# Question 1. Ignore bottom surface. Solution: Design variables: X = (R, H) Objective function: maximize volume, πr 2 H OR Minimize, f(x) = πr 2 H

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1 Question 1 (Problem 2.3 of rora s Introduction to Optimum Design): Design a beer mug, shown in fig, to hold as much beer as possible. The height and radius of the mug should be not more than 20 cm. The mug must be at least 5cm in radius. The surface area of the sides must not be greater than 900cm2 (ignore the area of the bottom of the mug and ignore the mug handle see fig). Formulate the optimal design problem. R H Ignore bottom surface Design variables: X = (R, H) Objective function: maximize volume, πr 2 H OR Minimize, f(x) = πr 2 H Subjected to: 1) H 20 OR g 1 (X) = H ) 5 R 20 OR g 2 (X) = R 20 0 g 3 (X) = 5 R 0 3) 2πRH 900 OR g 4 (X) = 2πRH ) H 0 1

2 Question 2 (Problem 2.16 of rora s Introduction to Optimum Design): Design of a two bar truss. Design a symmetric two bar truss (both members have the same cross section) shown in figure to support a load W. The truss consists of two steel tubes pinned together at one end and supported on the ground at the other. The span of the truss is fixed at s. formulate the minimum mass truss design problem using height and the cross sectional dimension as design variable. The design should satisfy the following constraint. 1. Because of the space limitation, the height of the truss must not exceed b 1, and must not be less than b The ratio of the mean diameter to thickness of the tube must not exceed b The compressive stress in the tubes must not exceed the allowable stress σ a for steel. 4. The height, diameter, and thickness must be chosen to safeguard against member buckling. Use the following data: W=10kN; span s=2m; b 1 = 5m; b 2 = 2m; b 3 = 90; allowable stress, σ a = 250 MPa; modulus of elasticity, E = 210 GPa; mass density, ρ = 7850 kg/m 3 ; factor of safety against buckling, FS = 2; 0.1 <=D<=2 (m); and 0.01 <=t<=0.1 (m). W Height, H t s D Section at W F 1 F 2 α L s F 1 and F 2 are the forces in the two truss bars. Let L be the length of each bar and s be the distance between two end points of bars (as shown in fig). Let t be the thickness of cross section. Due to the symmetry, F 1 = F 2 = F lso, ΣF y = 0 2Fcos(α) + W = 0 F = W/2cos(α) ssuming t << D, cross section area of truss bar, = πdt Length of bar.5 Stress in truss member, σ c = F/ = W/[2cos(α) πdt] For pinned pinned column, critical buckling load, P cr = π 2 EI/L 2 2

3 Where, moment of inertia, I = πd 3 t/8 and E is young s modulus Design variables: Objective function: X = (D,t) Minimize the mass of truss Minimize, f(x) = 2ρL Subjected to: 1) b 2 H b 1 OR g 1 (X) = H b 1 0 g 2 (X) = b 2 H 0 2) D/t b 3 OR g 3 (X) = D b 3 t 0 3) σ c σ a OR g 4 (X) = σ c σ a 0 4) F P cr /2 OR g 5 (X) = F P cr /2 0 5) 0.1 D 2 6) 0.01 t 0.1 3

4 Question 3 (Problem 2.17 of rora s Introduction to Optimum Design): beam of rectangular cross section as shown in figure is subjected to maximum bending moment of M and maximum sheer of V. The allowable bending and shearing stresses are σ a and τ a respectively. The bending stress in the beam is calculated as σ = 6M/bd 2 and average shear stress in the beam is calculated as τ = 3V/2bd Where d is the depth and b is the width of the beam. It is also desired that the depth of the beam shall not exceed twice its width. Formulate the design problem for minimum cross sectional area using the following data: M = 140kN, V=24kN σ a = 165MPa τ a = 50MPa d Bending stress, σ = 6M/bd 2 Shear stress, τ = 3V/2bd Design variables: X = (d, b) b Objective function: minimize the cross sectional area Minimize, f(x) = bd Subjected to: 1) d 2b OR g 1 (X) = d 2b 0 2) σ σ a OR g 2 (X) = σ σ a 0 3) τ τ a OR g 3 (X) = τ τ a 0 4) b, d 0 4

