Assume that the panel considered in Sample Problem 1 is actually an eight-ply [0/30/90/-30]s graphite–epoxy laminate. Assume that the laminate was initially flat and stress-free (i.e., ignore possible preexisting stresses/strains due to temperature and/or moisture changes). Determine the strains

Question

Assume that the panel considered in Sample Problem 1 is actually an eight-ply [latex][0/30/90/-30]_{s}[/latex] graphite–epoxy laminate. Assume that the laminate was initially flat and stress-free (i.e., ignore possible preexisting stresses/strains due to temperature and/or moisture changes). Determine the strains and stresses induced at each ply interface. Use material properties listed in Table 3 of Chap. 3, and assume that the thickness of each ply is 0.125 mm.

Table 3 Nominal Material Properties for Common Unidirectional Composites
Property	Glass/epoxy	Kevlar/epoxy	Graphite/epoxy
[latex]E_{11}[/latex]	55 GPa (8.0 Msi)	100 GPa (15 Msi)	170 GPa (25 Msi)
[latex]E_{22}[/latex]	16 GPa (2.3 Msi)	6 GPa (0.90 Msi)	10 GPa (1.5 Msi)
[latex]ν_{12}[/latex]	0.28	0.33	0.30
[latex]G_{12}[/latex]	7.6 GPa (1.1 Msi)	2.1 GPa (0.30 Msi)	13 GPa (1.9 Msi)
[latex]σ_{11}^{fT}[/latex]	1050 MPa (150 ksi)	1380 MPa (200 ksi)	1500 MPa (218 ksi)
[latex]σ_{11}^{fC}[/latex]	690 MPa (100 ksi)	280 MPa (40 ksi)	1200 MPa (175 ksi)
[latex]σ_{22}^{yT}[/latex]	45 MPa (5.8 ksi)	35 MPa (2.9 ksi)	50 MPa (7.25 ksi)
[latex]σ_{22}^{yC}[/latex]	120 MPa (16 ksi)	105 MPa (15 ksi)	100 MPa (14.5 ksi)
[latex]σ_{22}^{fT}[/latex]	55 MPa (7.0 ksi)	45 MPa (4.3 ksi)	70 MPa (10 ksi)
[latex]σ_{22}^{fC}[/latex]	140 MPa (20 ksi)	140 Msi (20 ksi)	130 MPa (18.8 ksi)
[latex]τ_{12}^{y}[/latex]	40 MPa (4.4 ksi)	40 MPa (4.0 ksi)	75 MPa (10.9 ksi)
[latex]τ_{12}^{f}[/latex]	70 MPa (10 ksi)	60 MPa (9 ksi)	130 MPa (22 ksi)
[latex]α_{11}[/latex]	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)
[latex]α_{22}[/latex]	25 μ/m °C (14 μin./in. °F)	58 μm/m °C (32 μin./in. °F)	27 μm/m °C (15 μin./in. °F)
[latex]β_{11}[/latex]	100 μm/m %M (100 μin./in. %M)	175 μm/m %M (175 μin./in. %M)	50 μm/m %M (50 μin./in. %M)
[latex]β_{22}[/latex]	1200 μm/m %M (1200 μin./in. %M)	1700 μm/m %M (1700 μin./in. %M)	1200 μm/m %M (1200 μin./in. %M)

Accepted Answer

From Sample Problem 1, the midplane strains and curvatures are:

[latex]ε°_{xx} = 0 μm/m,[/latex] [latex] \kappa _{xx} = -0.50 rad/m[/latex]

[latex]ε°_{yy} = -1300 μm/m,[/latex] [latex] \kappa _{yy} = 0.40 rad/m[/latex]

[latex]γ°_{xy} = 900 μrad,[/latex] [latex] \kappa _{xy} = -0.20 rad/m[/latex]

To determine ply interface positions, first note that the total laminate thickness is:

[latex]t = (8 plies)(0.125 mm) = 1.0 mm = 0.001 m[/latex]

A total of nine ply interface positions must be determined because there are eight plies in the laminate. Following the numbering scheme discussed in Sec. 4 and referring to Fig. 11, ply interface positions are:

[latex]z_{0}=-t/2=-(0.001 m)=-0.000500 m[/latex]
[latex]z_{1}=z_{0}+t_{1}=-0.000500 m+0.000125 m=-0.000375 m[/latex]
[latex]z_{2}=z_{1}+t_{2}=-0.000375 m+0.000125 m=-0.000250 m[/latex]
[latex]z_{3}=z_{2}+t_{3}=-0.000250 m+0.000125 m=-0.000125 m[/latex]
[latex]z_{4}=z_{3}+t_{4}=-.000125 m+0.000125 m=0.000000 m[/latex]
[latex]z_{5}=z_{4}+t_{5}=0.000000 m+0.000125 m=0.000125 m[/latex]
[latex]z_{6}=z_{5}+t_{6}=0.000125 m+0.000125 m=0.000250 m[/latex]
[latex]z_{7}=z_{6}+t_{7}=0.000250 m+0.000125 m=0.000375 m[/latex]
[latex]z_{8}=z_{7}+t_{7}=0.000375 m+0.000125 m=0.000500 m[/latex]

