As described in this section, effective moduli are calculated using various elements of the [abd] matrix. A six-ply symmetric laminate is considered in this problem. The total laminate thickness is t = 6(0.000125 m) = 0.000750 m. Using methods discussed in Section 6.6, the [ABD] matrix is determined to be
[ABD]=\left [ \begin{matrix} 72.2\times 10^6& 8.02\times 10^6 & 12.0\times 10^6 & 0 & 0 & 0 \\8.02\times 10^6 & 52.0\times 10^6 & 5.38\times 10^6 & 0& 0 & 0 \\12.0\times 10^6 & 5.38\times 10^6& 15.5\times 10^6 & 0& 0 & 0 \\ 0& 0 & 0& 4.23 &0.676 & 1.19 \\ 0 & 0 &0 &0.676 & 0.988 & 0.532 \\ 0 & 0& 0 & 1.19 & 0.532 & 1.03 \end{matrix} \right ]
Since the laminate is symmetric, all elements of the B_{ij} matrix are zero, as expected. We obtain the [abd] by inverting the [ABD] numerically:
[abd]=\left [ \begin{matrix} 16.0\times 10^{-9}& 1.23\times 10^{-9} & -12.0\times 10^{-9} & 0 & 0 & 0 \\-1.23\times 10^{-9} & 20.0\times 10^{-9} & -6.0\times 10^{-9} & 0& 0 & 0 \\-12.0\times 10^{-9} & -6.0\times 10^{-9}& 75.8\times 10^{-9} & 0& 0 & 0 \\ 0& 0 & 0& 3.51\times 10^{-1} &-2.92\times 10^{-2} & -3.92\times 10^{-1} \\ 0 & 0 &0 &-2.92\times 10^{-2} & 1.408 & -6.96\times 10^{-1} \\ 0 & 0& 0 & -3.92\times 10^{-1} &-6.96\times 10^{-1} & 1.79 \end{matrix} \right ]
The effective extensional modulii of the laminate can now be calculated using Equations 6.63 through 6.70:
\overline{E}_{xx}^{ex}=\frac{\overline{\sigma } _{xx}}{\varepsilon^o_{xx}} = \frac{(N_{xx}/t)}{(a_{11}N_{xx})} = \frac{1}{ta_{11}} (6.63)
\overline{\eta }_{xy,xx}=\frac{\varepsilon ^o _{xx}}{\gamma ^o_{xx}} = \frac{a_{16} N_{xx}}{a_{66}N_{xy}} = \frac{a_{16}}{a_{66}} (6.70a)
\overline{\eta }_{xy,yy}=\frac{\varepsilon ^o _{yy}}{\gamma ^o_{xy}} = \frac{a_{26}}{a_{66}} (6.70b)
\overline{E}_{xx}^{ex}=\frac{1}{ta_{11}}=\frac{1}{(0.000750)(16.0\times 10^{-9})}=83.3GPa
\overline{E}_{xx}^{ex}=\frac{1}{ta_{22}}=\frac{1}{(0.000750)(20\times 10^{-9})}=66.7GPa
\overline{G} _{xy}=\frac{1}{ta_{66}}=\frac{1}{(0.000750)(75.8\times 10^{-9})}=17.6GPa
\overline{\nu} _{xy}^{ex}=\frac{-a_{12}}{a_{11}} =\frac{-(-123\times 10^{-9})}{16.0\times 10^{-9}} =0.077
\overline{\nu}_{xy}^{ex}=\frac{-a_{12}}{a_{22}} =\frac{-(1.23\times 10^{-9})}{20\times 10^{-9}} =0.061
\overline{\eta }_{xx,xy}^{ex}=\frac{a_{16}}{a_{11}} =\frac{-12\times 10^{-9}}{16\times 10^{-9}} =-0.75
\overline{\eta }_{yy,xy}^{ex}=\frac{a_{26}}{a_{22}} =\frac{-6.0\times 10^{-9}}{20\times 10^{-9}} =-0.30
\overline{\eta }_{xy,xx}^{ex}=\frac{a_{16}}{a_{66}} =\frac{-12.0\times 10^{-9}}{75.8\times 10^{-9}} =-0.16
\overline{\eta }_{xy,yy}^{ex}=\frac{a_{26}}{a_{66}} =\frac{-6.0\times 10^{-9}}{75.8\times 10^{-9}} =-0.079
Effective flexural properties are found to be
\overline{E}_{xx}^{fl}=\frac{12}{d_{11}t^3} =\frac{12}{(0.351)(0.000750)^3} =81.0 GPa
\overline{E}_{yy}^{fl}=\frac{12}{d_{22}t^3} =\frac{12}{(1.408)(0.000750)^3} =20.2 GPa
\overline{\nu} _{xy}^{fl}=\frac{-d_{12}}{d_{11}} =\frac{2.92\times 10^{-2}}{0.351} =0.083
\overline{\nu} _{yx}^{fl}=\frac{-d_{12}}{d_{22}} =\frac{2.92\times 10^{-2}}{1.408} =0.021
\overline{\eta }_{xx,xy}^{fl}=\frac{d_{16}}{d_{11}} =\frac{-0.392}{0.351} =-1.12
\overline{\eta }_{yy,xy}^{fl}=\frac{d_{26}}{d_{22}} =\frac{-0.696}{1.408} =-0.49
Note that the values of extensional properties are quite different from analogous flexural properties.