5 Question 4 (Problem 2.17 of rora s Introduction to Optimum Design): Design a hollow circular beam shown in figure for two conditions when P = 50KN, the axial stress σ should be less then σ a, and when P = 0, deflection δ due to self weight should satisfy δ <= 0.001l. The limits for dimension are t = 0.10 to 1.0 cm, R = 2.0 to 20.0 cm. Formulate the minimum weight design problem and transcribe it into the standard form. Use the following the data: δ = 5wl 4 /384EI; w = self weight force/length (N/m); σ a = 250MPa; modulus of elasticity, E = 210GPa; mass density; ρ = 7800kg/m 3 ; σ = P/; gradational constant, g = 9.9m/s 2 ; moment of inertia, I = πr 3 t (m 4 ). P Beam δ P t l = 3m 2R Part 1: P = 50KN Part 2: P=0 Beam cross section area, = 2πRt xial stress, σ = P/ w = self weight force/length = ρg*2πrtl/l = 7800*9.9*2πRt Moment of inertia I = πr3t Displacement under self weight: δ = 5wl 4 /384EI Design variables: X = (R, t) Objective function: minimization of weight = ρg*2πrtl Minimize, f(x) = Subjected to: 1) σ σ a OR g 1 (X) = (σ/ σ a ) 1 0 2) δ l 0 OR g 2 (X) = (δ / l) 1 0 NOTE: IT IS GOOD TO WORK WITH NON DIMENSIONL NORMLIZED CONSTRINTS 3) 0.1 t 1 4) 2 R 20 5

6 Question 5 Consider a laminate made from graphite/epoxy with E 1 =128 GPa, E 2 =13 GPa, G 12 = 6.4 GPa, and Formulate the design problem of obtaining an 8 ply symmetric balanced laminate with maximum E x, such that G xy is at least 25 GPa, and is not bigger than 1. E 1 = 128 GPa, E 2 = 13 GPa, G 12 = 6.4 GPa and v 12 = 0.3 v 21 = E 2 * v 12 /E 1 = lso, Q 11 = E 1 /(1 v 12 * v 21 ) =128*10 9 /( ) = *10 9 Q 22 = 13.11*10 9 Q 12 = 3.93*10 9 Q 66 = 6.4*10 9 lso we can find the material invariants by: U 1 = (3Q Q Q Q 66 )/8 = 57.53*10 9 U 2 = (Q 11 Q 22 )/2 = 58.02*10 9 U 3 = (Q 11 + Q 22 2Q 12 4Q 66 )/8 = 13.6*10 9 U 4 = (Q 11 + Q Q 12 4Q 66 )/8 = 17.55*10 9 U 5 = (Q 11 + Q 22 2Q Q 66 )/8 = 20*10 9 Thickness of individual ply is assumed to be constant = t For balanced laminate we will have the following stacking sequence, (θ1/ θ1 / θ2/ θ2)s xy Design variables: X = (θ1, θ2) Objective function: Maximize E x = [( 11 /h)*( 22 /h) ( 22 /h) 2 ] / ( 22 /h) OR Minimize, f(x) = [( 11 /h)*( 22 /h) ( 22 /h) 2 ] / ( 22 /h) Subjected to: 1) G xy = ( 66 /h) 25*10 9 OR g 1 (X) = 1 [( 66 /h)/ 25*10 9 ] 0 2) v xy = 12 / 22 1 OR g 2 (X) = ( 12 / 22 ) 1 0 3) 0 θ1 90 4) 0 θ2 90 6

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