Strain Calculations. Strains are calculated using Eq. (12), and can be determined at any through-thickness position. Usually, strains of greatest interest are those induced at the ply interface locations. For example, strains present at the outer surface of ply 1 (i.e., strains present at [latex]z_{o}=-0.000500 m[/latex]) are:

[latex]\left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy} \end{matrix} ight\} = \left\{\begin{matrix} \varepsilon° _{xx} \ \varepsilon° _{yy} \ \gamma° _{xy} \end{matrix} ight\} + z \left\{\begin{matrix} \kappa _{xx} \ \kappa _{yy} \ \kappa _{xy} \end{matrix} ight\}[/latex] (12)

[latex]\left .\left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy} \end{matrix} ight\} ight|_{z = z_{0}} = \left\{\begin{matrix} \varepsilon° _{xx} \ \varepsilon° _{yy} \ \gamma° _{xy} \end{matrix} ight\} + z_{0} \left\{\begin{matrix} \kappa _{xx} \ \kappa _{yy} \ \kappa _{xy} \end{matrix} ight\} = \left\{\begin{matrix} 0 \ -1300 imes 10^{-6} m/m \900 imes 10^{-6} m/m\end{matrix} ight\} + (- 0.000500 m)\left\{\begin{matrix} -0.50 rad/m \ 0.40 rad/m \ -0.20 rad/m \end{matrix} ight\} [/latex]

[latex]\left .\left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy} \end{matrix} ight\} ight|_{z = z_{0}} = \left\{\begin{matrix} 250 μm/m \ -1500 μm/m \ 1000 μrad\end{matrix} ight\} [/latex]

Similarly, strains present at the interface between plies 1 and 2 (i.e., strains present at [latex]z_{1}=-0.000375 m)[/latex] are:

[latex]\left .\left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy} \end{matrix} ight\} ight|_{z = z_{1}} = \left\{\begin{matrix} \varepsilon° _{xx} \ \varepsilon° _{yy} \ \gamma° _{xy} \end{matrix} ight\} + z_{1} \left\{\begin{matrix} \kappa _{xx} \ \kappa _{yy} \ \kappa _{xy} \end{matrix} ight\} = \left\{\begin{matrix} 0 \ -1300 imes 10^{-6} m/m \900 imes 10^{-6} m/m\end{matrix} ight\} + (- 0.000375 m)\left\{\begin{matrix} -0.50 rad/m \ 0.40 rad/m \ -0.20 rad/m \end{matrix} ight\} [/latex]

[latex]\left .\left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy} \end{matrix} ight\} ight|_{z = z_{1}} = \left\{\begin{matrix} 188 μm/m \ -1450 μm/m \ 975 μrad \end{matrix} ight\} [/latex]

Strains present at all remaining interfaces are calculated in exactly the same fashion. Strains calculated at all ply interfaces are summarized in Table 1 and are plotted in Fig. 13. Note that all three strain components [latex] (ε_{xx}, ε_{yy}[/latex], and [latex] γ_{xy}[/latex])are predicted to be linearly distributed through the plate thickness. This linear distribution is a direct consequence of the Kirchhoff hypothesis, which is a good approximation as long as the plate is ‘‘thin.’’ In fact, identical strain distributions would be predicted for any thin plate subjected to the midplane strains and curvatures specified in Sample Problem 1. For example, we would predict the identical strains if an aluminum plate were under consideration rather than a laminated composite plate.

The strains listed in Table 1 and plotted in Fig. 13 are referenced to the global x–y coordinate system. As will be seen, knowledge of ply strains referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the strain tensor from one coordinate system to another was reviewed in Chap. 2 and, in particular, strains can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (44) of Chap. 2. In practice, strains are usually calculated at both the ‘‘top’’ and ‘‘bottom’’ interface for each ply. Example calculations for plies 1 and 2 are listed below: Ply 1. Because [latex] θ_{1}=0°[/latex], the x–y and 1–2 coordinate systems are coincident, and therefore the description of the strain tensor is identical in both coor-dinate systems. This can be confirmed through application of Eq. (44) Chap. 2:

Top interface:

