By Peter W. Christensen

Mechanical and structural engineers have regularly strived to make as effective use of fabric as attainable, e.g. by means of making buildings as mild as attainable but capable of hold the hundreds subjected to them. long ago, the hunt for extra effective constructions used to be a trial-and-error technique. despite the fact that, within the final 20 years computational instruments in accordance with optimization concept were constructed that give the chance to discover optimum constructions roughly immediately. a result of excessive price reductions and function profits which may be completed, such instruments are discovering expanding commercial use.

This textbook provides an advent to all 3 periods of geometry optimization difficulties of mechanical constructions: sizing, form and topology optimization. the fashion is particular and urban, concentrating on challenge formulations and numerical resolution tools. The therapy is designated adequate to let readers to put in writing their very own implementations. at the book's homepage, courses should be downloaded that additional facilitate the training of the cloth covered.

The mathematical must haves are saved to a naked minimal, making the publication compatible for undergraduate, or starting graduate, scholars of mechanical or structural engineering. practising engineers operating with structural optimization software program might additionally make the most of interpreting this book.

**Read or Download An Introduction to Structural Optimization PDF**

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**Extra resources for An Introduction to Structural Optimization**

**Example text**

8. 5λ. L1 (x1 ,λ) L2 (x2 ,λ) Differentiation gives ∂L1 = 2x1 + λ − 6, ∂x1 ∂L2 = 2x2 + λ + 2. 4 Lagrangian Duality 49 Fig. 14), we find the x, denoted x ∗ , that minimizes L for any given λ ≥ 0. 16) λ ∴ x1∗ = 3 − , if 4 ≤ λ ≤ 6 2 ∴ x2∗ = −2, if λ ≥ 2 never satisfied since λ ≥ 0 λ ∴ x2∗ = −1 − , if 0 ≤ λ ≤ 2. 17) 50 3 Basics of Convex Programming Fig. 18) if 0 ≤ λ ≤ 2 if 2 ≤ λ ≤ 4 if 4 ≤ λ ≤ 6 if λ ≥ 6. Note that ϕ is continuously differentiable (ϕ(2) = 0, ϕ(4) = −5, ϕ(6) = −11, ϕ (2) = − 52 , ϕ (4) = − 52 , ϕ (6) = − 72 ).

This cannot be the actual solution, however, since x1∗ 1. We therefore put x1∗ = 1 and try to find the optimum x2∗ . Again, ∂L2 /∂x2 = 0 gives x2∗ = −1 − λ/2. The dual objective function is ϕ(λ) = − λ2 3 − λ + 4, 4 2 which is maximized for λ∗ = 0. This results in the optimal solution x1∗ = 1 and x2∗ = −1 which is a feasible point. 14) are satisfied, so that L has really been minimized for x ∗ . Since we have also maximized ϕ, the x ∗ obtained is the optimum solution of (P)3 . 7 We will solve the two-bar truss problem on page 14 once again, this time using Lagrangian duality.

If f : S → R and g : S → R are convex and h : S → R is strictly convex, then αf is convex, where α ≥ 0 is an arbitrary scalar, f + g is convex and f + h is strictly convex. If both the objective function and the feasible set of (P) are convex, the problem is said to be convex. The lemma above then states that (P) is convex if the objective function and all constraint functions gi , i = 1, . . , l, are convex. As previously mentioned, local minima are also global minima for convex problems. 1).