Chapter 1 (1).pdf

R. Zakhama
Structural Optimisation course
Chapter 1: Introduction

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Chap1: Introduction
I. Generalities
I.1 Engineering approach
•In the past the process of determining the optimal design of
structures such as bridges, cars, or airplanes relied heavily on
the engineer’s intuition and experience (fig.I.1). This type of
procedure is very expensive and inaccurate.
Pre-design
Analysis
Checking
Final design
Modification
Fig.I.1: Classical design approach.

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Chap1: Introduction
I.1 Engineering approach
•The introduction of mathematical optimisation theory into
structural design in the early 1960’s helped to reduce
prototyping costs in manufacturing. Since then research has
been carried out, and is intensifying, to find ways to automate
the design process (fig.I.2). This increase in research has been
made possible in particular by the development of the Finite
Element Method (FEM) promoting fast structural analyses.
Pre-design
Optimiser
Final design
Analysis
FigI.2. : Automatic design approach

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Chap1: Introduction
I.2 Structural optimisation classes
structural optimisation techniques are classified into the three
large families
•Size optimisation, see figure I.3: only the cross-section or the
thickness of the different elements of the structure can be
modified while the shape and the topology are fixed.
FigI.3. : Size optimisation

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Chap1: Introduction
•Shape optimisation, see figure I.4: the shape of the structure
can be changed while the topology is fixed. Shape optimisation
is characterised by modifying the border domain regarding the
original structure or by changing the transversal dimensions,
but modifying the connectivity or the nature of the structural
members is not allowed.
FigI.4. : Shape optimisation (K. Lee, 2000)

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Chap1: Introduction
•Topology optimisation, see figure I.5: fundamental
modifications of a structural nature can be achieved. In this
case, the modifications of connectivity or structural members
are possible. Topology optimisation commonly leads to
significant improvements in performance.
FigI.4. : Topology optimisation (R. Zakhama et al., 2009)

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Chap1: Introduction
II. General formulation of an optimisation design problem
•Goal of the design engineer is to develop the best possible
system, consistent with the resources allocated for the project,
to perform a prescribed job.
“If you don’t do the best you can with what you
happened to have got, you will never do the best you
might have done with what you should have had”
• Optimum Design: Determination of the best feasible
combination of system variables according to a pre-selected
quantitative measure of effectiveness.

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Chap1: Introduction
II.1 Example: airtank design
head
shell
FigII.1. : Airtank design (Papalambos & Wilde, 2000)
s = shell thickness
r = inside radius
h = head thickness
l = shell length
head
shell
s = shell thickness
r = inside radius
h = head thickness
l = shell length
head
shell

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Chap1: Introduction
Find : dimensions
Minimise : metal volume
Subject to : constraints on:
Total air volume (≥)
Head thickness (≥)
Shell thickness (≥)
Shell length (≤ & ≥)
outer radius (≤)
head
shell

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Chap1: Introduction
head
shell
Find : x = [h, l, r, s]
Minimise :
Subject to :
Physical relevance: x∈χ h, l, r, s > 0
( ) ( )
[ ]l
r
s
r
h
s
r 2
2
2
2 −
+
+
+ π
π
3
7
2
10
12
.
2 cm
l
r ⋅
≥
π
13
.
0
≥
r
h
01
.
0
≥
r
s
cm
l
cm 750
10 ≤
≤
cm
150
≤
+ s
r

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Chap1: Introduction
head
shell
Find : x = [h, l, r, s]
Minimise :
Subject to :
Physical relevance: x∈χ h, l, r, s > 0
( ) ( )
[ ]l
r
s
r
h
s
r 2
2
2
2 −
+
+
+ π
π
0
1
10
12
.
2 7
2
1 ≤
+
⋅
−
=
l
r
g
π
0
1
13
.
0
2 ≤
+
−
=
r
h
g
0
1
01
.
0
3 ≤
+
−
=
r
s
g
0
1
10
4 ≤
+
−
=
l
g
0
1
750
5 ≤
−
=
l
g
0
1
150
6 ≤
−
+
=
s
r
g

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Chap1: Introduction
II.2 Elements of Problem Formulation
To find the best possible solution for a given problem, the
optimisation problem must be defined in terms of design
variables.
Formulation steps:
a) Identify design variables (optimisation parameters)
(e.g., displacements, dimensions, fibres orientations,…)
b) Define objective function (merit function)
(e.g., weight, volume, potential energy,… )
c) Identify constraints
(e.g., stress, dimensions, critical loads,…)

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Chap1: Introduction
• Design variables (set of variables that describe the system):
• x = x1, x2, . . . . . , xn; or x = (x1, x2, . . . . . . . , xn)T
• types of design variables (DV’s)
physical nature: dimensional, shape, material, topology
numerical nature: continuous, discrete valued, integer,
binary (0/1)
• Selection of DV’s may not be unique.
• DV’s should be independent of each
other.
• For design variables use as many
independent parameters as possible.
FigII.2. : Tube cross-section

