Question 6.4: Determine the [ABD] and [abd] matrices for a [30/0/90]T grap...
Determine the [ABD] and [abd] matrices for a [30/0/90]_{T} graphite-epoxy laminate. Use material properties listed for graphite-epoxy in Table 3 of Chap. 3, and assume that each ply has a thickness of 0.125 mm.
| Table 3 Nominal Material Properties for Common Unidirectional Composites | |||
| Property | Glass/epoxy | Kevlar/epoxy | Graphite/epoxy |
| E_{11} | 55 GPa (8.0 Msi) | 100 GPa (15 Msi) | 170 GPa (25 Msi) |
| E_{22} | 16 GPa (2.3 Msi) | 6 GPa (0.90 Msi) | 10 GPa (1.5 Msi) |
| ν_{12} | 0.28 | 0.33 | 0.30 |
| G_{12} | 7.6 GPa (1.1 Msi) | 2.1 GPa (0.30 Msi) | 13 GPa (1.9 Msi) |
| σ_{11}^{fT} | 1050 MPa (150 ksi) | 1380 MPa (200 ksi) | 1500 MPa (218 ksi) |
| σ_{11}^{fC} | 690 MPa (100 ksi) | 280 MPa (40 ksi) | 1200 MPa (175 ksi) |
| σ_{22}^{yT} | 45 MPa (5.8 ksi) | 35 MPa (2.9 ksi) | 50 MPa (7.25 ksi) |
| σ_{22}^{yC} | 120 MPa (16 ksi) | 105 MPa (15 ksi) | 100 MPa (14.5 ksi) |
| σ_{22}^{fT} | 55 MPa (7.0 ksi) | 45 MPa (4.3 ksi) | 70 MPa (10 ksi) |
| σ_{22}^{fC} | 140 MPa (20 ksi) | 140 Msi (20 ksi) | 130 MPa (18.8 ksi) |
| τ_{12}^{y} | 40 MPa (4.4 ksi) | 40 MPa (4.0 ksi) | 75 MPa (10.9 ksi) |
| τ_{12}^{f} | 70 MPa (10 ksi) | 60 MPa (9 ksi) | 130 MPa (22 ksi) |
| α_{11} | 6.7 μ/m °C
(3.7 μin./in. °F) |
3.6 μm/m °C
(-2.0 μin./in. °F) |
0.9 μm/m °C
(-0.5 μin./in. °F) |
| α_{22} | 25 μ/m °C
(14 μin./in. °F) |
58 μm/m °C
(32 μin./in. °F) |
27 μm/m °C
(15 μin./in. °F) |
| β_{11} | 100 μm/m %M
(100 μin./in. %M) |
175 μm/m %M
(175 μin./in. %M) |
50 μm/m %M
(50 μin./in. %M) |
| β_{22} | 1200 μm/m %M
(1200 μin./in. %M) |
1700 μm/m %M
(1700 μin./in. %M) |
1200 μm/m %M
(1200 μin./in. %M) |
A side view of the laminate is shown in Fig. 17. The total laminate thickness t = 3 (0.125 mm) = 0.375 mm. Because 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 because 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 = – (0.375mm)/2 = -0.1875mm = -0.0001875mz_{1} = z_{0} + t_{1} = -0.1875mm + 0.125mm = -0.0625mm = -0.0000625m
z_{2} = z_{1} + t_{2} = -0.0625mm + 0.125mm = 0.0625mm = 0.0000625m
z_{3} = z_{2} + t_{3} = 0.0625mm + 0.125mm = 0.1875mm = 0.0001875m
We will also require the transformed reduced stiffness matrix for each ply. Elements of the [\bar{Q} ]_{k} matrices are calculated using Eq. (31) of Chap. 5^{*} and are equal to:
