1. Optimal 3-member Truss Design Submitted By :- Course Instructor:- Avinash Kumar (10105017) Prof. K. Deb Piyush Rai (10105070) (ME 752)

2. Coming Next ….  Problem Formulation  Objective function definition  Constraints definition  KKT conditions formulation  Optimal Solution to the problem  Penalty function method  Augmented Lagrangian Method  Results

3. Problem Formulation P θ L LL Fig. 1. A 3- member truss. δ 2 S 1 =Psin θ δ 1 S 2 = Pcos θ A1A1 A2A2 A3A3 1 2 3

4. Objective minimize the weight of the structure (W) optimal values of cross-sectional areas A 1, A 2, A 3 Minimize weight,w = f(A 1, A 2, A 3 ) subjected to g 1 : σ 1 ≤ σ 1 (allowable) g 2 : σ 2 ≤ σ 2 (allowable) g 3 : σ 3 ≤ σ 3 (allowable) g 4 : δ 1 ≤ δ 1 (allowable) g 5 : δ 2 ≤ δ 2 (allowable) and A i ≥ 0 ; i= 1,2,3. Stress constraints Displacement constraints

5. Mathematical formulation Material properties :- Mild steel (ρ = 7860 kg/m 3,E=2.1*10 11 Pa) Taking L=10 m, P = 50 KN, Max. allowable stress, σ all. = 220 MPa, δ 1 (all.) =δ 2 (all. ) = 50mm. Objective function : Minimize W = ρg(10√2 A 1 +10A 2 +10√2 A 3 ) subjected to,

6. KKT Conditions 1) ( Equilibrium Equation ) Finding out the gradients of objective function & all constraints, we get :- where,,

8. 2) u j g j (x)=0 (Complementary Slackness Condition) u 1 g 1 (x)=0 u 2 g 2 (x)=0 u 3 g 3 (x)=0 u 4 g 4 (x)=0 u 5 g 5 (x)=0 3) u j ≥ 0 ; j = 1,2,…..,5. (Non-negativity )

9. Optimal Solution to the problem Comparison of Penalty function method and Augmented Lagrangian Method :- R=1 For an initial value of R=1, Augmented Lagrangian method is better than penalty function method. The optimal solution we get :- A 1 = 0.1323 m 2, A 2 = 0.0492 m 2, A 3 = 0.000494 m 2 Minimum weight of the structure, W min. = 19.155 tonnes. Method Initial GuessMinimum values of A (m 2 )Function value (N) No. of function evaluations A1A1 A2A2 A3A3 A1A1 A2A2 A3A3 Penalty Function 1110.13710.04910.000521187918.11318 Augmented Lagrangian 1110.13230.04920.000494182715.81104

10. R=10 Method Initial GuessMinimum values of A (m 2 )Function value No. of function evaluations A1A1 A2A2 A3A3 A1A1 A2A2 A3A3 Penalty Function 1110.2123 0.03874 0.03179295972.11304 Augmented Lagrangian 1110.20430.033210.03082281994.21148 Here also, in terms of function evaluations, Augmented lagrangian method is better than the Penalty function method. The optimal solution for R=10 is :- A 1 = 0.2123 m 2, A 2 = 0.03874 m 2, A 3 = 0.03179 m 2 Minimum weight of the structure, W min. = 28.74 tonnes.

11. Results 1.For the 3-member structure, and for the selected values of maximum allowable stress, applied loads,etc., the Augmented Lagrangian method is better than Penalty function method. 2.For a large no. of variables, Penalty function method is better than Augmented Lagrangian method in terms of no. of function evaluations.

12. References 1.Haug & Arora, Applied Optimal Design. 2.K.Deb, Optimization for Engg. Design.

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