Optimal 3-member Truss Design

Submitted By :- Course Instructor:-Avinash Kumar (10105017) Prof. K. DebPiyush Rai (10105070) (ME 752)

Coming Next …. Problem Formulation

Objective function definition Constraints definition

KKT conditions formulation Optimal Solution to the problem

Penalty function method Augmented Lagrangian Method

Results

Problem Formulation

P

θ

L

L L

Fig. 1. A 3- member truss.

δ2

S1 =Psinθ

δ1

S2 = Pcosθ

A1

A2A3

12

3

Objective

• minimize the weight of the structure (W)• optimal values of cross-sectional areas A1 , A2 , A3

Minimize weight,w = f(A1 , A2 , A3)

subjected to g1 : σ1 ≤ σ1 (allowable)

g2 : σ2 ≤ σ2 (allowable)

g3 : σ3 ≤ σ3 (allowable)

g4 : δ1 ≤ δ1 (allowable)

g5 : δ2 ≤ δ2 (allowable)

and Ai ≥ 0 ; i= 1,2,3.

Stress constraints

Displacement constraints

Mathematical formulationMaterial properties :- Mild steel (ρ = 7860 kg/m3 ,E=2.1*1011 Pa)Taking L=10 m , P = 50 KN, Max. allowable stress, σall. = 220 MPa, δ1 (all.)=δ2 (all.) = 50mm.

Objective function : Minimize W = ρg(10√2 A1+10A2+10√2 A3) subjected to,

KKT Conditions

1) (Equilibrium Equation)

Finding out the gradients of objective function & all constraints , we get :-

where,

,

2) ujgj(x)=0 (Complementary Slackness Condition) u1g1(x)=0 u2g2(x)=0 u3g3(x)=0 u4g4(x)=0 u5g5(x)=0 3) uj ≥ 0 ; j = 1,2,…..,5. (Non-negativity )

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 :-

A1 = 0.1323 m2 , A2 = 0.0492 m2 , A3 = 0.000494 m2

Minimum weight of the structure, Wmin. = 19.155 tonnes.

MethodInitial Guess Minimum values of A (m2) Function

value(N)

No. of function

evaluationsA1 A2 A3 A1 A2 A3

Penalty Function 1 1 1 0.1371 0.0491 0.000521 187918.1 1318

Augmented Lagrangian 1 1 1 0.1323 0.0492 0.000494 182715.8 1104

• R=10

MethodInitial Guess Minimum values of A (m2) Function

valueNo. of

function evaluationsA1 A2 A3 A1 A2 A3

Penalty Function 1 1 1 0.2123 0.03874 0.03179 295972.1 1304

Augmented Lagrangian 1 1 1 0.2043 0.03321 0.03082 281994.2 1148

Here also, in terms of function evaluations, Augmented lagrangian method is better than the Penalty function method. The optimal solution for R=10 is :-

A1 = 0.2123 m2 , A2 = 0.03874 m2 , A3 = 0.03179 m2 Minimum weight of the structure, Wmin. = 28.74 tonnes.

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.

References1. Haug & Arora, Applied Optimal Design .2. K.Deb, Optimization for Engg. Design.

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