An edge view of the laminate is shown in Figure 6.17. The total laminate thickness t = 3(0.125 mm) = 0.375 mm. Since all three plies are of the same material, the thickness of each ply is identical: t_1 = t_2 = t_3 = 0.125 mm. Note that since an odd number of plies are used, the origin of the x−y−z coordinate system exists at the midplane of ply 2. The ply interface coordinates can be calculated as
z_0 = −t/2 = −(0.375 mm)/2 = −0.1875 mm = −0.0001875 m
z_1 = z_0 + t_1 = −0.1875 mm + 0.125 mm = −0.0625 mm = −0.0000625 m
z_2 = z_1 + t_2 = −0.0625 mm + 0.125 mm = 0.0625 mm = 0.0000625 m
z_3 = z_2 + t_3 = 0.0625 mm + 0.125 mm = 0.1875 mm = 0.0001875 m
We will also require the transformed reduced stiffness matrix for each ply. Elements of the [\overline{Q} ]_k matrices are calculated using Equations 5.31* and are equal to
\overline{Q}_{11} = {Q}_{11}\cos ^4\theta +2(Q_{12}+2Q_{66}) \cos ^2\theta\sin ^2\theta+Q_{22}\sin ^4\theta \\[0.5cm] \overline{Q}_{12} = \overline{Q}_{21} ={Q}_{12}\left(\cos ^4\theta+\sin ^4\theta \right) + (Q_{11}+Q_{22}-2Q_{66}) \cos ^2\theta\sin ^2\theta \\[0.5cm] \overline{Q}_{16} = \overline{Q}_{61} = (Q_{11}-Q_{12}-2Q_{66}) \cos ^3 \theta\sin \theta – (Q_{22}-Q_{12}-2Q_{66}) \cos \theta \sin ^3 \theta \\[0.5cm] \overline{Q}_{22} = {Q}_{11}\sin ^4\theta +2(Q_{12}+2Q_{66}) \cos ^2\theta\sin ^2\theta + Q_{22}\cos ^4\theta \\[0.5cm] \overline{Q}_{26} = \overline{Q}_{62} = (Q_{11}-Q_{12}-2Q_{66}) \cos \theta\sin ^3\theta – (Q_{22}-Q_{12}-2Q_{66}) \cos ^3 \theta \sin \theta \\[0.5cm] \overline{Q}_{66} = (Q_{11}+Q_{22}-2Q_{12}-2Q_{66}) \cos ^2 \theta\sin ^2 \theta + Q_{66}\left(\cos ^4\theta+\sin ^4\theta \right) (5.31)
For ply #1 (the 30° ply):
\left[\overline{Q} \right]_{30°ply}=\left [ \begin{matrix} \overline{Q}_{11} &\overline{Q}_{12} & \overline{Q}_{16} \\ \overline{Q}_{12} & \overline{Q}_{22} & \overline{Q}_{26} \\\overline{Q}_{16} & \overline{Q}_{26} & \overline{Q}_{66} \end{matrix} \right ]= \left [ \begin{matrix} 107.6\times 10^9 &26.06\times 10^9 & 48.13\times 10^9 \\ 26.06\times 10^9 & 27.22\times 10^9 & 21.52\times 10^9 \\ 48.13\times 10^9 &21.52\times 10^9 & 36.05\times 10^9 \end{matrix} \right ](Pa)
For ply #2 (the 0° ply):
\left[\overline{Q} \right]_{0°ply}=\left [ \begin{matrix} \overline{Q}_{11} &\overline{Q}_{12} & \overline{Q}_{16} \\ \overline{Q}_{12} & \overline{Q}_{22} & \overline{Q}_{26} \\\overline{Q}_{16} & \overline{Q}_{26} & \overline{Q}_{66} \end{matrix} \right ]= \left [ \begin{matrix} 170.9\times 10^9 & 3.016\times 10^9 & 0 \\ 3.016\times 10^9 & 10.05\times 10^9 & 0\\ 0 &0 & 13.00 \times 10^9 \end{matrix} \right ](Pa)
For ply #3 (the 90° ply):
\left[\overline{Q} \right]_{90°ply}=\left [ \begin{matrix} \overline{Q}_{11} &\overline{Q}_{12} & \overline{Q}_{16} \\ \overline{Q}_{12} & \overline{Q}_{22} & \overline{Q}_{26} \\\overline{Q}_{16} & \overline{Q}_{26} & \overline{Q}_{66} \end{matrix} \right ]= \left [ \begin{matrix} 10.05\times 10^9 &3.016\times 10^9 & 0 \\ 3.016\times 10^9 & 170.9\times 10^9 & 0\\ 0 &0 & 13.00\times 10^9 \end{matrix} \right ](Pa)
We can now calculate each member of the A_{ij}, B_{ij}, and D_{ij} matrices, in accordance with Equations 6.27a, 6.27b, and 6.34, respectively.
