Choice design variables

1. Choice design variables • The design variables we have considered for optimization define dimensions or angles or in more general terms the magnitude of a variable or the number of items. • For laminate design, we will also run into choice design variables that will define which of several options we will choose. • For Example 4.1.1, we had to choose between three laminates and this is formulated with choice design variables in Example 4.1.2. • Later, the main use would be to choose ply angles from a limited set of 0, 90, 45 or -45.

2. Example from geometry • Choose between a circle, a square, and an equilateral triangle of unit area, to get the smallest perimeter. • We can define a single variable that will take the values 1,2 or 3, but it turns out that it is advantageous to use three variables each taking the values 0 or 1, that is binary variables. • We will use three choice variables . A value of 1 indicates that we choose the corresponding option and a value of zero that we do not choose it. • To ensure that we choose only one option we add satisfy an equality constraint . • We add the defining dimension d (diameter or side) as our fourth variable.

3. Optimization formulation • Areas and perimeters • Area with choice variables • Optimization formulation

4. Solver initial screen

5. Solver solution

6. Answer report

7. Example 4.1.1 revisited • Graphite/epoxy • Two load conditions • Allowable strains: Normal strains 0.4%, shear strain 0.006 • Try • Ply thickness is t=0.005 in.

8. Strain constraints and moduli • Normal strain constraint leads to • Shear strain constraints leads to • For all zero laminate =18.5Msi, =0.93Msi • For laminate =3.176Msi, =4.858Msi • For laminate =7.552Msi, =2.894Msi. Do these values make sense? • Example 4.1.2 formulates the problem using choice design variables

9. Use of choice variables in 4.1.2 • Laminates • Design variables • Choice constraint =1 • Objective function • Young’s modulus • Solve with Solver.