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Unformatted text preview: EGM6365 Project #1 Optimization of Space Structure
A space frame structure as shown in the figure consists of 25 truss members. Initially, all
members have the same circular crosssections with diameter d = 2.0 in. At nodes 1 and 2,
a constant force F = 60,000 lbf is applied in the positive ydirection. Four nodes (7, 8, 9,
and 10) are fixed on the ground. The frame structure is made of a steel material whose
properties are Young’s modulus E = 3×107 psi, Poisson’s ration ν = 0.3, yield stress σY =
36,000 psi, and mass density ρ = 7.3×10−4 lb⋅sec2/in4. The safety factor N = 1.5 needs to
be used for the yield stress. Due to the manufacturing constraint, the diameter of the
frame can be continuously changed between 0.1 in and 2.5 in. (1) Solve the initial truss structure using truss finite element MATLAB program. Provide
a plot that shows labels for elements and nodes along with boundary conditions.
Provide a stress contour plot and a table of stress at each element.
(2) Using ‘fmincon’ function in MATLAB, minimize the structural weight by changing
the crosssectional diameter of each truss element, while all members are safe under
the given yield stress and the safety factor. You can use the symmetric geometry of
the structure. Identify zeroforce members. For zeroforce members, use the lower
bound of the crosssectional diameter. Provide a stress contour plot at the optimum
design along with a table of stress at each element. Provide structural weights and
diameters of each member at initial and optimum designs. Submit your part of
MATLAB program. Output
max
Iter
Fcount
f(x)
constraint
Stepsize
1
26
3.64062
0.5
1
2
53
3.1317
218.2
1
3
80
2.30249
670
1
4
107
1.87707
6570
1
5
134
1.78052
925.4
1
6
161
1.7421
25.14
1
7
188
1.67351
2.776e17
1
8
215
1.59223
575.7
1
9
242
1.59204
6.498
1
10
269
1.58079
0.03891
1
11
296
1.57686
0.001086
1
12
323
1.57437
0.00619
1
13
351
1.57146
0.05981
0.5
14
378
1.56778
0.05448
1
15
405
1.56405
0.8109
1
16
432
1.56406
3.385e06
1
Hessian modified
Optimization terminated successfully:
Search direction less than 2*options.TolX and
maximum constraint violation is less than options.TolCon
x = 0.1000
2.1274
0.9803
2.2250
fval = 0.1000
2.1274
0.9803
2.2250 0.1000
0.1000
0.9803
2.2250 0.1000
0.1000
0.1000
2.2250 0.1000
1.6246
0.1000 Directional
derivative
Pro
0.53
0.951
0.503
0.112
0.0415
0.213
0.0835
0.000583
0.0174
Hessia
0.00404
0.00307
Hessia
0.0167
Hessia
0.00901
0.00373
6.9e06
Hessia
1.22e07 2.1274
1.6246
0.1000 2.1274
0.9803
0.1000 1.5641 Weight = fval * 384 = 600.6 lbf
output =
iterations:
funcCount:
stepsize:
algorithm:
firstorderopt:
cgiterations: 16
434
1
'mediumscale: SQP, QuasiNewton, linesearch' Lagrange Multiplier
1.0e05 *
0
0
0.5551
0.5551
0.1997
0.1997
0.7585
0.7585 0
0
0.1997
0.7585 0
0
0
0.7585 0
0.2281
0 0.5551
0.2282
0 0.5551
0.1997
0 File project1.m
x0 = [2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2];
lb=[0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1,0.1
,0.1,0.1,0.1,0.1,0.1,0.1];
ub=[2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5,2.5
,2.5,2.5,2.5,2.5,2.5,2.5];
option=optimset('Display','iter');
[x, fval, flag,output,lambda] =
fmincon('objprj',x0,,,,,lb,ub,'conprj',option); File objprj.m
% Objective function is the mass of the structures
function f = objfun(x) nodes=[ 1.5,
0,
1.5, 1.5,
4,
4,
4,
4, 8; 1.5,
0, 8; 1.5, 1.5, 4;
4; 1.5, 1.5, 4; 1.5, 1.5, 4;
0; 4,
4,
0; 4,
4,
0;
0]*12;
%in/ft A = pi;
E = 3E7;
den = 7.3E4; % in^2
% psi %node 1, node 2,area, E
members = [ 1,2,A,E; 1,4,A,E;
2,4,A,E; 2,5,A,E;
4,5,A,E; 3,4,A,E;
4,9,A,E; 5,8,A,E;
6,9,A,E; 6,10,A,E; 2,3,A,E;
1,3,A,E;
6,5,A,E;
4,7,A,E;
3,7,A,E; 1,5,A,E;
1,6,A,E;
3,10,A,E;
3,8,A,E;
4,8,A,E; 2,6,A,E;
3,6,A,E;
6,7,A,E;
5,10,A,E;
5,9,A,E]; area = 0.25 * pi * (x.^2);
f = 0;
%mass
for i = 1:25
n1 = members(i,1);
n2 = members(i,2);
length = sqrt((nodes(n1,1)nodes(n2,1))^2 + (nodes(n1,2)nodes(n2,2))^2 +
(nodes(n1,3)nodes(n2,3))^2);
f = f + den * length * area(i);
end File conprj.m
function [c, ceq] = confun(x)
% 25member space structure
nodes=[ 1.5,
0, 8; 1.5,
0, 8; 1.5, 1.5, 4;
1.5, 1.5, 4; 1.5, 1.5, 4; 1.5, 1.5, 4;
4,
4,
0; 4,
4,
0; 4,
4,
0;
4,
4,
0]*12;
%in/ft
E = 3E7;
A = 0.25 * pi * (x.^2);
%node 1, node 2,area, E
members = [ 1,2,A(1) ,E;
2,4,A(6) ,E;
4,5,A(11),E;
4,9,A(16),E;
6,9,A(21),E; % psi 1,4 ,A(2) ,E;
2,5 ,A(7) ,E;
3,4 ,A(12),E;
5,8 ,A(17),E;
6,10,A(22),E; 2,3,A(3) ,E;
1,3,A(8) ,E;
6,5,A(13),E;
4,7,A(18),E;
3,7,A(23),E; 1,5 ,A(4) ,E; 2,6 ,A(5) ,E;
1,6 ,A(9) ,E; 3,6 ,A(10),E;
3,10,A(14),E; 6,7 ,A(15),E;
3,8 ,A(19),E; 5,10,A(20),E;
4,8 ,A(24),E; 5,9 ,A(25),E]; active_nodes = [1,2,3,4,5,6];
loads = [1,0,60000,0;2,0,60000,0]; %node, fi,fj,fk dof = 3;
%FE solution for the nodal displacements
[displacements, stress] = solve_fem(nodes,members,active_nodes,loads,dof);
c = abs(stress)  36000/1.5;
ceq=; ...
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This note was uploaded on 06/07/2011 for the course EGM 6365 taught by Professor Staff during the Spring '08 term at University of Florida.
 Spring '08
 Staff
 Optimization

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