[latex]\left\{\begin{matrix} \varepsilon _{x^{\prime }x^{\prime }} \ \varepsilon _{y^{\prime } y^{\prime }} \ \frac{\gamma _{x^{\prime }y^{\prime }}}{2} \end{matrix} ight\}= \left[\begin{matrix} \cos ^{2}( heta) & \sin ^{2}( heta) & 2 \cos ( heta) \sin( heta) \ \sin ^{2} ( heta) & \cos ^{2} ( heta)&-2 \cos ( heta) \sin ( heta) \ - \cos ( heta) \sin ( heta) & \cos ( heta) \sin ( heta)& \cos^{2} ( heta)- \sin^{2}( heta) \end{matrix} ight] \left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \frac{\gamma _{xy}}{2} \end{matrix} ight\}[/latex] (44)

Table 1 Ply Interface Strains in a[latex] [0/-30/90/30_{s}[/latex] Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1
z-coordinate (mm)	[latex]ε_{xx} (μm/m)[/latex]	[latex]ε_{yy} (μm/m)[/latex]	[latex] γ_{xy} (μrad)[/latex]
-0.500	250	-1500	1000
-0.375	188	-145	975
-0.250	125	-1400	950
-0.125	62	-1350	925
0.0	0	-1300	900
0.125	-62	-1250	875
0.250	-125	-1200	850
0.375	-188	-1150	850
0.500	-250	-1100	800
Strains are referenced to the x–y coordinate system.

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}} = \left[\begin{matrix} \cos ^{2} heta _{1} & \sin ^{2} heta _{1} & 2 \cos heta _{1} \sin heta _{1} \ \sin ^{2} heta _{1} & \cos ^{2} heta _{1} &-2 \cos heta _{1} \sin heta _{1}\ - \cos heta _{1} \sin heta _{1} & \cos heta _{1} \sin heta _{1} & \cos^{2} heta _{1} - \sin^{2} heta _{1} \end{matrix} ight] \left . \left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}} = \left[\begin{matrix} 1 & 0 & 0 \ 0 & 1& 0 \0 & 0 & 1 \end{matrix} ight] \left . \left\{\begin{matrix} 250 \mu m/m \ -1500 \mu m/m \ (1000 \mu rad)/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}} =\left . \left\{\begin{matrix} 250 \mu m/m \ -1500 \mu m/m \ 1000 \mu rad \end{matrix} ight\} ight|^{ply 1}_{z = z_{0}}[/latex]

Bottom interface:

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}} = \left[\begin{matrix} \cos ^{2} heta _{1} & \sin ^{2} heta _{1} & 2 \cos heta _{1} \sin heta _{1} \ \sin ^{2} heta _{1} & \cos ^{2} heta _{1} &-2 \cos heta _{1} \sin heta _{1}\ - \cos heta _{1} \sin heta _{1} & \cos heta _{1} \sin heta _{1} & \cos^{2} heta _{1} - \sin^{2} heta _{1} \end{matrix} ight]\left . \left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}} = \left[\begin{matrix} 1 & 0 & 0 \ 0 & 1& 0 \0 & 0 & 1 \end{matrix} ight] \left . \left\{\begin{matrix} 188 \mu m/m \ -1450\mu m/m \ (975 \mu rad)/2 \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12} \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}} =\left . \left\{\begin{matrix} 188 \mu m/m \ -1450\mu m/m \ 975 \mu rad \end{matrix} ight\} ight|^{ply 1}_{z = z_{1}}[/latex]

Ply 2. In this case, [latex]θ_{2}=30°[/latex] and consequently the description of strain in the x–y and 1–2 coordinate systems differs substantially. Applying Eq. (44) of Chap. 2, we have:
Top interface:

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} \cos ^{2} heta _{2} & \sin ^{2} heta _{2} & 2 \cos heta _{2} \sin heta _{2} \ \sin ^{2} heta _{2} & \cos ^{2} heta _{2} &-2 \cos heta _{2} \sin heta _{2}\ - \cos heta _{2} \sin heta _{2} & \cos heta _{2} \sin heta _{2} & \cos^{2} heta _{2} - \sin^{2} heta _{2} \end{matrix} ight] \left . \left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} \cos ^{2} (30°)& \sin ^{2} (30°) & 2 \cos (30°) \sin (30°) \ \sin ^{2} (30°) & \cos ^{2} (30°) &-2 \cos (30°) \sin (30°)\ - \cos (30°) \sin (30°) & \cos (30°) \sin (30°) & \cos^{2} (30°) - \sin^{2} (30°) \end{matrix} ight]\ ×\left . \left\{\begin{matrix} 188 μm/m \ -1450 μm/m \ (975 μrad )/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} 0.750 & 0.250 & 0.866 \ 0.250 & 0.750 &-0.866 \ -0.433 & 0.433 & 0.500 \end{matrix} ight] \left . \left\{\begin{matrix} 188 μm/m \ -1450 μm/m \ (975 μrad )/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} =\left . \left\{\begin{matrix} 200 μm/m \ -1463 μm/m \ -931 μrad \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

