Question 14.7: Design a rectangular flat panel of length 1200 mm and width ...

Taking a factor of safety of 1.25, the force resultants acting on the panel are

N_{xx}=\frac{5000\times 1000\times 1.25}{800}=7812.5 \ N/mm \\ N_{yy}=\frac{3000\times 1000\times 1.25}{1200}=3125 \ N/mm \\ N_{xy}=\frac{1200\times 1000\times 1.25}{1200}=1250 \ N/mm

For these force resultants and the given material properties, carpet plots for laminate moduli and strengths are generated and given in Figures 14.13 through 14.18. The laminates in these figures consist of symmetric combinations of 0°, ±45°, and 90° plies at different proportions.

Now, let us choose the ply sequence based on stiffness and strength requirements. The maximum normal and in-plane shear strains that can be allowed are

(\varepsilon _{xx})_{max}=\frac{8}{1200}=6667\times 10^{-6} \\ (\varepsilon _{yy})_{max}=\frac{10}{800}=12,500\times 10^{-6} \\ (\gamma _{xy})_{max}=\frac{1\times \pi}{180}=17,453\times 10^{-6}

Let the laminate thickness be t. Then, the overall normal and in-plane shear stresses are

(\sigma_{xx})_{tot}=\frac{7812.5}{t} \\ (\sigma_{yy}) _{tot} =\frac{3125}{t} \\ (\tau_{xy})_{tot}=\frac{1250}{t}

The overall normal and in-plane shear moduli required are obtained as follows:

(E_{xx})_{tot}=\frac{7812.5/t}{6667\times 10^{-6}}=\frac{1.172\times 10^6}{t} \\ (E_{yy})_{tot}=\frac{3125/t}{12,500\times 10^{-6}}= \frac{0.25\times 10^6}{t} \\ (G_{xy})_{tot}=\frac{1250/t}{17,453\times 10^{-6}}=\frac{0.072\times 10^6}{t}

We can see that the required axial stiffness is 4.7 times the required transverse stiffness and the required transverse modulus is about 3.5 times the required inplane shear modulus. Obviously, it demands a high proportion of 0° plies. Let us consider a laminate consisting of plies at 0°, ±45°, and 90° at the ratio 80:10:10.
From Figures 14.13 through 14.15, we see that 80% 0° plies together with 10% ±45° and 10% 90° plies provide (E_{xx})_{tot}=108 \ GPa,(E_{yy})_{tot}=22 \ GPa , \ \text{ and } \ (G_{xy})_{tot}=7.8 \ GPa. The corresponding strengths are X^T\approx 680 \ MPa,Y^T=270 \ MPa, \text{ and } \ S\approx 110 \ MPa.

Now, based on stiffness requirements, the required laminate thickness is obtained as the maximum of the following:

\frac{1.172\times 10^6}{108\times 10^3}=10.9و \ \frac{0.25\times 10^6}{22\times 10^3} =11.4, \ \text{ and } \ \frac{0.072\times 10^6}{7.8\times 10^3} =9.2

And, based on strength requirements, the required laminate thickness is obtained as the maximum of the following:

\frac{7812.5}{680}=11.5, \ \frac{3125}{270}=11.6, \ \text{ and } \ \frac{1250}{110} =11.4

Let us then consider a laminate of total thickness 12 mm, in which individual thicknesses for different ply orientations are as follows: 0°→9.6 mm (64 plies), +45°→0.6 mm (4 plies), −45°→0.6 mm (4 plies), and 90°→1.2 mm (8 plies).
We have chosen the number of plies and their orientations. However, the ply sequence is yet to be chosen. Note that, for a given proportion of plies, ply sequence does not influence in-plane normal and shear deformations and lamina stresses, as long as symmetry is maintained. However, shear buckling characteristics can be improved by providing ±45° plies on the outer faces. Let us then choose the following ply sequence for our panel:

[45^\circ _2/-45^\circ _2/0^\circ _{16}/90^\circ _4/0^\circ _{16}]_s

For this ply sequence and the given material properties, the reduced stiffness matrix and transformed reduced stiffness matrices are given by

[Q]=[\bar{Q}]^{(0^\circ )}=\begin{bmatrix} 130,407.524&1630.094&0\\1630.094&6520.376&0\\0&0&5015.674 \end{bmatrix} \\ [\bar{Q}]^{(45^\circ )}=\begin{bmatrix} 40,062.696&30,031.348&30,971.787 \\ 30,031.348&40,062.696&30,971.787 \\ 30,971.787&30,971.787&33,416.928 \end{bmatrix} \\ [\bar{Q}]^{(-45^\circ )}=\begin{bmatrix} 40,062.696&30,031.348&-30,971.787 \\ 30,031.348&40,062.696&-30,971.787 \\ -30,971.787 &-30,971.787&33,416.928 \end{bmatrix} \\ [\bar{Q}]^{(90^\circ )}=\begin{bmatrix} 6520.376&1630.094&0 \\ 1630.094&130,407.524&0 \\ 0&0&5015.674 \end{bmatrix}