A_{ij}=\sum\limits_{k=1}^{n}{\left\{\overline{Q}_{ij} \right\}_k(z_k -z_{k-1}) } (6.27a)
B_{ij}=\frac{1}{2} \sum\limits_{k=1}^{n}{\left\{\overline{Q}_{ij} \right\}_k(z_k^2 -z_{k-1}^2) } (6.27b)
D_{ij}=\frac{1}{3} \sum\limits_{k=1}^{n}{\left\{\overline{Q}_{ij} \right\}_k(z_k^3 -z_{k-1}^3) } (6.34)
• Using Equation 6.27a, element A_{11} is calculated as follows:
A_{11}= \sum\limits_{k=1}^{3}{\left\{\overline{Q}_{11} \right\}_k(z_k -z_{k-1}) }
A_{11}=\left\{\overline{Q}_{11} \right\}_1(z_1 -z_0)+\left\{\overline{Q}_{11} \right\}_2(z_2 -z_1) +\left\{\overline{Q}_{11} \right\}_3 (z_3 – z_2)
A_{11}=\left\{107.6\times 10^9 \right\}(-0.000625 + 0.0001875) \\ +\left\{170.6\times 10^9 \right\}(0.0000625 + 0.0000625) \\ +\left\{10.05\times 10^9 \right\} (0.0001875-0.0000625)
A_{11} =36.07 \times 10^9 Pa–m
The remaining elements of the A_{ij} matrix are found in similar fashion:
A_{ij}=\left [ \begin{matrix} 36.07 & 4.012 & 6.016 \\ 4.012 & 26.02 & 2.690 \\ 6.016 & 2.690 & 7.756 \end{matrix} \right ]\times 10^6(Pa−m)
• Using Equations 6.27b, element B_{11} is calculated as follows:
B_{11}= \frac{1}{2}\sum\limits_{k=1}^{3}{\left\{\overline{Q}_{11}\right\}_k(z_k^2 – z_{k -1}^2) }
B_{11}=\frac{1}{2}\left[\left\{\overline{Q}_{11} \right\}_1(z_1^2 -z_0^2)+\left\{\overline{Q}_{11} \right\}_2(z_2^2 -z_1^2)\left\{\overline{Q}_{11} \right\}_3(z_3^2 -z_2^2) \right]
B_{11}=\frac{1}{2} \left[\left\{107.6\times 10^9\right\}\left\{(-0.0000625)^2 – (-0.0001875)^2\right\} \\ +\left\{170.9\times 10^9\right\}\left\{(0.0000625)^2-(-0.0000625)^2\right\} \\ +\left\{10.05\times 10^9\right\}\left\{(0.0001875)^2 – (0.0000625)^2\right\} \right]
B_{11}=-1.524\times 10^3 Pa − m^2
The remaining elements of the B_{ij} matrix are found in similar fashion:
B_{ij}=\left [ \begin{matrix} -1.524 & -0.3601 & -0.7521 \\ -0.3601 & 2.245 & -0.3362 \\ -0.7521 & -0.3362 & -0.3601\end{matrix} \right ]\times 10^3(Pa−m^2)
In passing, in this example, it appears that B_{12} is numerically equal to B_{66}. This is not true, in general. In this problem the apparent numerical equivalence is due to the fact that only 4 significant digits have been listed. Nevertheless, for laminates produced using a single material system it is often (but not always) the case that B_{12} ≈ B_{66}. This common occurrence can be traced to the fact the functional form and magnitude of \overline{Q}_{12} and \overline{Q}_{66} are similar (see Equations 5.31). Since B_{12} and B_{66}are directly related to \overline{Q}_{12} and \overline{Q}_{12} , respectively, their values are often nearly identical. Also, in the next section it will be seen that all elements within the B_{ij} matrix are zero for symmetric laminates. Hence, for symmetric laminates these two terms are, in fact, numerically equal, that is, B_{12} = B_{66} = 0 for symmetric laminates.