Bottom interface:

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}} = \left[\begin{matrix} \cos ^{2} heta _{2} & \sin ^{2} heta _{2} & 2 \cos heta _{2} \sin heta _{2} \ \sin ^{2} heta _{2} & \cos ^{2} heta _{2} &-2 \cos heta _{2} \sin heta _{2}\ - \cos heta _{2} \sin heta _{2} & \cos heta _{2} \sin heta _{2} & \cos^{2} heta _{2} - \sin^{2} heta _{2} \end{matrix} ight] \left . \left\{\begin{matrix} \varepsilon _{xx} \ \varepsilon _{yy} \ \gamma _{xy}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}} = \left[\begin{matrix} \cos ^{2} (30°)& \sin ^{2} (30°) & 2 \cos (30°) \sin (30°) \ \sin ^{2} (30°) & \cos ^{2} (30°) &-2 \cos (30°) \sin (30°)\ - \cos (30°) \sin (30°) & \cos (30°) \sin (30°) & \cos^{2} (30°) - \sin^{2} (30°) \end{matrix} ight] \ ×\left . \left\{\begin{matrix} 125 μm/m \ -1400 μm/m \ (950 μrad )/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12}/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}} = \left[\begin{matrix} 0.750 & 0.250 & 0.866 \ 0.250 & 0.750 &-0.866 \ -0.433 & 0.433 & 0.500 \end{matrix} ight] \left . \left\{\begin{matrix} 125 μm/m \ -1400 μm/m \ (950 μrad )/2 \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}}[/latex]

[latex]\left . \left\{\begin{matrix} \varepsilon _{11} \ \varepsilon _{22} \ \gamma _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}} =\left . \left\{\begin{matrix} 155 μm/m \ -1430 μm/m \ -846 μrad \end{matrix} ight\} ight|^{ply 2}_{z = z_{2}}[/latex]

Ply strains referenced to the local 1–2 coordinate systems at all interface locations are summarized in Table 2 and plotted in Fig. 14. Comparing Figs. 12 and 13, it is apparent that the through-thickness strain distributions no longer appear linear or continuous when referenced to the 1–2 coordinate system. This is of course illusionary, in the sense that strains appear to be dis-continuous only because the coordinate system used to describe the through-thickness strain is varied from one ply to the next.

Table 2 Ply Interface Strains in a [latex][0/-30/90/30]_{s}[/latex] Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1
Ply number	z-coordinate (mm)	[latex]ε_{11} (μm/m)[/latex]	[latex]ε_{22} (μm/m)[/latex]	[latex] γ_{12} μrad[/latex]
Ply 1	-0.500	250	-1500	1000
Ply 1	-0.375	188	-1450	975
Ply 2	-0.375	200	-1463	-931
Ply 2	-0.250	155	-1430	-846
Ply 3	-0.250	-1400	125	-950
Ply 3	-0.125	-1350	63	-925
Ply 4	-0.125	-691	-596	1686
Ply 4	0.000	-715	-585	1576
Ply 5	0.000	-715	-585	1576
Ply 5	0.125	-738	-574	1466
Ply 6	0.125	-1250	-62	-875
Ply 6	0.250	-1200	-125	-850
Ply 7	0.250	-26	-1299	-506
Ply 7	0.375	-71	-1267	-421
Ply 8	0.375	-188	-1150	825
Ply 8	0.500	-250	-1100	800
Strains are referenced to the 1–2 coordinate system local to individual plies.

Stress Calculations. Because strains are now known at all ply interface positions, we can calculate stresses at these locations using Eq. (30) of Chap. 5, with ΔT=ΔM=0. During these calculations, we will require the trans-formed reduced stiffness matrix for each ply. Using graphite–epoxy material properties from Table 2 of Chap. 3 and Eqs. (11) and (31) of Chap. 5, we find:

[latex]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} ight\} }dz[/latex] (30)

[latex]Q_{11} = \frac{E_{11}^{2}}{E_{11} ν_{12}^{2} E_{22}}[/latex] [latex]Q_{12} = Q_{21}= \frac{ν_{12}E_{11}E_{22}}{E_{11} ν_{12}^{2} E_{22}}\ Q_{22}= \frac{E_{11}E_{22}}{E_{11} ν_{12}^{2} E_{22}}[/latex] [latex] Q_{16} = G_{12}[/latex] (11)

