Question 6.5: A [30/0/90]T graphite-epoxy laminate is subjected to the fol...
A [30/0/90]_{T} graphite-epoxy laminate is subjected to the following stress and moment resultants:
N_{xx} = 50 kN/m N_{yy} = -10 kN/m N_{xy} = 0 N/m
M_{xx} = 1 N – m/m M_{yy} = -1 N – m/m M_{xy} = 0 N – m/m
Determine the following quantities caused by these stress and moment resultants:
(a) Midplane strains and curvatures
(b) Ply strains relative to the x–y coordinate system
(c) Ply stresses relative to the x–y coordinate system.
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) |
Note that this is the same laminate considered in Example Problem 4. A side view of the laminate appears in Fig. 17.
(a) Midplane strains and curvatures. The [abd] matrix for this laminate was calculated as a part of Example Problem 4. Hence, midplane strains and curvature may be obtained through application of Eq. (38), which becomes:
\left\{\begin{matrix} \varepsilon _{xx}^{o} \\ \varepsilon _{yy}^{o} \\ \gamma _{xy}^{o} \\ \kappa _{xx}\\\kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\} = \left[\begin{matrix} a_{11} & a_{12} & a_{16} & b_{11}& b_{12 } & b_{16} \\ a_{12} & a_{22} & a_{26} & b_{12}& b_{22} & b_{26} \\ a_{16} & a_{26} & a_{66} & b_{16} & b_{26} & b_{66} \\ b_{11}& b_{12 } & b_{16} & d_{11} & d_{12} & d_{16}\\ b_{12}& b_{22} & b_{26} & d_{12}& d_{22} & d_{26} \\b_{16} & b_{26} & b_{66} & d_{16} & d_{26} & d_{66} \end{matrix} \right] \left\{\begin{matrix} N_{xx} \\ N_{yy} \\ N_{xy} \\ M_{xx} \\M_{yy} \\M_{xy} \end{matrix} \right\} (38)
\left\{\begin{matrix} \varepsilon _{xx}^{o} \\ \varepsilon _{yy}^{o} \\ \gamma _{xy}^{o} \\ \kappa _{xx}\\\kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\} = \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] \times \left\{\begin{matrix} 50 \times 10^{3} \\ -10 \times 10^{-3} \\ 0 \\ 1 \\-1 \\0 \end{matrix} \right\}
Completing this matrix multiplication, we obtain:
\left\{\begin{matrix} \varepsilon _{xx}^{o} \\ \varepsilon _{yy}^{o} \\ \gamma _{xy}^{o} \\ \kappa _{xx}\\\kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\} = \left\{\begin{matrix} 2039 μm/m \\ -518 μm/m \\ – 55 μrad\\ 14.48 m^{-1}\\0.096 m^{-1} \\- 1.323 m^{-1} \end{matrix} \right\}(b) Ply strains relative to the x–y coordinate system. Ply strains may now be calculated using Eq. (12). For example, strains present at the outer surface of ply 1 (i.e., strains present at z_{0} = -0.0001875 m) are:
\left\{\begin{matrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \gamma _{xy} \end{matrix} \right\} = \left\{\begin{matrix} \varepsilon° _{xx} \\ \varepsilon° _{yy} \\ \gamma° _{xy} \end{matrix} \right\} + z \left\{\begin{matrix} \kappa _{xx} \\ \kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\} (12)
\left. \left\{\begin{matrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \gamma _{xy} \end{matrix} \right\} \right|_{z = z_{0}}= \left\{\begin{matrix} \varepsilon° _{xx} \\ \varepsilon° _{yy} \\ \gamma° _{xy} \end{matrix} \right\} + z_{0} \left\{\begin{matrix} \kappa _{xx} \\ \kappa _{yy} \\ \kappa _{xy} \end{matrix} \right\}= \left\{\begin{matrix} 2038 \times 10^{-6} m/m \\ -518 \times 10^{-6} m/m \\ -55 \times 10^{-6} m/m \end{matrix} \right\} + ( – 0.0001875 m) \left\{\begin{matrix} 14.48 rad/m \\ 0.096 rad/m \\ – 1.328 rad/m \end{matrix} \right\}
\left. \left\{\begin{matrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \gamma _{xy} \end{matrix} \right\} \right|_{z = z_{0}}= \left\{\begin{matrix} -677 μm/m \\ -536 μm/m \\ 194 μrad\end{matrix} \right\}
Strains calculated at the remaining ply interface positions are summarized in Table 5.
(c) Ply stresses relative to the x–y coordinate system. The [\overline{Q}] matrix for all plies was calculated as a part of Example Problem 4. Ply stresses may now be calculated using Eq. (30) of Chap. 5, with ΔT = ΔM = 0. The stresses present at the outer surface of ply 1 (i.e., at z=z_{0}) are:
M_{xx}= \int_{-t/2}^{t/2}{\left\{z \overline{Q}_{11} \varepsilon _{xx}^{o}+ z \overline{Q}_{12} \varepsilon _{yy}^{o} + z \overline{Q}_{16} \gamma _{xx}^{o} + z^{2} \overline{Q}_{11} \kappa _{xx} + z^{2} \overline{Q}_{12} \kappa _{yy} + z^{2} \overline{Q}_{16} \kappa _{xy}\right\} }dz (30)
| Table 5 Ply Interface Strains in a [30/0/90] Graphite-Epoxy Laminate Caused by the Stress and Moment Resultants Specified in Example Problem 5 | |||
| z-coordinate (mm) | \varepsilon _{xx} (\mu m/m) | \varepsilon _{yy} (\mu m/m) | \gamma _{xy} (\mu rad) |
| -0.1875 | -677 | -536 | 194 |
| -0.0625 | 1133 | -524 | 28 |
| 0.0625 | 2943 | -512 | -137 |
| 0.1875 | 4753 | -500 | -303 |
| Strains are referenced to the x–y coordinate system. | |||
\left. \left\{\begin{matrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau _{xy} \end{matrix} \right\} \right|_{z = z_{0}}^{ply 1} = \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] \left\{\begin{matrix} -677 × 10^{-6} \\ -536 × 10^{-6} \\ 194 × 10^{-6}\end{matrix} \right\}
\left. \left\{\begin{matrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau _{xy} \end{matrix} \right\} \right|_{z = z_{0}}^{ply 1} = \left\{\begin{matrix} -77.5 MPa \\ -28.1 MPa \\ -37.1 MPa \end{matrix} \right\}
Stresses calculated at remaining ply interface positions are summarized in Table 6.
| Table 6 Ply Interface Stresses in a [30/0/90]T Graphite-Epoxy Laminate Caused by the Stress and Moment Resultants Specified in Example Problem 5 | ||||
| Ply number | z-coordinate (mm) | \sigma _{xx} (Mpa) | \sigma _{yy} (Mpa) | \tau _{xy} (Mpa) |
| Ply 1 | -0.1875 | -77.5 | -28.1 | -37.1 |
| -0.0625 | 109.7 | 15.9 | 44.3 | |
| Ply 2 | -0.0625 | 192.1 | -1.85 | 0.366 |
| 0.0625 | 501.5 | 3.73 | -1.78 | |
| Ply 3 | 0.0625 | 28.0 | -78.6 | -1.78 |
| 0.1875 | 46.3 | -71.1 | -3.93 | |
| Stresses are referenced to the x–y coordinate system. | ||||
