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Question 12.1: A [(±45/0)2]s graphite–epoxy laminate is cured at 175°C and ...

A [(±45/0)_2]_s graphite–epoxy laminate is cured at 175°C and then cooled to room temperature (20°C). After cooling, the flat laminate is trimmed to in-plane dimensions of 300 × 150 mm and mounted in an assembly that provides type S4 simple supports along all four edges. The x-axis is defined parallel to the 300 mm edge (i.e., a = 0.3 m, b = 0.15 m). The laminate is then subjected to a uniform transverse load q(x,y) = 30 kPa. No in-plane loads are applied (i.e., N_{xx} = N_{yy} = N_{xy} = 0). Determine the maximum out-of-plane displacement based on a Ritz analysis and plot the out-of-plane displacement field. Use the properties listed for graphite–epoxy in Table 3.1, and assume each ply has a thickness of 0.125 mm.

TABLE 3.1
Typical Properties of Common Unidirectional Composites

Property Glass/ Epoxy Kevlar/ Epoxy Graphite/ Epoxy
E_{11} 55 GPa 100 GPa 170 GPa
(8.0 Msi) (15 Msi) (25 Msi)
E_{22} 16 GPa 6 GPa 10 GPa
(2.3 Msi) (0.90 Msi) (1.5 Msi)
ν_{12} 0.28 0.33 0.3
G_{12} 7.6 GPa 2.1 GPa 13 GPa
(1.1 Msi) (0.30 Msi) (1.9 Msi)
 \sigma _{11}^{fT} 1050 MPa 1380 MPa 1500 MPa
(150 ksi) (200 ksi) (218 ksi)
\sigma _{11}^{fC} 690 MPa 280 MPa 1200 MPa
(100 ksi) (40 ksi) (175 ksi)
\sigma _{22}^{ff} 45 MPa 35 MPa 50 MPa
(5.8 ksi) (2.9 ksi) (7.25 ksi)
\sigma _{22}^{fC} 120 MPa 105 MPa 100 MPa
(16 ksi) (15 ksi) (14.5 ksi)
\tau ^f_{22} 40 MPa 40 MPa 90 MPa
(4.4 ksi) (4.0 ksi) (13.1 ksi)
\alpha _{11} 6.7 μm/m−°C −3.6 μm/m−°C −0.9 μm/m−°C
(3.7 μin./in.\boxtimes  °F) (−2.0 μin./in.−°F) (−0.5 μin./in.−°F)
\alpha _{22} 25 μm/m−°C 58 μm/m−°C 27 μm/m−°C
(14 μin./in.−°F) (32 μin./in.−°F) (15 μin./in.−°F)
\beta _{11} 100 μm/m−%M 175 μm/m−%M 50 μm/m−%M
(100 μin./in.−%M) (175 μin./in.−%M) (50 μin./in.−%M)
\beta _{22} 1200 μm/m−%M 1700 μm/m−%M 1200 μm/m−%M
(1200 μin./in.−%M) (1700 μin./in.−%M) (1200 μin./in.−%M)
Ply 0.125 mm 0.125 mm 0.125 mm
Thickness (0.005 in.) (0.005 in.) (0.005 in.)
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Based on the properties listed in Table 3.1 for graphite–epoxy, the [ABD] matrix for a [(±45/0)_2]_s laminate is

[ABD]=\left [ \begin{matrix} 145.2\times 10^6 & 35.3\times 10^6 & 0 & 0 &0 & 0 \\ 35.3\times 10^6 & 64.8\times 10^6 & 0 & 0 & 0 & 0 \\ 0 & 0 & 50.2\times 10^6 & 0 & 0 & 0 \\ 0 & 0 & 0 & 22.3 &7.97 &2.20 \\ 0 & 0 &0 & 7.97 & 14.3 & 2.20 \\ 0 & 0 & 0 & 2.20 & 2.20 & 10.8 \end{matrix} \right ]

where the units of A_{ij} are Pa\boxtimes m and the units of D_{ij} are Pa \boxtimes m^3. Notice that neither D_{16} nor D_{26} equal zero, and hence the laminate is generally orthotropic. The 12-ply laminate has a total thickness t = 1.5 mm and aspect ratio R = a/b = 2.0.

The computer program SYMM (described in Section 12.6) can be used to perform the required Ritz analysis. Several analyses were performed using increasing values of M (and N), to evaluate whether the solution has converged to a reasonably constant value. Solutions were obtained using values of M (and N) ranging from 1 through 10 (i.e., analyses were performed in which the number of terms used to describe the displacement field ranged from 1 through 100). Maximum predicted displacement is plotted as a function of M and N in Figure 12.3. As indicated, the maximum displacement converges to a value of 8.03 mm when M = N = 10.

A contour plot of out-of-plane displacements predicted using M = N = 10 is shown in Figure 12.4. As would be expected, the maximum displacement occurs at the center of the plate (i.e., at x = 150 mm, y = 75 mm). Careful examination of these contours will reveal that the contours are very slightly distorted. This distortion (which is barely discernible in Figure 12.4) occurs because the plate is generally orthotropic. That is, for a [(±45/0)_2]_s laminate D_{16}, D_{26} ≠ 0. However, for this problem the magnitudes of D_{16} and D_{26} (relative to D_{11} and D_{22}) are very small. Specifically, for the laminate considered in this problem D_{16}/D_{11} = D_{26} D_{11} = 0.0986 and D_{16}/D_{22} = D_{26} D_{22} = 0.153. Consequently, distortion of out-of-plane displacements is very slight. The out-of-plane displacement induced by a uniform transverse load applied to a laminate with relatively higher values of D_{16} and D_{26} is considered in Example Problem 12.3. As will be seen, the distortion of displacement contours is much more pronounced in that case, due to the relatively higher values of D_{16} and D_{26}.

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Related Answered Questions

A [25°]12 graphite–epoxy laminate is trimmed to in-plane dimensions of 300 × 150 mm and mounted in an assembly that provides type S4 simple supports along all four edges. The laminate is then subjected to a uniform transverse load q(x,y) = 30 kPa. No in-plane loads are applied

Verified Answer:
Note that the plate has an aspect ratio R = 150/30...

The [(±45/0)2]s graphite–epoxy laminate described in Example Problem 12.1 is again subjected to type S4 simple supports along all four edges and a uniform transverse load q(x,y) = 30 kPa. However, the plate is now also subjected to uniform in-plane stress resultants Nxx = Nyy. Use a Ritz analysis

Verified Answer:
As was the case for Example Problem 12.1, solution...