[latex]\overline{Q}_{11} = Q_{11} cos^{4} heta + 2(Q_{12} + 2Q_{66})cos^{2} heta sin^{2} heta + Q_{22} sin^{4} heta \\overline{Q}_{12} =\overline{Q}_{21} = Q_{12} ( cos^{4} heta + sin^{4} heta) +(Q_{11} +Q_{22} -4 Q_{66})cos^{2} heta sin^{2} heta \ \overline{Q}_{16} =\overline{Q}_{61} = (Q_{11}- Q_{12} -2Q_{66})cos^{3} heta sin heta- (Q_{22}-Q_{12}-2Q_{66})cos heta sin^{3} heta\ \overline{Q}_{22} = Q_{11}sin^{4} heta + 2(Q_{12} + 2Q_{66})cos^{2} heta sin^{2} heta + Q_{22} cos^{4} heta \ \overline{Q}_{26} =\overline{Q}_{62} = (Q_{11} - Q_{12} - 2Q_{66})cos heta sin^{3} heta -(Q_{22} -Q_{12} - 2Q_{66})cos^{3} heta sin heta \ \overline{Q}_{66} = (Q_{11} +Q_{22} -2Q_{12} -2Q_{66}) cos^{2} heta sin^{2} heta +Q_{66} ( cos^{4} heta + sin^{4} heta) [/latex] (31)

Table 2 ASTM Test Standards Related to Failure of Polymeric Composites (see also Table 1)
Designation	Title
D3479	Standard Test Method for Tension–Tension Fatigue of Polymer Matrix Composite Materials
D5766	Standard Test Method for Open Hole Tensile Strength of Polymer Matrix Composite Laminates
D2344	Standard Test Method for Apparent Interlaminar Shear Strength of Parallel Fiber Composites by Short-Beam Method
D5528	Standard Test Method for Mode I Interlaminar Fracture Toughness of Unidirectional Fiber-Reinforced Polymer Matrix Composites
E1922	Standard Test Method for Translaminar Fracture Toughness of Laminated Polymer Matrix Composite Materials
D2290	Standard Test Method for Apparent Tensile Strength of Ring or Tubular Plastics and Reinforced Plastics by Split Disk Method

Table 1 ASTM Test Standards for Determining Elastic Moduli of Polymeric Composites and Related Standards
Designation	Title
D3039	Standard Test Method for Tensile Properties of Polymer Matrix Composite Materials
D5450	Standard Test Method for Transverse Tensile Properties of Hoop Wound Polymer Matrix Composite Cylinders
D695	Standard Test Method for Compressive Properties of Rigid Plastics
D3410	Standard Test Method for Compressive Properties of Polymer Matrix Composite Materials with Unsupported Gage Section by Shear Loading
D5467	Standard Test Method for Compressive Properties of Unidirectional Polymer Matrix Composites Using a Sandwich Beam
D5449	Standard Test Method for Transverse Compressive Properties of Hoop Wound Polymer Matrix Composite Cylinders
D3518	Standard Practice for In-Plane Shear Response of Polymer Matrix Composite Materials by Tensile Test of a +45° Laminate
D5379	Standard Test Method for Shear Properties of Composite Materials by the V-Notched Beam Method
D4255	Standard Guide for Testing In-Plane Shear Properties of Composite Laminates
D5448	Standard Test Method for In-Plane Shear Properties of Hoop Wound Polymer Matrix Composite Cylinders
Related standards
D5687	Standard Guide for Preparation of Flat Composite Panels with Processing Guidelines for Specimen Preparation
D638	Standard Test Method for Tensile Properties of Plastics
D882	Standard Test Method for Tensile Properties of Thin Plastic Sheeting
D4018	Standard Test Methods for Properties of Continuous Filament Carbon and Graphite Fiber Tows
D2343	Standard Test Method for Tensile Properties of Glass Fiber Strands, Yarns, and Rovings Used in Reinforced Plastics

For 0° plies:

[latex] [\bar{Q} ]_{0° plies} = \left[\begin{matrix} 170.9 imes 10^{9} &3.016 imes 10^{9} & 0 \ 3.016 imes 10^{9} & 10.05 imes 10^{9} & 0 \ 0 & 0 & 13.00 imes 10^{9} \end{matrix} ight] (Pa) [/latex]

For 30° plies:

[latex] [\bar{Q} ]_{30° plies} = \left[\begin{matrix} 170.6 imes 10^{9} &26.06 imes 10^{9} & 48.3 imes 10^{9} \ 26.06 imes 10^{9} & 27.22 imes 10^{9} & 21.52 imes 10^{9}\ 48.3 imes 10^{9} & 21.52 imes 10^{9} & 36.05 imes 10^{9} \end{matrix} ight] (Pa) [/latex]

For 90° plies:

[latex] [\bar{Q} ]_{90° plies} = \left[\begin{matrix} 10.05 imes 10^{9} &3.016 imes 10^{9} & 0 \ 3.016 imes 10^{9} & 170.9 imes 10^{9} & 0 \ 0 & 0 & 13.00 imes 10^{9} \end{matrix} ight] (Pa) [/latex]

