Slide 1

Optimality criterion methodsMethods 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.

For unconstrained minimization, we often solve for the optimum from the optimality condition that the first derivatives are zero. For constrained problems we have similar methods.

We will first discuss the fully-stressed-design (FSD) approach that is based on the intuitive idea that when we design with only stress constraints, every element that is not constrained by minimum gage should be at the stress limit.

Then we will briefly discuss dual methods that search for the optimum in the lagrange multiplier domain.

Finally, we will look at the case where we optimize a function subject to a single inequality constraint. For that case it is possible to derive a simple optimality condition and a simple resizing method.1Fully stressed designIf 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.

The fully stressed design (FSD) criterion stipulates that a structure designed only subject to stress constraints will be optimal (minimum weight) when every point is either at the stress limit or limited by a bound constraint.

While this is generally true for a single material, it is very easy to construct examples where it does not work with two materials. Section 9.1 of Haftka and Gurdals Elements of Structural Optimization has a simple example of two rods in parallel that show the difficulty. One rod has lower density and higher yield stress, but the fully stressed design approach tends to push towards utilizing the second rod that has inferior material.

There is also an example of a 10-bar truss that was invented by Schmit, where one element of the truss has a higher yield stress than the other 9, which causes the optimum design not to be fully stressed.2Single inequality constraintOptimization problem

LagrangianOptimality condition

Ratio can be viewed as cost-effectiveness of design variable.

For a single inequality constraint, we can use the Karush-Kuhn Tucker conditions to obtain a resizing rule. We start by reminding ourselves of the KKT optimality condition, and see that it implies that at the optimum the ratio of all the derivatives of the objective function to the derivatives of the constraints should be the same and equal to the Lagrange multiplier.

The ratio of the derivatives may be viewed as the cost effectiveness of the design variables. The numerator measures the cost of changing the design variable, and the denominator its effectiveness in changing the constraints. If a variable has poor cost effectiveness we should use less of it and keep doing it until they all have the same effectiveness.3Iterative procedureIncrease variables that are cost effective and reduce less cost effective ones

Lambda found so that constraint remains criticalIn top.m this is done by bisection methodHowever, linear or reciprocal approximation usually works too.

A simple iterative procedure works when the derivatives of the objective function are positive, as it is for sizes. The procedure increases the design variables that are cost effective and decreases the ones that are not, as shown by the equation. Eta is a damping parameter. If it is large convergence is slow and smooth. If it is small, it will converge faster, but possibly oscillate or even fail to converge.

The Lagrange multiplier is found from the condition that the constraint remains critical, that is, its value remains zero. This can be done by a simple numerical procedure, such as bi-section, but in the textbook (Section 9.3) it is done by using the reciprocal approximation.4Using the reciprocal approximationThe approximation

Then with some algebra get

We use a linear approximation in the reciprocals of the design variables, called the reciprocal approximation, which is written in the first set of equation. Then substituting the expression for xi from the previous slide, and setting g=0, gives us the expression for lambda.5Container design exampleOpen top container needs minimum volume of 125m3. Sides cost $10/m2, bottom and ends $15/m2. Minimize cost.

Then

What solution do you estimate?

The container problem is intended to show that one can use the iterative procedure for non-structural problems, as long as variables have the property that increasing their values incurs additional costs but improve constraint satisfaction.

The objective function is the cost, and the constraint is on the volume. X1 and x2 are the dimensions of the bottom and x3 is the height. All the variables have the same effect on the constraint but unequal contributions to the objective function. If we start with all being the same, then the ratios of the derivatives are like 45:35:50, and so we can expect that at the optimum x2>x1>x36Effect of etaEta=0.3Eta=3

Right figure produced with [x,f,e,lambda,g]=container(x0,3,10);plot(x');legend('x1','x2','x3'); xlabel('iterations')unction [x,f,e,lambda,g]=container(x0,eta,iter)%Example 9.3.2 structural optimization, modified 2/24/14 for g 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; endend

The OC (optimality criteria) routine in the top function, implements the optimality criterion with eta=2. It also has move limits of 0.2, meaning that we cannot change the density by more than 0.2. Because all the design variables have a derivative of 1 on the volume, we see only the derivative of the compliance. Also, for compliance minimization we reverse the objective function and constraint. The objective is a structural response, and the constraint is resources, so we divide by the Lagrange multiplier instead of multiplying.

The loop is in order to find the Lagrange multiplier that will satisfy the volume fraction. We start by allowing it to vary between 0 and 100,000, which is a wide range, but each iteration reduces it by a factor of 2. So the requirement of convergence to 1.e-4, which is a reduction by 9 orders of magnitude, should take less than 30 iterations.10