A procedure to design optimum composite plates using implicit decision trees

Abstract

This paper presents a novel methodology to minimize the weight of a composite plate, taking into account both its structural integrity and its manufacturing constraints. This optimization problem has been abstracted, and reduced to, a graph problem where finding the optimum is equivalent to finding the shortest path in the graph. This graph is an implicit ternary decision tree and path finding is accomplished with a parallelized breadth-first search procedure. As a result of this search, the procedure is able to provide both a global optimum and a Pareto front. The implementation of an implicit decision tree as a way to optimize a laminate stacking sequence is a novel idea, in spite that both issues have been tackled in numerous publications separately. Its benefits are clearly stated in this work.

Appendix 1

In appendix 1 a demonstration of $\overrightarrow{{\ \left\Vert \varepsilon \right\Vert}^{\acute{\mkern6mu}}}<\overrightarrow{\left\Vert \varepsilon \right\Vert }$ is shown with a counter-example.

Supposing the opposite, in particular considering the case where:

$$ \overrightarrow{{\ \left\Vert \varepsilon \right\Vert}^{\acute{\mkern6mu}}}>\overrightarrow{\left\Vert \varepsilon \right\Vert } $$

(14a)

$$ \overrightarrow{\varepsilon^{\acute{\mkern6mu}}}=\overrightarrow{\varepsilon}+\alpha \overrightarrow{\varepsilon}\ being\ 0<\alpha $$

(14b)

‘α’ being a positive scalar, then the following must be true:

$$ \left(\begin{array}{c}{\varepsilon}_x\\ {}{\varepsilon}_y\\ {}{\gamma}_{x y}\end{array}\right)-\left(\begin{array}{c}{\varepsilon^{\acute{\mkern6mu}}}_x\\ {}{\varepsilon^{\acute{\mkern6mu}}}_y\\ {}{\gamma^{\acute{\mkern6mu}}}_{x y}\end{array}\right)=-\alpha \left(\begin{array}{c}{\varepsilon}_x\\ {}{\varepsilon}_y\\ {}{\gamma}_{x y}\end{array}\right) $$

(15)

Then, from (10), it follows:

$$ \begin{array}{l}\left(\begin{array}{c}\hfill 0\hfill \\ {}\hfill 0\hfill \\ {}\hfill 0\hfill \end{array}\right)=\left\{\left[\left(1+\alpha \right){\left(\begin{array}{c}\hfill {\varepsilon}_x\hfill \\ {}\hfill {\varepsilon}_y\hfill \\ {}\hfill {\gamma}_{x y}\hfill \end{array}\right)}^T\right]\boldsymbol{A}\left[-\alpha \left(\begin{array}{c}\hfill {\varepsilon}_x\hfill \\ {}\hfill {\varepsilon}_y\hfill \\ {}\hfill {\gamma}_{x y}\hfill \end{array}\right)\right]\right\}-\lambda \hfill \\ {} being\ \lambda, \alpha >0\hfill \end{array} $$

(16)

Then factoring the above (16):

$$ \begin{array}{l}\left(\begin{array}{c}\hfill 0\hfill \\ {}\hfill 0\hfill \\ {}\hfill 0\hfill \end{array}\right)=\left\{\left(-\alpha -{\alpha}^2\right)\left[\boxed{{\left(\begin{array}{c}\hfill {\varepsilon}_x\hfill \\ {}\hfill {\varepsilon}_y\hfill \\ {}\hfill {\gamma}_{x y}\hfill \end{array}\right)}^T\mathbf{A}\left(\begin{array}{c}\hfill {\varepsilon}_x\hfill \\ {}\hfill {\varepsilon}_y\hfill \\ {}\hfill {\gamma}_{x y}\hfill \end{array}\right)}\right]\right\}-\lambda \hfill \\ {} being\ \lambda, \alpha >0\hfill \end{array} $$

(17)

Since A is symmetric and positively definite the squared term in (17) is a positive scalar and this (17) cannot be cancelled. Thus, it can be concluded that the starting point ((14)) is wrong.