Then, for the chosen ply sequence, the laminate extensional stiffness and compliance matrices are obtained as

[A]=\begin{bmatrix} 1,307,811.917&53,642.633&0\\53,642.633&267,159.662&0\\0&0&94,269.59 \end{bmatrix} \\ [A^\ast ]= =\begin{bmatrix} 0.771&-0.155&0 \\ -0.155&3.774&0\\0&0&10.608 \end{bmatrix} \times 10^{-6}

Laminate middle surface strains are

\begin{Bmatrix} \varepsilon ^0_{xx} \\ \varepsilon ^0_{yy} \\ \gamma _{xy} \end{Bmatrix} =\begin{bmatrix} 0.771&-0.155&0 \\ -0.155&3.774&0 \\ 0&0&10.608 \end{bmatrix} \times \begin{Bmatrix} 7812.5 \\ 3125 \\ 1250 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 5539 \\ 10,583 \\ 13,260 \end{Bmatrix} \times 10^{-6}

The laminate curvatures are zero, which means

\begin{Bmatrix} \varepsilon _{xx} \\ \varepsilon _{yy} \\ \gamma _{xy} \end{Bmatrix} =\begin{Bmatrix} 5539 \\ 10,583 \\ 13,260 \end{Bmatrix} \times 10^{-6}

at all locations in the laminate thickness
Deformations are

Axial deformation = 1200\times 5539\times 10^{-6}=6.6 \ mm

Transverse deformation = 800\times 10,583\times 10^{-6}=8.5 \ mm

In-plane shear deformation = \frac{180\times 13,260\times 10^{-6}}{\pi}=0.76^\circ

Global stresses in the plies are

\begin{Bmatrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau_{xy} \end{Bmatrix}^{(0^\circ )} =\begin{bmatrix} 130,407.524&1630.094&0 \\ 1630.094&6520.376&0 \\ 0&0&5015.674 \end{bmatrix} \begin{Bmatrix} 5539 \\ 10,583 \\ 13,260 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix}739.6 \\ 78.0 \\ 66.5 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau_{xy} \end{Bmatrix}^{(45^\circ )} =\begin{bmatrix} 40,062.696&30,031.348&30,971.787 \\ 30,031.348&40,062.696&30,971.787 \\ 30,971.787 &30,971.787&33,416.928 \end{bmatrix} \begin{Bmatrix} 5539 \\ 10,583 \\ 13,260 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 950.4 \\ 1001.0 \\ 942.4 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau_{xy} \end{Bmatrix} ^{(-45^\circ )}=\begin{bmatrix} 40,062.696&30,031.348&-30,971.787 \\ 30,031.348&40,062.696&-30,971.787 \\ -30,971.787&-30,971.787&33,416.928 \end{bmatrix} \begin{Bmatrix} 5539 \\ 10,583 \\ 13,260 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 129.0 \\ 179.6 \\ -56.2 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{xx} \\ \sigma _{yy} \\ \tau_{xy} \end{Bmatrix} ^{(90^\circ )}=\begin{bmatrix} 6520.376&1630.094&0 \\ 1630.094&130,407.524&0 \\ 0&0&5015.674 \end{bmatrix} \begin{Bmatrix} 5539 \\ 10,583 \\ 13,260\end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 53.4 \\ 1389.1 \\ 66.5 \end{Bmatrix} \ MPa

Then, the local stresses in the plies are obtained as follows:

\begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \tau_{12} \end{Bmatrix}^{(0^\circ )} =\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \begin{Bmatrix} 739.6 \\ 78.0 \\ 66.5 \end{Bmatrix} =\begin{Bmatrix} 739.6 \\ 78.0 \\ 66.5 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \tau_{12} \end{Bmatrix}^{(45^\circ )}=\begin{bmatrix} 0.5&0.5&1 \\ 0.5&0.5&-1 \\ -0.5&0.5&0 \end{bmatrix} \begin{Bmatrix} 950.4 \\ 1001.0 \\ 942.4 \end{Bmatrix} =\begin{Bmatrix} 1918.1 \\ 33.3 \\ 25.3 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \tau_{12} \end{Bmatrix}^{(-45^\circ )}=\begin{bmatrix} 0.5&0.5&-1 \\ 0.5&0.5&1 \\ 0.5&-0.5&0 \end{bmatrix} \begin{Bmatrix} 129.0 \\ 179.6 \\ -56.2 \end{Bmatrix} =\begin{Bmatrix} 210.5 \\ 98.1 \\ -25.3 \end{Bmatrix} \ MPa \\ \begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \tau_{12} \end{Bmatrix}^{(90^\circ )} = \begin{bmatrix}0&1&0 \\ 1&0&0 \\ 0&0&-1 \end{bmatrix} \begin{Bmatrix} 53.4 \\ 1389.1 \\ 66.5 \end{Bmatrix}=\begin{Bmatrix} 1389.1 \\ 53.4 \\ -66.5 \end{Bmatrix} \ MPa

The deformations are within the limits and the stresses are within the material allowables. Also, the stresses in the individual plies are all tensile and thus buckling is unlikely.