M_{xx} = \varepsilon _{xx}^{o} \left\{(\overline{Q}_{11})_{1} \int_{z_{o}}^{z_{1}}{z dz} + (\overline{Q}_{11})_{2} \int_{z_{1}}^{z_{2}}{z dz} +(\overline{Q}_{11})_{3} \int_{z_{2}}^{z_{3}}{z dz} + … + (\overline{Q}_{11})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z dz} + (\overline{Q}_{11})_{n} \int_{z_{n}-1}^{z_{n}}{z dz}\right\} \varepsilon _{yy}^{o} \left\{(\overline{Q}_{12})_{1} \int_{z_{o}}^{z_{1}}{z dz} + (\overline{Q}_{12})_{2} \int_{z_{1}}^{z_{2}}{z dz} +(\overline{Q}_{12})_{3} \int_{z_{2}}^{z_{3}}{z dz} + … + (\overline{Q}_{12})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z dz} + (\overline{Q}_{12})_{n} \int_{z_{n}-1}^{z_{n}}{z dz}\right\} γ _{xy}^{o} \left\{(\overline{Q}_{16})_{1} \int_{z_{o}}^{z_{1}}{z dz} + (\overline{Q}_{16})_{2} \int_{z_{1}}^{z_{2}}{z dz} +(\overline{Q}_{16})_{3} \int_{z_{2}}^{z_{3}}{z dz} + … + (\overline{Q}_{16})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z dz} + (\overline{Q}_{16})_{n} \int_{z_{n}-1}^{z_{n}}{z dz}\right\} \kappa _{xx} \left\{(\overline{Q}_{11})_{1} \int_{z_{o}}^{z_{1}}{z^{2} dz} + (\overline{Q}_{11})_{2} \int_{z_{1}}^{z_{2}}{z^{2} dz} +(\overline{Q}_{11})_{3} \int_{z_{2}}^{z_{3}}{z^{2} dz} + … + (\overline{Q}_{11})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z^{2} dz} + (\overline{Q}_{11})_{n} \int_{z_{n}-1}^{z_{n}}{z^{2} dz}\right\} \kappa _{yy} \left\{(\overline{Q}_{12})_{1} \int_{z_{o}}^{z_{1}}{z ^{2}dz} + (\overline{Q}_{12})_{2} \int_{z_{1}}^{z_{2}}{z^{2} dz} +(\overline{Q}_{12})_{3} \int_{z_{2}}^{z_{3}}{z^{2} dz} + … + (\overline{Q}_{12})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z ^{2}dz} + (\overline{Q}_{12})_{n} \int_{z_{n}-1}^{z_{n}}{z^{2} dz}\right\}\kappa _{xy} \left\{(\overline{Q}_{16})_{1} \int_{z_{o}}^{z_{1}}{z^{2} dz} + (\overline{Q}_{16})_{2} \int_{z_{1}}^{z_{2}}{z^{2} dz} +(\overline{Q}_{16})_{3} \int_{z_{2}}^{z_{3}}{z^{2} dz} + … + (\overline{Q}_{16})_{n-1} \int_{z_{n} -2}^{z_{n}-1}{z^{2} dz} + (\overline{Q}_{16})_{n} \int_{z_{n}-1}^{z_{n}}{z^{2} dz}\right\} (31)
For ply 1 (the 30° ply):
[\bar{Q} ]_{30° ply} = \left[\begin{matrix} \bar{Q} _{11} & \bar{Q}_{12} & \bar{Q}_{16} \\ \bar{Q}_{12} & \bar{Q}_{22} & \bar{Q}_{26} \\ \bar{Q}_{16} & \bar{Q}_{26} & \bar{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):
[\bar{Q} ]_{0° ply} = \left[\begin{matrix} \bar{Q} _{11} & \bar{Q}_{12} & \bar{Q}_{16} \\ \bar{Q}_{12} & \bar{Q}_{22} & \bar{Q}_{26} \\ \bar{Q}_{16} & \bar{Q}_{26} & \bar{Q}_{66} \end{matrix} \right]= \left[\begin{matrix} 107.9 \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):
[\bar{Q} ]_{90° ply} = \left[\begin{matrix} \bar{Q} _{11} & \bar{Q}_{12} & \bar{Q}_{16} \\ \bar{Q}_{12} & \bar{Q}_{22} & \bar{Q}_{26} \\ \bar{Q}_{16} & \bar{Q}_{26} & \bar{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 (Eq. (27a), (27b), and (34), respectively.