For -30° plies:

[latex] [\bar{Q} ]_{30° plies} = \left[\begin{matrix} 170.6 imes 10^{9} &26.06 imes 10^{9} & -48.3 imes 10^{9} \ 26.06 imes 10^{9} & 27.22 imes 10^{9} & -21.52 imes 10^{9}\- 48.3 imes 10^{9} & -21.52 imes 10^{9} & 36.05 imes 10^{9} \end{matrix} ight] (Pa) [/latex]

Stresses present at the outer surface of ply 1 (i.e., strains present at [latex]z_{0}=-0.000500 m[/latex]) can now be calculated:

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{0}}^{ply 1} = \left .\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} ight] ight|_{z=z_{0}}^{ply 1} \left .\left\{\begin{matrix} ε _{xx} \ ε_{yy} \γ_{xy} \end{matrix} ight\} ight|_{z=z_{0}}[/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{0}}^{ply 1} = \left[\begin{matrix}170.9 imes 10^{9} & 3.016 imes 10^{9} &0 \ 3.016 imes 10^{9}& 10.05 imes 10^{9} & 0 \ & 0 & 13.00 imes 10^{9}\end{matrix} ight] \left\{\begin{matrix} 250 \mu m/m \ -1500 \mu m/m \1000 \mu rad \end{matrix} ight\} [/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{0}}^{ply 1} = \left\{\begin{matrix}38.2 MPa\ -14.3 MPa \13 MPa \end{matrix} ight\} [/latex]

To calculate stresses at the interface between plies 1 and 2 (i.e., at [latex]z_{1} =-0.000375 m[/latex]), we must specify whether we are interested in the stresses within ply 1 or ply 2. That is, according to our idealized model, a ply interface is treated as a plane of discontinuity in material properties. Ply 1 ‘‘ends’’ at [latex]z =z_{1}^{(-)}[/latex], whereas ply 2 ‘‘begins’’ at [latex]z=z_{1}^{(+)}[/latex]. Hence, the stresses within ply 1 at [latex]z =z_{1}^{(-) }[/latex] are:

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 1} = \left .\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} ight] ight|_{z=z_{1}}^{ply 1} \left .\left\{\begin{matrix} ε _{xx} \ ε_{yy} \γ_{xy} \end{matrix} ight\} ight|_{z=z_{1}}[/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 1} = \left[\begin{matrix}170.9 imes 10^{9} & 3.016 imes 10^{9} &0 \ 3.016 imes 10^{9}& 10.05 imes 10^{9} & 0 \ & 0 & 13.00 imes 10^{9}\end{matrix} ight] \left\{\begin{matrix} 188 \mu m/m \ -1450 \mu m/m \975 \mu rad \end{matrix} ight\} [/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 1} = \left\{\begin{matrix}27.8 MPa\ -14.0 MPa \12.7 MPa \end{matrix} ight\} [/latex]

The stresses within ply 2 (a 30° ply) at [latex]z=z_{1}^{(+)}[/latex] are:

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 2} = \left .\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} ight] ight|_{z=z_{1}}^{ply 2} \left .\left\{\begin{matrix} ε _{xx} \ ε_{yy} \γ_{xy} \end{matrix} ight\} ight|_{z=z_{1}}[/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 1} = \left[\begin{matrix}170.6 imes 10^{9} & 26.96 imes 10^{9} &48.3 imes 10^{9}\ 26.06 imes 10^{9}& 27.22 imes 10^{9} & 21.52 imes 10^{9} \ 48.3 imes 10^{9}& 21.52 imes 10^{9} & 36.05 imes 10^{9}\end{matrix} ight] \left\{\begin{matrix} 188 \mu m/m \ -1450 \mu m/m \975 \mu rad \end{matrix} ight\} [/latex]

[latex]\left .\left\{\begin{matrix} \sigma _{xx} \ \sigma_{yy} \ au _{xy} \end{matrix} ight\} ight|_{z=z_{1}}^{ply 2} = \left\{\begin{matrix}29.5 MPa\ -13.6 MPa \13.0 MPa \end{matrix} ight\} [/latex]

Stresses are calculated at all remaining ply interfaces in exactly the same fashion. Ply interface stresses are summarized in Table 3 and are plotted in Fig. 15. Obviously, stresses are not linearly distributed through the thickness of the laminate, even when referenced to the global x–y coordinate system. In general, all stress components exhibit a sudden discontinuous change at all ply interface positions. The abrupt change in stresses at ply interfaces is due to the discontinuous change in the [latex] [\overline{Q}] [/latex] matrix from one ply to the next. In turn, the discontinuous change in [latex] [\overline{Q}][/latex] occurs because the fiber angle (in general) changes from one ply to the next. Indeed, in this example problem, the same fiber angle occurs in only two adjacent plies (namely, plies 4 and 5, both of which have a fiber angle of -30°), and inspection of Fig. 15 shows that the interface between plies 4 and 5 is the only interface for which the stresses do not change abruptly.

