Question 6.S.P.8: Determine the effective extensional and flexural moduli, the...
Determine the effective extensional and flexural moduli, thermal expansion coefficients, and moisture expansion coefficients for a [30/0/90]_{s} 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) |
As described in this section, effective moduli are calculated using various elements of the [abd] matrix. A six-ply symmetrical laminate is considered in this problem. The total laminate thickness is t=6(0.000125
m)=0.000750 m. Using methods discussed in Sec. 6, the [ABD] matrix is determined to be:
Because the laminate is symmetrical, all elements of the B_{ij} matrix are zero, as expected. We obtain the [abd] by inverting the [ABD] numerically:
[abd] = \left[\begin{matrix} 16.0 \times 10^{-9} & -1.23 \times 10^{-9} & -12.0 \times 10^{-9} &0 &0&0\\ -1.23 \times 10^{-9} & 20.0 \times 10^{-9}& -6.0 \times 10^{-9} &0 &0&0\\-12.0 \times 10^{-9} & -6.0 \times 10^{-9} & 75.8 \times 10^{-9} &0 &0&0 \\ 0 &0&0 & 3.51 \times 10^{-1} & -2.92 \times 10^{-2} & -3.92 \times 10^{-1} \\ 0 &0&0 & -2.92 \times 10^{-2}& 1.408& -6.96 \times 10^{-1} \\ 0 &0&0 & -3.92 \times 10^{-1} & -6.96 \times 10^{-1} & 1.79\end{matrix} \right]The effective extensional moduli of the laminate can now be calculated using Eqs. (63)–(70b):
\overline{E}_{xx}^{ex} =\frac{\overline{\sigma _{xx}}}{ν_{xx}^{o}} = \frac{(N_{xx}/t)}{(a_{11}N_{xx})} = \frac{1}{ta_{11}} (63)
\overline{ν}_{xy}^{ex} =\frac{ν _{yy}^{o}}{ν _{xx}^{o}} = \frac{-a_{12}N_{xx}}{a_{11}N_{xx}} = \frac{-a_{12}}{a_{11}} (64)
\overline{\eta }_{xx,xy}^{ex} =\frac{\gamma _{xy}^{o}}{ν_{xx}^{o}} = \frac{-a_{16}N_{xx}}{a_{11}N_{xx}} = \frac{a_{16}}{a_{11}} (65)
ν_{xx}^{o}= a_{12} N_{yy} (66a)
ν _{yy}^{o}= a_{22} N_{yy} (66b)
\gamma _{xy}^{o} = a_{26 } N_{yy}(66c)
\overline{E}_{yy}^{ex} = \frac{1}{ta_{22}} (67 a)
\overline{ν}_{yx}^{ex} = \frac{-a_{12}}{a_{22}} (67b)
\overline{\eta }_{yy,xy}^{ex} = \frac{a_{26}}{a_{22}} (67c)
ε_{xx}^{o}= a_{16} N_{xy} (68a)
ε _{yy}^{o}= a_{26} N_{xy} (68b)
\gamma _{xy}^{o} = a_{66 } N_{xy}(68c)
\overline{G}_{xy}^{ex} =\frac{\overline{\tau }_{xy}}{\gamma _{xy}^{o}} = \frac{(N_{xy}/t)}{a_{66}N_{xy}}= \frac{1}{ta_{66}} (69)
\overline{\eta }_{xy,xx}^{ex} = \frac{ν_{xx}^{o}}{\gamma _{xy}^{o}}= \frac{a_{16}N_{xx}}{a_{66}N_{xx}}= \frac{a_{16}}{a_{66}} (70 a)
\overline{\eta }_{xy,yy}^{ex} = \frac{ν _{yy}^{o}}{\gamma _{xy}^{o}}=\frac{a_{26}}{a_{66}} (70 b)
\overline{E}_{xx}^{ex} = \frac{1}{ta_{11}} = \frac{1}{( 0.000750) (16.0 × 10^{-9} )} =83.3 GPa
\overline{E}_{yy}^{ex} = \frac{1}{ta_{22}} = \frac{1}{( 0.000750) (20 × 10^{-9} )} =66.7 GPa