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Chap1: Introduction
•The choice of design variables can be critical to the success of
the optimisation process. In particular it is important to make
sure that the choice of design variables is consistent with the
analysis model (fig II.2).
FigII.3. : Optimised shape of a hole in a plate, (a) initial
design, (b) final design. (T. Haftka & Z. Gürdal)

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Chap1: Introduction
• Objective function (merit function)
measure of goodness of the system that is being designed:
• f(x) = f(x1, x2, . . . . . . . , xn)
• to be Maximised or Minimised
• types of objective function
physical nature: cost, profit, weight, comfort,
deformations, stress
mathematical nature: linear or nonlinear

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Chap1: Introduction
• Constraints: define the boundaries of the feasible design space
• constraints on design variables
- direct limitations on DV's (side constraints);
- relations between dv’s (variable linking); e.g.,
• constraints on system behavior
- limits on system output; e.g. stress, displacement, buckling load limits.
- physical laws governing the system; conservation of mass, energy.
• inequality constraints
• equality constraints

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Chap1: Introduction
II.3 Mathematical problem formulation
Minimize
subject to

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Chap1: Introduction
II.3 Mathematical problem formulation
• Special Cases
• Problem is to maximise a function instead of minimisation,
Max f(x) = Min –f(x)
Max f(x) = Min 1/f(x)
• Optimiser cannot handle equality constraints,
h(x) = 0 g1(x)= h(x) ≤ 0 and g2(x)= − h(x) ≤ 0
• Optimiser cannot handle design variables with negative
values, say, -a ≤ x ≤ b where a and b are positive numbers
- Introduce a new variable x’= x + a where 0 ≤ x’
- Introduce two new variables x+−x- = x

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Chap1: Introduction
II.3 Graphical illustration
Graphical illustration of an optimisation problem is possible for
one, two or at most three design variable problems. A graphical
method not only gives a solution, but also helps us to
understand the nature of the problems.
• Plot the constraint equations
• Identify the feasible design space
• Plot objective function contours
• Locate optimum by inspection

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Chap1: Introduction
• Example: airtank 1-D
x = s (shell thickness as design variable)
r,h,l constant parameters
0
1
01
.
0
s.t. 3 ≤
+
−
=
r
x
g
( ) ( )
[ ]l
r
x
r
h
x
r 2
2
2
x
2
min −
+
+
+ π
π
0
1
150
6 ≤
−
+
=
x
r
g
0
>
x
0
1
01
.
0
s.t. 3 ≤
+
−
=
r
x
g
( ) ( )
[ ]l
r
x
r
h
x
r 2
2
2
x
2
min −
+
+
+ π
π
0
1
150
6 ≤
−
+
=
x
r
g
0
1
01
.
0
s.t. 3 ≤
+
−
=
r
x
g
( ) ( )
[ ]l
r
x
r
h
x
r 2
2
2
x
2
min −
+
+
+ π
π
0
>
x
0
1
150
6 ≤
−
+
=
x
r
g
0
1
01
.
0
s.t. 3 ≤
+
−
=
r
x
g
( ) ( )
[ ]l
r
x
r
h
x
r 2
2
2
x
2
min −
+
+
+ π
π

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Chap1: Introduction
f
x
f(x)
g
•x
g6(x)
0
feasible
infeasible
g3(x)
FigII.4. : Graphical illustration of the objective and the
constraints

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Chap1: Introduction
f
•x
g
•x
0
f(x)
g6(x)
g3(x)
g6=0
g3=0
FigII.5. : Feasible domain

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Chap1: Introduction
• Constrained versus unconstrained optimum
f
x
f(x)
g=0
• constrained optimum
• bounded optimum
f
x
f(x)
g1=0
g2=0
• unconstrained optimum
• interior optimum
FigII.6. : Constrained and unconstrained optimum

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Chap1: Introduction
• Multiple local minima
f
x
local global
f(x) f
x
local
global
f(x)
FigII.7. : Local and global minima
Global (Absolute) Minimum
f*=f(x*) ≤ f(x)
for all xL ≤ x ≤ xU
strict global minimum
f*=f(x*) < f(x)
Local (Relative) Minimum
f*=f(x*) ≤ f(x)
for || x - x* || < δ

