is

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

Lastly, from Eq. (24.62)

\frac{\bar{\tau}_{x y}}{\gamma_{x y}}=\frac{1}{t a_{33}}=G_{x y}  (24.62)

The shear coupling coefficient, m_{x} is, from Eq. (24.58), zero. Note that the equivalent elastic constants are identical in value to the given elastic constants for the ply. This could have been deduced initially since, for an isotropic ply, the elastic constants are independent of laminate thickness. However, the example serves to illustrate the method.

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