\overline{G}_{xy}^{ex} = \frac{1}{ta_{66}} = \frac{1}{( 0.000750) (75.8 × 10^{-9} )} =17.6 GPa
\overline{ν}_{xy}^{ex} = \frac{-a_{12}}{a_{11}} = \frac{-(-1.23 × 10^{-9} )}{16.0 × 10^{-9} } =0.077
\overline{ν}_{yx}^{ex} = \frac{-a_{12}}{a_{22}} = \frac{-(-1.23 × 10^{-9} )}{20 × 10^{-9} } =0.061
\overline{\eta }_{xx,xy}^{ex} = \frac{a_{16}}{a_{11}} = \frac{-12 × 10^{-9} }{16× 10^{-9} } =-0.75
\overline{\eta }_{yy,xy}^{ex} = \frac{a_{26}}{a_{22}} = \frac{-6.0 × 10^{-9} }{20 × 10^{-9} } =-0.30
\overline{\eta }_{xy,xx}^{ex} = \frac{a_{16}}{a_{66}} = \frac{-12.0 × 10^{-9} }{75.8 × 10^{-9} } =-0.16
\overline{\eta }_{xy,yy}^{ex} = \frac{a_{26}}{a_{66}} = \frac{-6.0 × 10^{-9} }{75.8 × 10^{-9} } =-0.079
Effective flexural properties are found to be:
\overline{E}_{xx}^{fl} = \frac{12}{d_{11}t^{3}} = \frac{12}{( 0.351) (0.000750)^{3}} =81.0 GPa\overline{E}_{yy}^{fl} = \frac{12}{d_{22}t^{3}} = \frac{12}{(1.408) (0.000750)^{3}} =20.2 GPa
\overline{ν}_{xy}^{fl} = \frac{-d_{12}}{d_{11}} = \frac{2.92 × 10^{-2} }{0.351} =0.083
\overline{ν}_{yx}^{fl} = \frac{-d_{12}}{d_{22}} = \frac{2.92 × 10^{-2}}{1.408} =0.021
\overline{\eta }_{xx,xy}^{fl} = \frac{d_{16}}{d_{11}} = \frac{-0.392 }{0.351} =-1.12
\overline{\eta }_{yy,xy}^{fl} = \frac{d_{26}}{d_{22}} = \frac{-0.696}{1.408} =-0.49
Note that the values of extensional properties are quite different from anal-ogous flexural properties.
The thermal stress resultants associated with a given change in temper-ature must be determined in order to calculate the effective thermal expansion coefficients. Numerically speaking, any change in temperature can be used, but for present purposes, a unit change in temperature will be assumed (i.e., ΔT=1). Using (42a–(42f), the thermal stress resultants associated with ΔT=1 are:
N_{xx}^{M} = \Delta M \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{11}\beta _{xx} + \overline{Q}_{12}\beta _{yy} + \overline{Q}_{16}\beta _{xy} ]_{k}[z_{k} – z_{k-1}]\right\} } (42a)
N_{yy}^{M} \equiv \Delta M \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{12}\beta _{xx} + \overline{Q}_{22}\beta _{yy} + \overline{Q}_{26}\beta _{xy} ]_{k}[z_{k} – z_{k-1}]\right\} } (42b)
N_{xy}^{M} \equiv \Delta M \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{16}\beta _{xx} + \overline{Q}_{26}\beta _{yy} + \overline{Q}_{66}\beta _{xy} ]_{k}[z_{k} – z_{k-1}]\right\} } (42c)
M_{xx}^{M} = \frac{\Delta M}{2} \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{11}\beta _{xx} + \overline{Q}_{12}\beta _{yy} + \overline{Q}_{16}\beta _{xy} ]_{k}[z_{k}^{2} – z_{k-1}^{2}]\right\} } (42d)
