Question 6.6: A [30/0/90]T graphite-epoxy laminate is cured at 175°C and t...

Note that this is the same laminate considered in Sample Problem 4. A side view of the laminate appears in Fig. 17.
(a) Midplane strains and curvatures. The laminate has experienced a change in temperature ΔT = (20–175) = -155°C and, consequently, is sub-jected thermal stress and moment resultants. However, no external loads are applied and there has been no change in moisture content; therefore, the stress and moment resultants and the moisture stress and moment results are zero:

he effective thermal expansion coefficients for each ply are calculated using Eq. (25) of Chap. 5, repeated here for convenience:

α_{xy} = 2 \cos(θ) \sin(θ) (α_{11}  – α_{22} )      (5.25)

From Table 3 of Chap. 3, the thermal expansion coefficients for graphite-epoxy (relative to the 1–2 coordinate system) are a_{11} = -0.9  μm/m °C and a_{22} = 27  μm/m °C . Therefore:

For ply 1 (the 30° ply):

α_{xx}^{(1)} = (-0.9  μm/m -°C ) \cos^{2} (30°)+ ( 27  μm/m -°C) \sin^{2} (30°)
=  6.08 μm/m -°C

For ply 2 (the 0° ply):

α_{xx}^{(2)} = (-0.9  μm/m -°C ) \cos^{2} (0°)+ ( 27  μm/m -°C) \sin^{2} (0°)
=  -0.9 μm/m -°C

For ply 3 (the 90° ply):

α_{xx}^{(3)} = (-0.9  μm/m -°C )\cos^{2} (90°)+ ( 27  μm/m -°C) \sin^{2} (90°)
=  27.0 μm/m -°C

Both the ply interface positions as well as the \overline{Q_{ij}} matrices for each ply were calculated as a part of Example Problem 4. Hence, we now have all the information needed to calculate the thermal stress and moment resultants, using Eqs. (41a)–(41f ). For example, Eq. (41a) is evaluated as follows:

N^{T}_{xx} = \Delta T \sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{11} \alpha _{xx}+ \overline{Q}_{12} \alpha _{yy} + \overline{Q}_{16} \alpha _{xy}\right]_{k}[z_{k} – z_{k-1}] \right\} } (41 a)
N^{T}_{yy} \equiv \Delta T\sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{12} \alpha _{xx}+ \overline{Q}_{22} \alpha _{yy} + \overline{Q}_{26} \alpha _{xy}\right]_{k}[z_{k} – z_{k-1}] \right\} } (41 b)
N^{T}_{xy} \equiv \Delta T \sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{16} \alpha _{xx}+ \overline{Q}_{26} \alpha _{yy} + \overline{Q}_{66} \alpha _{xy}\right]_{k}[z_{k} – z_{k-1}] \right\} } (41 c)

N^{T}_{xx} \equiv \frac{\Delta T}{2} \sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{11} \alpha _{xx}+ \overline{Q}_{12} \alpha _{yy} + \overline{Q}_{16} \alpha _{xy}\right]_{k}[z_{k}^{2} – z_{k-1}^{2}] \right\} } (41 d)
N^{T}_{yy} \equiv\frac{\Delta T}{2}\sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{12} \alpha _{xx}+ \overline{Q}_{22} \alpha _{yy} + \overline{Q}_{26} \alpha _{xy}\right]_{k}[z_{k}^{2} – z_{k-1}^{2}] \right\} } (41 e)
N^{T}_{xy} \equiv \frac{\Delta T}{2} \sum\limits_{k = 1}^{n}{\left\{\left[\overline{Q}_{16} \alpha _{xx}+ \overline{Q}_{26} \alpha _{yy} + \overline{Q}_{66} \alpha _{xy}\right]_{k}[z_{k}^{2} – z_{k-1}^{2}] \right\} } (41 f)

The remaining thermal stress and moment resultants are calculated in similar fashion, eventually resulting in:

We can now calculate midplane strains and curvature using Eq. (45), 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_{xx}^{T} + N_{xx}^{M}\\ N_{yy}+ N_{yy}^{T} + N_{yy}^{M} \\ N_{xy}+ N_{xy}^{T} + N_{xy}^{M} \\ M_{xx} +M_{xx}^{T} + M_{xx}^{M} \\M_{yy}+M_{yy}^{T} + M_{yy}^{M} \\M_{xy} +M_{xy}^{T} + M_{xy}^{M} \end{matrix} \right\}     (45)

(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_{o}=-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)

Strains calculated at the remaining ply interface positions are summa-rized in Table 7.

(c) Ply stresses relative to the x–y coordinate system. Ply stresses may now be calculated using Eq. (30) of Chap. 5, with ΔM=0. The stresses present at the outer surface of ply 1 are (i.e., at z=z_{0}):

\left\{\begin{matrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau _{xy} \end{matrix} \right\} = \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} \varepsilon _{xx} – \Delta T \alpha _{xx}-\Delta M \beta _{xx} \\ \varepsilon _{yy}- \Delta T \alpha _{yy}-\Delta M \beta _{yy} \\ \gamma _{xy} – \Delta T \alpha _{xy} -\Delta M \beta _{xy}\end{matrix} \right\} (30)

Stresses calculated at the remaining plies and ply interface positions are summarized in Table 8.