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Chap1: Introduction
x1 = r, x2 = l (2 design variables)
s,h constant parameters
0
, 2
1 >
x
x
0
1
10
12
.
2
s.t. 7
2
2
1
1 ≤
+
⋅
−
=
x
x
g
π
( ) ( )
[ ] 2
2
1
2
1
2
1
x
2
min x
x
s
x
h
s
x −
+
+
+ π
π
0
1
13
.
0 1
2 ≤
+
−
=
x
h
g
0
1
01
.
0 1
3 ≤
+
−
=
x
s
g
0
1
10
2
4 ≤
+
−
=
x
g
0
1
750
2
5 ≤
−
=
x
g
0
1
150
1
6 ≤
−
+
=
s
x
g

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Chap1: Introduction
The objective function contours is plotted by setting the
objective function equal to a constant and plotting the curves
corresponding to various values of this constant.
x2
x1
f=6
f=7
f=8 f=9
f=10
∇f
FigII.8. : Objective function contours

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Chap1: Introduction
x2
x1
g1=0
g1<0
feasible
x2
x1
g1
g5
g6
g3
g2
g4
FigII.9. : Constraints representation

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Chap1: Introduction
A design which satisfies all the constraints gj({x}) ≤ 0 is a
feasible region. The set of values of the design variables that
satisfy the equation gj({x}) = 0 forms a surface in the design
space. It is a surface in the sense that it cuts the space into two
regions: one where gj > 0 and the other where gj < 0.
The set of all feasible designs form the
feasible region (see fig. II.10) .
FigII.10. : Feasible region
x2
x1
g1
feasible
region
g6
g3
g2
g4
F
g5

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Chap1: Introduction
The optimal design, by definition, the one which is feasible and
has minimal/maximal value of the objective function.
FigII.11. : Optimum solution
x2
x1
g1
optimum g6
g3
g2
g4
F
g5

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Chap1: Introduction
• Constraint activity theorem
gj(x) ≤ 0 is active at the optimum implies that
1. If gj(x) ≤ 0 is left out, the location of the optimum changes
2. The constraint is satisfied with strict equality: gj(x) = 0
For the airtank 2-D example, the constraints g5 and g1 are active
but g2 is not.

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Chap1: Introduction
II.4 Example
To illustrate the formulation of design variables, consider the three-bar example
shown in figure II.12. All three bars are made of material having Young’s modulus
E=200000, and that we want to minimise the volume of the structure subject to the
constraint that the stress in any members does not exceed 20 in tension and 15 in
compression. A lower bound on cross-sectional area is fixed to zero. The structure is
subjected to two load cases, P1=20 and P2=20, respectively.
1. Identify the design variables of the optimisation problem.
2. Express the stresses in the bars with respect to the design variables.
3. Formulate the truss optimisation problem.
4. Sketch the constraints in the two-dimensional
design. Identify the feasible and infeasible
domains; plot the contours of the objective
function and locate the optimum solution.
100 100
100
x
y
1
u
2
u
P1=20
P2=20
(1) (2) (3)
A1 A2 A3
FigII.12. : Three-bar truss

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Chap1: Introduction
II.4 Example
1.The design variables of this problem are the section area of the truss members: A1 ,
A2 and A3. Due to symmetry of loading and geometry, the number of design
variables is reduced to two X1= A1 = A3 and X2 = A2 and only one loading condition
may be considered.
2.The equilibrium equations in x- and y- directions:
in matrix form, the equilibrium can be written as:

direction)
-
(x
0
2
2
2
1
3
1
=
+
+
−
P
s
s
direction)
-
(y
0
2
2
2
1
3
2
1
=
+
−
−
−
P
s
s
s

s
B
F
s
s
s
P
T






















−
=






3
2
1
1
2
1
1
2
1
2
1
0
2
1
1
1
2

s
B
F T
=
s1 s2 s3
P1
FigII.13. : Forces on the cut-
out free node

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Chap1: Introduction
II.4 Example
Hooke’s law and the geometry condition will give us:
 the internal forces vector s, can be expressed as:
Since δ = Bu, the bar forces are obtained as: s = DBu
The equilibrium equation becomes:
 F = BTs = BTDBu = Ku
The stiffness matrix is easily calculated as:
We obtain the displacements of the free node by solving the linear system Ku = F
3
..
1
, =
= i
EA
s
l
i
i
i
i
δ










=
=
1
2
1
0
0
0
2
0
0
0
2
ere
wh
X
X
X
l
E
D
D
s δ






+
=
2
0
0
2 2
1
1
X
X
X
l
E
K

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Chap1: Introduction
II.4 Example
The stresses can be written as:
Some straightforward calculations give us the bar stresses as:
( )





+
=






2
/
1
/
1
2
1
1
1
2
1
X
X
X
E
l
P
u
u










=
=
= −
−
1
2
1
1
1
0
0
0
0
0
0
A
where
X
X
X
DBu
A
s
A
σ
2
2
2
2
1
2
1
1
2
1
1
X
X
X
X
X
P
+
+
=
σ
2
2
2
2
1
2
1
1
1
2
X
X
X
X
P
+
=
σ
2
2
2
1
2
1
2
1
3
X
X
X
X
P
+
−
=
σ