M_{yy}^{M} = \frac{\Delta M}{2} \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{12}\beta _{xx} + \overline{Q}_{22}\beta _{yy} + \overline{Q}_{26}\beta _{xy} ]_{k}[z_{k}^{2} – z_{k-1}^{2}]\right\} } (42e)
M_{xy}^{M} = \frac{\Delta M}{2} \sum\limits_{k=1}^{n}{\left\{[\overline{Q}_{16}\beta _{xx} + \overline{Q}_{26}\beta _{yy} + \overline{Q}_{66}\beta _{xy} ]_{k}[z_{k}^{2} – z_{k-1}^{2}]\right\} } (42f)
N_{xx}^{T}|_{ΔT=1} = 52.3 N/m N_{yy}^{T}|_{ΔT=1} = 94.9 N/m N_{xy}^{T}|_{ΔT=1} = -36.9 N/m
The effective thermal expansion coefficient α_{xx} can now be calculated in accordance with Eq. (73a):
\overline{α}_{xx} = \frac{1}{ ΔT}[a_{11} N_{xx}^{T} + a_{12} N_{yy}^{T} + a_{16} N_{xy}^{T}]\\ \overline{α}_{yy} = \frac{1}{ ΔT}[a_{12} N_{xx}^{T} + a_{22} N_{yy}^{T} + a_{26} N_{xy}^{T}]\\ \overline{α}_{xy} = \frac{1}{ ΔT}[a_{16} N_{xx}^{T} + a_{26} N_{yy}^{T} + a_{66} N_{xy}^{T}]\overline{α}_{xx} = \frac{1}{ ΔT}[α_{11} N_{xx}^{T} + α_{12} N_{yy}^{T} + α_{16} N_{xy}^{T}]
\overline{α}_{xx} = \frac{1}{ (1)}[16.0 × 10^{-9}(52.3) -1.23 × 10^{-9} (94.9) \\-12.0 × 10^{-9} (-36.9) ]
\overline{α}_{xx} = 1.16 μm/m – °C
Using an equivalent procedure:
\overline{α}_{yy} = 2.06 μm/m – °C \overline{α}_{xy} = -4.00 μrad / °C
Finally, moisture stress resultants associated with a given change in moisture content must be determined in order to calculate the effective moisture expansion coefficients. Numerically speaking, any change moisture content can be used, but for present purposes, a unit change in content will be assumed (i.e., ΔM=1). Using Eq. (43), the moisture stress resultants associated with ΔM=1% are:
N_{xx} = A_{11} \varepsilon _{xx}^{o} + A_{12} \varepsilon _{yy}^{o}+ A_{16} \gamma _{xy}^{o} + B_{11} \kappa _{xx} + B_{12} \kappa _{yy} + B_{16} \kappa _{xy} – N_{xx}^{T} – N_{xx}^{M}(43a)
N_{xx}^{M}|_{ΔM=1} = 32,800 N/m N_{yy}^{M}|_{ΔM=1} = 33,800 N/m N_{xy}^{M}|_{ΔM=1} = -930 N/m
Applying Eqs. (76a) and (76b), we find:
\overline{\beta }_{xx} = \frac{1}{ ΔM}[a_{11} N_{xx}^{M} + a_{12} N_{yy}^{M} + a_{16} N_{xy}^{M}]\\\overline{\beta }_{yy} = \frac{1}{ ΔM}[a_{12} N_{xx}^{M} + a_{22} N_{yy}^{M} + a_{26} N_{xy}^{M}]\\\overline{\beta }_{xy} = \frac{1}{ ΔM}[a_{16} N_{xx}^{M} + a_{26} N_{yy}^{M} + a_{66} N_{xy}^{M}] (76a)
N_{xx}^{M} = \Delta M[A_{11} \overline{\beta }_{xx} +A_{12} \overline{\beta }_{yy} + A_{16} \overline{\beta }_{xy}]\\ N_{yy}^{M} = \Delta M[A_{12} \overline{\beta }_{xx} +A_{22} \overline{\beta }_{yy} + A_{26} \overline{\beta }_{xy}]\\N_{xy}^{M} = \Delta M[A_{16} \overline{\beta }_{xx} +A_{26} \overline{\beta }_{yy} + A_{66} \overline{\beta }_{xy}] (76b)
\overline{β}_{xx} = 494 μm/m \%M \overline{β}_{yy} = 643 μm/m \%M \overline{β}_{xy} = -667 μrad \%M