It has been mentioned that the linear strain distributions shown in Fig. 13 would be the same for any thin plate, regardless of the material the plate is made of. The same statement cannot be made for stress distributions. In general, through-thickness stress distributions for isotropic plates (e.g., an isotropic aluminum plate) are linear and continuous, unless high nonlinear stresses occur, in which case the stress distribution may not be linear but will nevertheless be continuous. In contrast, the stress distributions in laminated composite plates are usually discontinuous. The only conditions under which a linear and continuous stress distribution is encountered is when: (a) the laminate is subjected to elastic stress/strain levels, and (b) when the [latex] [\overline{Q}][/latex] matrix does not vary from one ply to the next (i.e., for unidirectional laminates in which the fiber angle does not vary from one ply to the next).

Knowledge of ply stresses referenced to the local 1–2 coordinate system (defined by the fiber angle within each ply) is often required. Transformation of the stress tensor from one coordinate system to another was reviewed in Chap. 2 and, in particular, stresses can be rotated from the x–y coordinate system to the 1–2 coordinate system using Eq. (20) of Chap. 2. Typically, stresses are calculated at both the ‘‘top’’ and ‘‘bottom’’ interfaces for all plies. For example, rotation of the ply stresses that exist within ply 2 at the interface between plies 1 and 2 (i.e., at [latex]z=z_{1} = -0.375 mm[/latex]) proceeds as follows:

Table 3 Ply Interface Stresses in a [latex][0/30/90/30]_{s}[/latex] Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1
Ply number	z-coordinate (mm)	[latex]\sigma _{xx} (MPa)[/latex]	[latex]\sigma _{yy} (MPa)[/latex]	[latex] au _{xy} (MPa)[/latex]	[latex]\Theta (MPa)[/latex]	[latex]\Phi (MPa)^{2}[/latex]
Ply 1	-0.500	38.2	-14.3	13.0	23.9	-715
Ply 1	-0.375	27.8	-14.0	12.7	13.8	-550
Ply 2	-0.375	29.3	-13.6	13.0	15.7	-567
Ply 2	-0.250	22.7	-14.4	10.1	8.3	-429
Ply 3	-0.250	-2.97	-239.	12.4	-242	556
Ply 3	-0.125	-3.44	-231	12.0	-234	651
Ply 4	-0.125	-73.0	-55.0	59.4	-128	487
Ply 4	0.000	-77.2	-54.7	60.4	-132	575
Ply 5	0.000	-77.2	-54.7	60.4	-132	575
Ply 5	0.125	-81.4	-54.5	61.4	-136	666
Ply 6	0.125	-4.40	-214	11.4	-218	812
Ply 6	0.250	-4.90	-205	11.0	-210	884
Ply 7	0.250	-3.82	-17.6	-1.20	-21.4	65.8
Ply 7	0.375	-10.4	-18.4	-4.03	-28.8	175
Ply 8	0.375	-35.5	-12.1	10.7	-47.6	315
Ply 8	0.500	-46.0	-11.8	10.4	-57.8	435
Stresses are referenced to the x–y coordinate system.

[latex]\left\{\begin{matrix} \sigma _{x^{\prime }x^{\prime }} \ \sigma _{y^{\prime } y^{\prime }} \ au _{x^{\prime }y^{\prime }} \end{matrix} ight\}= \left[\begin{matrix} \cos ^{2}( heta) & \sin ^{2}( heta) & 2 \cos ( heta) \sin( heta) \ \sin ^{2} ( heta) & \cos ^{2} ( heta)&-2 \cos ( heta) \sin ( heta) \ - \cos ( heta) \sin ( heta) & \cos ( heta) \sin ( heta)& \cos^{2} ( heta)- \sin^{2}( heta) \end{matrix} ight]\ imes \left\{\begin{matrix} \sigma _{xx} \ \sigma _{yy} \ au _{xy} \end{matrix} ight\}[/latex] (20)

