2. Optimality criterion methods Methods that use directly an optimality criterion in order to resize the structure. Fully stress design and stress ratio resizing. Dual methods. Simple resizing rule based on Lagrange multiplier for single constraint.

3. Fully stressed design If we have a structure designed only under stress constraints, FSD assumes that at the optimum every point is at the stress limit for at least one load case, unless at that point the sizing variable is limited by a bound constraint. This is generally true for a single material, but Section 9.1 of Elements of Structural Optimization has two examples of failure when there are two materials.

4. Single inequality constraint Optimization problem Lagrangian Optimality condition Ratio can be viewed as cost-effectiveness of design variable.

5. Iterative procedure Increase variables that are cost effective and reduce less cost effective ones Lambda found so that constraint remains critical In top.m this is done by bisection method However, linear or reciprocal approximation usually works too.

6. Using the reciprocal approximation The approximation Then with some algebra get

7. Container design example Open top container needs minimum volume of 125m 3. Sides cost $10/m 2, bottom and ends $15/m 2. Minimize cost. Then What solution do you estimate?

8. Effect of eta Eta=0.3Eta=3

9. Eta=1 x, and f e’s and g

10. Problems container We started the iterations with an initial design that was in the ball park of the optimum. What kind of initial design will make the iteration diverge? What is the logarithmic derivative of the optimum cost with respect to the required volume (if we change it from 125)? Note that you do not need to use any number from the problem or the solution to answer. Use the answer of the previous problem to show that you can calculate the Lagrange multiplier of $8.333/m 3 from the optimum cost of $1560 and required volume of 125m 3.

11. Top OC function function [xnew]=OC(nelx,nely,x,volfrac,dc) l1 = 0; l2 = 100000; move = 0.2; while (l2-l1 > 1e-4) lmid = 0.5*(l2+l1); xnew = max(0.001,max(x-move,min(1.,min(x+move,x.*sqrt(- dc./lmid))))); if sum(sum(xnew)) - volfrac*nelx*nely > 0; l1 = lmid; else l2 = lmid; end