• Using Equations 6.34, element D_{11} is calculated as follows:
D_{11}=\frac{1}{3} \sum\limits_{k=1}^{3}{\left\{\overline{Q}_{11} \right\}_k(z_k^3 – z_{k-1}^3) }
D_{11}=\frac{1}{3}\left[\left\{\overline{Q}_{11} \right\}_1(z_1^3 -z_0^3)+\left\{\overline{Q}_{11} \right\}_2(z_2^3 -z_1^3)\left\{\overline{Q}_{11} \right\}_3(z_3^3 -z_2^3) \right]
D_{11}=\frac{1}{3}\left[\left\{107.6\times 10^9 \right\}\left\{(-0.0000625)^3-(-0.0001875)^3 \right\} \\ +\left\{170.9\times 10^9\right\} \left\{(0.0000625)^3 -(-0.0000625)^3\right\} \\ + \left\{10.05\times 10^9 \right\}\left\{(0.0001875)^3 -(0.0000625)^3\right\} \right]
D_{11}=0.2767 Pa–m^3
The remaining elements of the D_{ij} matrix are found in similar fashion:
D_{ij}=\left [ \begin{matrix} 0.2767 &0.0620 & 0.1018 \\ 0.0620 & 0.4208 & 0.0455 \\ 0.1018 & 0.0455 & 0.1059 \end{matrix} \right ](Pa − m^3)
The [ABD] matrix can now be assembled:
[ABD]=\left [ \begin{matrix} 36.07\times 10^6 & 4.012\times 10^6 & 6.0164\times 10^6 & −1524 & −360.1 & −752.1 \\ 4.012 \times 10^6& 26.02\times 10^6 & 2.690\times 10^6 & -360.1 & 2245 & -336.2 \\ 6.016\times 10^6 & 2.690\times 10^6 & 7.756\times 10^6 & -752.1 & -336.2 & -360.1 \\ -1524 & -360.1 & -7521 & 0.2767 &0.0620 & 0.1018 \\ -360.1 & 2245 & -3362 & 0.0620 & 0.4208 & 0.0455 \\ -752.1 & -336.2 & -360.1 &0.1018 & 0.0455& 0.1059 \end{matrix} \right ]
The [abd] matrix is obtained by inverting the [ABD] matrix, and is found to be
[abd]=\left [ \begin{matrix} 3.757\times 10^{-8} &-1.964\times 10^{-9} & -1.038\times 10^{-8} & 1.440\times 10^{-4} & 3.905\times 10^{-6} & 8.513\times 10^{-6} \\ -1.964\times 10^{-9} & 1.037\times 10^{-7} & -4.234\times 10^{-8}& -1.866\times 10^{-5} & -6.361\times 10^{-4} & 4.6288\times 10^{4} \\ -1.038 \times 10^{-5} & -4.234\times 10^{-4} & 2.004\times 10^{-7} &3.661\times 10^{-4} & 3.251\times 10^{-4} & -1.851\times 10^{-5} \\ 1.440\times 10^{-4} & -1.866\times 10^{-5} & 3661\times 10^{-4} & 7.064 & -3.122\times 10^{-2} & -4.572 \\ 3.905\times 10^{-6} & -6.361\times 10^{-4} & 3.251\times 10^{-4} & -3.122\times 10^{-2} & 6.429 & -3.620 \\ 8.513\times 10^{-5} & 4.628\times 10^4 & -1.851\times 10^{-5} & -4.572 & -3.620 & 17.41 \end{matrix} \right ]
^\ast [\overline{Q} ] matrix for a 30°graphite–epoxy ply was calculated as a part of Example Problem 5.6.
