On symmetry and non-uniqueness in exact topology optimization

Abstract

The aim of this article is to initiate an exchange of ideas on symmetry and non-uniqueness in topology optimization. These concepts are discussed in the context of 2D trusses and grillages, but could be extended to other structures and design constraints, including 3D problems and numerical solutions. The treatment of the subject is pitched at the background of engineering researchers, and principles of mechanics are given preference to those of pure mathematics. The author hopes to provide some new insights into fundamental properties of exact optimal topologies. Combining elements of the optimal layout theory (of Prager and the author) with those of linear programming, it is concluded that for the considered problems the optimal topology is in general unique and symmetric if the loads, domain boundaries and supports are symmetric. However, in some special cases the number of optimal solutions may be infinite, and some of these may be non-symmetric. The deeper reasons for the above findings are explained in the light of the above layout theory.

Appendix: A simple demonstration of statical determinacy of stress-based optimal trusses with a single load

In order to transform a piece-wise linear cost function into a linear one, e.g. Bendsoe and Sigmund (2003) use separate non-negative variables for the tensile and compressive member forces $({q_i^+ ,q_i^-})$. This is a useful alternative to Hemp’s (1973) method of expressing the same using inequalities and slack functions. Then our truss problem takes on the standard LP form:

$$ \label{eq6} \min \sum\limits_{i=1}^m {\frac{L_i }{\sigma _p }\big({q_i^+ +q_i^-}\big)} $$

(7)

$$ \label{eq7} \mbox{s.t}.\quad \mathbf{B}({\mathbf{q}^+-\mathbf{q}^-})=\mathbf{f}, $$

(8)

$$ \label{eq8} q_i^+ \ge 0,\quad q_i^- \ge 0\quad ( {i=1,...,m}), $$

(9)

where L _i are the member lengths, σ _p the permissible stress, B the static (joint) equilibrium equations, f the external loads, and m the number of truss members.

Since the number of variables in this formulation is 2m, for any basic solution we need 2m equality constraints (e.g. Strang 1980, Chapter 8), some of which were originally inequalities. Let the number of static equations be r, and the number of zero member forces k.

For each zero member force, (9) gives two equalities $({q_i^+ =0, q_i^- =0})$, and for any nonzero member force it gives one of these two. Hence we have for a basic solution

$$\label{eq9} 2m=r+2k+({m-k}), $$

(10)

implying that the number of zero member forces are

$$\label{eq10} k=m-r. $$

(11)

This means that the number (n) of remaining members with nonzero cross sections will be

$$\label{eq11} n=m-({m-r})=r. $$

(12)

Any optimal solution of an LP problem must be either

• (i) a basic feasible solution, satisfying (12) above, or

• (ii) the convex combination of basic feasible solutions.

For any statically determinate truss, the number of truss members equals the number of active equilibrium equations, hence for the considered problems by (12) the optimal basic feasible solutions are statically determinate, and any other optimal solution must be a convex combination of such statically determinate solutions.

Example

For the problem in Fig. 26a and b, we have m = 5 members in the ground structure, r = 2 active equilibrium equations (zero sum of horizontal and vertical forces at the loaded joint), k = 3 members dropping out (as by (11)), and n = 2 members in the optimal solution (as by (12)).

Fig. 26

Examples demonstrating statical determinacy of optimal basic feasible solutions. a Ground structures and loading, b optimal solution

Note

A layout with r members could be partially redundant and partially unstable, but this would be statically inadmissible and therefore not in the feasible set.

As an example, consider the ten-bar truss problem in Fig. 27. The ground structure with ten bars (m = 10) is shown in Fig. 27a, the number of joint equilibrium equations is r = 8. Then by (12) we have for any basic solution n = 8 nonvanishing members. The layouts in both Fig. 27b and c are basic, but the one in (b) is feasible and the one in (c) is non-feasible, because it violates the equilibrium equations. It was explained by Strang (1980) and demonstrated on examples, that many basic solutions are outside the feasible region, but the Simplex method does not consider these, because it always moves from one basic feasible solution to another one (of lower cost) in each iteration.

Fig. 27

a Ground structure for a ten-bar truss, b basic feasible solution, c basic nonfeasible solution