[latex]\left . \left\{\begin{matrix} \sigma _{11} \ \sigma_{22} \ au _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} \cos ^{2} heta _{2} & \sin ^{2} heta _{2} & 2 \cos heta _{2} \sin heta _{2} \ \sin ^{2} heta _{2} & \cos ^{2} heta _{2} &-2 \cos heta _{2} \sin heta _{2}\ - \cos heta _{2} \sin heta _{2} & \cos heta _{2} \sin heta _{2} & \cos^{2} heta _{2} - \sin^{2} heta _{2} \end{matrix} ight]\left . \left\{\begin{matrix} \sigma _{xx} \ \sigma _{yy} \ au _{xy} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \sigma _{11} \ \sigma_{22} \ au _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} \cos ^{2} (30°)& \sin ^{2} (30°) & 2 \cos (30°) \sin (30°) \ \sin ^{2} (30°) & \cos ^{2} (30°) &-2 \cos (30°) \sin (30°)\ - \cos (30°) \sin (30°) & \cos (30°) \sin (30°) & \cos^{2} (30°) - \sin^{2} (30°) \end{matrix} ight] × \left . \left\{\begin{matrix} 29.3 MPa \ -13.6 MPa \ 13.0 MPa \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \sigma _{11} \ \sigma_{22} \ au _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left[\begin{matrix} 0.750& 0.250 & 0.866 \ 0.750& 0.250 &-0.866\ - 0.433& 0.433 &0.500 \end{matrix} ight] \ \left . \left\{\begin{matrix} 29.3 MPa \ -13.6 MPa \ 13.0 MPa \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

[latex]\left . \left\{\begin{matrix} \sigma _{11} \ \sigma_{22} \ au _{12} \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}} = \left . \left\{\begin{matrix} 29.8 MPa \ -14.1 MPa \ -12.1 MPa \end{matrix} ight\} ight|^{ply 2}_{z = z_{1}}[/latex]

Ply interface stresses referenced to local 1–2 coordinate systems are summarized in Table 4 and plotted in Fig. 16. Once again, stresses are not linearly distributed through the thickness of the laminate, and instead exhibit a sudden discontinuous change at all ply interface positions.
Stress invariants can be used to confirm that the ply stresses referenced to the x–y coordinate system and listed in Table 3 are equivalent to the ply stresses referenced to the 1–2 coordinate system, as listed in Table 4. The concept of ‘‘stress invariants’’ was discussed in Chap. 2. The stress invariants for the case of plane stress are given by Eq. (22) of Chap. 2, repeated here for convenience:

Table 4 Ply Interface Stresses in a [latex][0/-30/90/30]_{s}[/latex] Graphite-Epoxy Laminate Subjected to the Midplane Strains and Curvatures Discussed in Sample Problem 1
Ply number	z-coordinate (mm)	[latex]\sigma _{11} (MPa)[/latex]	[latex]\sigma _{yy} (MPa)[/latex]	[latex] au _{xy} (MPa)[/latex]	[latex]\Theta (MPa)[/latex]	[latex]\Phi (MPa)^{2}[/latex]
Ply 1	-0.500	38.2	-14.3	13.0	23.9	-715
Ply 1	-0.375	27.8	-14.0	12.7	13.8	-550
Ply 2	-0.375	29.8	-14.1	-12.1	15.7	-567
Ply 2	-0.250	22.2	-13.9	-11.0	8.3	-429
Ply 3	-0.250	-239	-2.97	-12.4	-242	556
Ply 3	-0.125	-231	-3.44	-12.0	-234	651
Ply 4	-0.125	-120	-8.08	21.9	-128	487
Ply 4	0.000	-124	-8.04	20.5	-132	575
Ply 5	0.000	-124	-8.04	20.5	-132	575
Ply 5	0.125	-128	-8.00	19.1	-136	666
Ply 6	0.125	-214	-4.40	-11.4	-218	812
Ply 6	0.250	-205	-4.88	-11.0	-210	884
Ply 7	0.250	-8.31	-13.1	-6.58	-21.4	65.8
Ply 7	0.375	-15.9	-13.0	-5.47	-28.8	175
Ply 8	0.375	-35.5	-12.1	10.7	-47.6	315
Ply 8	0.500	-46.0	-11.8	10.4	-57.8	435
Stresses are referenced to the 1–2 coordinate system.

First stress invariant =[latex]\Theta = \sigma _{xx} + \sigma _{yy} [/latex]

Second stress invariant = [latex] \Phi = \sigma _{xx} \sigma _{yy} - au _{xy}^{2} [/latex] (repeated) (2.22)

Third stress invariant = [latex] \Psi = 0[/latex]

For plane stress conditions, the third stress invariant always equals zero, and so [latex] \Psi[/latex] cannot be used to evaluate whether two plane stress states are equivalent. The first and second stress invariants, [latex]\Theta[/latex] and [latex] \Phi[/latex], respectively, have been calculated using the ply stress components referenced to both the x–y and 1–2 coordinate systems. Values calculated for [latex] \Theta[/latex] and [latex] \Phi[/latex] are included in the last two columns of both Tables 3 and 4. Identical values are obtained in all cases, indicating the equivalence of the ply stress states described using the two different coordinate systems.

Question 6.S.P.3: Assume that the panel considered in Sample Problem 1 is actu...

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