The thermal stress resultants associated with a given change in temperature must be determined in order to calculate the effective thermal expansion coefficients. Numerically speaking, any change in temperature can be used, but for present purposes a unit change in temperature will be assumed (i.e., ΔT = 1). Using Equations 6.41, the thermal stress resultants associated with ΔT = 1 are
N_{xx}^T\equiv \Delta T\sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{11} \alpha _{xx} +\overline{Q}_{12} \alpha _{yy}+\overline{Q}_{16} \alpha _{xy}\right] _{k}\left[z_{k}-z_{k-1}\right] \right\} } (6.41a)
N_{yy}^T\equiv \Delta T\sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{12} \alpha _{xx} +\overline{Q}_{22} \alpha _{yy}+\overline{Q}_{26} \alpha _{xy}\right] _{k}\left[z_{k}-z_{k-1}\right] \right\} } (6.41b)
N_{xy}^T\equiv \Delta T\sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{16} \alpha _{xx} +\overline{Q}_{26} \alpha _{yy}+\overline{Q}_{66} \alpha _{xy}\right] _{k}\left[z_{k}-z_{k-1}\right] \right\} } (6.41c)
M_{xx}^T\equiv \frac{\Delta T}{2} \sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{11} \alpha _{xx} +\overline{Q}_{12} \alpha _{yy}+\overline{Q}_{16} \alpha _{xy}\right] _{k}\left[z^2_{k}-z^2_{k-1}\right] \right\} } (6.41d)
M_{yy}^T\equiv \frac{\Delta T}{2} \sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{12} \alpha _{xx} +\overline{Q}_{22} \alpha _{yy}+\overline{Q}_{26} \alpha _{xy}\right] _{k}\left[z^2_{k}-z^2_{k-1}\right] \right\} } (6.41e)
M_{xy}^T\equiv \frac{\Delta T}{2} \sum\limits_{k=1}^{n}{\left\{\left[\overline{Q}_{16} \alpha _{xx} +\overline{Q}_{26} \alpha _{yy}+\overline{Q}_{66} \alpha _{xy}\right] _{k}\left[z^2_{k}-z^2_{k-1}\right] \right\} } (6.41f)
N_{xx}^T|_{\Delta T=1}=52.3N/m N_{yy}^T|_{\Delta T=1}=94.9N/m
N_{xy}^T|_{\Delta T=1}=36.9N/m
The effective thermal expansion coefficient \overline{\alpha }_{xx} can now be calculated in accordance with Equation 6.73a:
\overline{\alpha }_{xx}=\frac{1}{\Delta T}\left[a_{11}N_{xx}^T+a_{12}N_{yy}^T+a_{16}N_{xy}^T\right]
\overline{\alpha }_{yy}=\frac{1}{\Delta T}\left[a_{12}N_{xx}^T+a_{22}N_{yy}^T+a_{26}N_{xy}^T\right]
\overline{\alpha }_{xy}=\frac{1}{\Delta T}\left[a_{16}N_{xx}^T+a_{26}N_{yy}^T+a_{66}N_{xy}^T\right] (6.73a)
\overline{\alpha }_{xx}=\frac{1}{(1)}\left[16.0\times 10^{-9}(52.3)-1.23\times (94.9)\times 10^{-9} – 12.0\times 10^{-9}\times (-36.9)\right]
\overline{\alpha }_{xx}=1.16 μrad/m−° C
Using an equivalent procedure:
\overline{\alpha }_{yy}=2.06 μm/m−° C
\overline{\alpha }_{xy}=-4.00 μm/m−° C
Finally, moisture stress resultants associated with a given change in moisture content must be determined in order to calculate the effective moisture expansion coefficients. Numerically speaking, any change moisture content can be used, but for present purposes a unit change in content will be assumed (i.e., ΔM = 1). Using Equations 6.42, the moisture stress resultants associated with ΔM = 1 are
N_{xx}^M|_{\Delta M=1}=32800 N/m
N_{yy}^M|_{\Delta M=1}=33800 N/m N_{xy}^M|_{\Delta M=1}=-930N/m
Applying Equations 6.76a, we find
\overline{\beta }_{xx}=\frac{1}{\Delta M}\left[a_{11}N_{xx}^M + a_{12}N_{yy}^M + a_{16}N_{xy}^M\right]
\overline{\beta }_{yy}=\frac{1}{\Delta M}\left[a_{12}N_{xx}^M + a_{22}N_{yy}^M + a_{26}N_{xy}^M\right]
\overline{\beta }_{xy}=\frac{1}{\Delta M}\left[a_{16}N_{xx}^M + a_{26}N_{yy}^M + a_{66}N_{xy}^M\right] (6.76a)
\overline{\beta }_{xx}=494μm/m −\%M
\overline{\beta }_{yy}=643μm/m −\%M \overline{\beta }_{xy}=-667μrad/\%M