A_{ij} = \sum\limits_{k =1}^{n}{\left\{\overline{Q}_{ij} \right\} } _{k} (z_{k} – z_{k-1}) (27a)
B_{ij} = \frac{1}{2} \sum\limits_{k =1}^{n}{\left\{\overline{Q}_{ij}\right\}_{k} } (z_{k}^{2} – z_{k-1}^{2}) (27b)
D_{ij} = \frac{1}{3} \sum\limits_{k =1}^{n}{\left\{\overline{Q}_{ij}\right\}_{k} } (z_{k}^{3} – z_{k-1}^{3}) (34)
• Using Eq. (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 × 10^{9}\right\} } (-.0000625 + 0.0001875) + {\left\{107.9 × 10^{9} \right\} } × (0.0000625 + 0.0000625) + {\left\{10.05 × 10^{9} \right\} } × (0.0001875 – 0.0000625)
A_{11} = 36.07 × 10^{6} Pa – m
The remaining elements of the A_{ij} matrix are found in similar fashion:
A = \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 Eq. (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\{\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})]
B_{11} = \frac{1}{2} [\left\{107.6 × 10^{9}\right\} \left\{ ( -.0000625)^{2} + ( -0.0001875)^{2}\right\} + \left\{170.9 × 10^{9}\right\} \left\{ ( 0.0000625)^{2} – ( -0.0000625)^{2}\right\} + \left\{10.05 × 10^{9}\right\} \left\{ ( 0.0001875)^{2} – ( 0.0000625)^{2}\right\}]
B_{11} = -1.524 × 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 appearsthat 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 four significant digits have been used. 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 Eq.(31) of Chap. 5). Because B_{12} and B_{66} are directly related to \overline{Q}_{12} and \overline{Q}_{66}, respectively, their values are often nearly identical. Also, in Sec. 6.2, it will be seen that all elements within the B_{ij} matrix are zero for symmetrical laminates. Hence, for symmetrical laminates, these two terms are, in fact, numerically equal, that is, B_{12}=B_{66}=0 for symmetrical laminates.
• Using Eq. (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\{\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})]
D_{11} = \frac{1}{3} [\left\{107.6 × 10^{9}\right\} \left\{ ( -.0000625)^{3} – ( -0.0001875)^{3}\right\} + \left\{170.9 × 10^{9}\right\} + \left\{ ( 0.0000625)^{3} – ( -0.0000625)^{3}\right\} + \left\{10.05 × 10^{9}\right\} \left\{ ( 0.0001875)^{3} – ( 0.0000625)^{3}\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 & 2.513& 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} & 6.016 \times 10^{6} \\ 4.012 \times 10^{6} & 26.02 \times 10^{6} & 2.690 \times 10^{6} \\ 6.016 \times 10^{6} & 2.690 \times 10^{6} & 7.756 \times 10^{6}\\ -1524 &-360.1 & -752.1\\-360.1 & 2245 & -336.2\\ -752.1& -336.2 &-360.1\end{matrix} \begin{matrix} -1524 & -360.1 & -752.1 \\ -360.1 & 2245 & -336.2 \\ -752.1 & -336.2 & -360.1\\ 0.2767 & 0.0620& 0.1018\\0.0620 & 2.513 & 0.0455\\ 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} \\ -1.964 \times 10^{-9} & 1.037 \times 10^{-7} & -4.234 \times 10^{-8} \\ -1.038 \times 10^{-8} & -4.234 \times 10^{-8} & 2.004\times 10^{-7}\\ 1.440 × 10^{-4}&-1.866 × 10^{-5}& 3.661 × 10^{-4}\\3.905 × 10^{-6}& -6.361 × 10^{-4} & 3.251× 10^{-4}\\ 8.513 × 10^{-5} & 4.268 × 10^{4} & -1.851 × 10^{-5}\end{matrix} \begin{matrix} 1.440 × 10^{-4}&3.905 × 10^{-6}& 8.513 × 10^{-5}\\-1.866 × 10^{-5}& 6.361 × 10^{-4} & 4.268 × 10^{4}\\ 3.661 × 10^{-4} & 3.251 × 10^{-4} & -1.851 × 10^{-5}\\7.064 & -3.122 × 10^{-2} & -4.572\\ -3.122 × 10^{-2} & 6.429 & -3.620\\-4.572 & -3.620 & 17.41\end{matrix} \right]