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Chap1: Introduction
II.4 Example
3. The optimisation problem formulation is written as following:
Explicitly, the optimisation problem formulation for the three bars truss structure
can be written as:
volume
truss
:
min V
V(X)
X
[ ]
2
1,X
X
X =
,
,
i
(X)
σ
σ i 3
2
1
,
s.t. =
≤
≤ σ
X 0
≥
( )
2
2
l
l
X
l
X
l
X
min 2
1
3
1
2
2
1
1 X
X
V(X)
X
+
=
+
+
=
[ ]
2
1,X
X
X =
0
15
,
0
20
s.t. 1
2
1
1 ≤
−
−
≡
≤
−
≡ (X)
σ
g
(X)
σ
g
0
,
0 2
8
1
7 ≤
−
≡
≤
−
≡ X
g
X
g
0
15
,
0
20 2
4
2
3 ≤
−
−
≡
≤
−
≡ (X)
σ
g
(X)
σ
g
0
15
,
0
20 3
6
3
5 ≤
−
−
≡
≤
−
≡ (X)
σ
g
(X)
σ
g

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Chap1: Introduction
II.4 Example
Only constraints which may affect the design must be considered, Since σ1 and σ2
will always be positive, and σ3 negative, we may consider only the stress
constraints:
The optimisation problem becomes:
0
15
0
20
0
20
3
2
1
≤
−
−
≤
−
≤
−
σ
σ
σ
( )
2
2
l
min 2
1 X
X
V(X)
X
+
=
[ ]
2
1,X
X
X =
0
20
2
2
2
s.t. 2
1
2
1
1
2
1
1 ≤
−
+
+
≡
X
X
X
X
X
P
g
0
20
2
2
2
2
1
2
1
1
1
2 ≤
−
+
≡
X
X
X
X
P
g
0
15
2
2
2
1
2
1
2
1
3 ≤
−
+
≡
X
X
X
X
P
g
0
,
0 2
5
1
4 ≤
−
≡
≤
−
≡ X
g
X
g

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Chap1: Introduction
II.4 Example
4. Graphical illustration :
The optimal design is at point A: X1=0.788 X2=0.410 min V = 263.9
X2
X1
1.0
2.0
3.0
0.5 1.0 1.5
feasible
region
intersection
point
constrained
design
unconstrained
design
X2
1.0
2.0
3.0
X1
0.5 1.0 1.5
objective function
contours
optimal
design A
FigII.14. : Design space and objective function contours

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Chap1: Introduction
III. Mathematical tools and notations
III.1 Vectors, matrices
X: column vector of Rn , XT =(x1, x2, … xn)
I : identity matrix
H : matrix (n, n), HT its transpose
– H .HT is symmetric
– det(H) = determinant of H
– if det(H) = 0, H is a singular matrix
– if H is non-singular, a unique matrix exists H-1 = inverse matrix of H, such that:
H.H-1 = H–1 .H = I
– scalar product of 2 vectors :
XT.X noted too (X, X)
– norm of vector X of Rn :
‖X‖ = (XT.X)1/2 = (∑ xi
2)1/2
– X1, X2…, Xp vectors called orthogonal if:
(Xi)T .Xj = 0 for all i≠j

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Chap1: Introduction
III.1 Vectors, matrices
– in addition, they are called orthonormal if:
(Xi)T.Xi = 1 for all i
– a symmetric square matrix H is called positive semidefinite if :
X ∈ Rn , XT .H. X ≥ 0 for all X ≠ 0,
– positive definite if the inequality is strict

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Chap1: Introduction
III.2 Differentiability
– Gradient Vector (the vector of first derivatives):
Geometrically, the gradient vector at point xp is normal to the tangent
plane to the function at that point, and points in the direction of
maximum increase in the function.

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Chap1: Introduction
III.2 Differentiability
– Hessian Matrix (the matrix of second derivatives):
Hessian matrix is always a symmetric matrix
The two-times differentiable function f(x) is (strictly) convex, if its
Hessian matrix H is positive semidefinite (definite) for all x ∈ χ. That
means all eigenvalues λi of H are (strictly) positive.

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Chap1: Introduction
III.3 Taylor series expansion
– for a function with one variable, Taylor series expansion about xp
– for a function with n variables
– change in the value of the function in moving from xp to a neighbouring
point d = x – xp distance away from it

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Chap1: Introduction
III.4 Convex set
A set A ⊂
⊂
⊂
⊂ Rn is called convex if for every pair of points within the set, every
point on the straight line segment that joins them is also within the set A :
∀X∈A, ∀Y ∈ A , ∀α such that 0 ≤ α ≤ 1, we have : αx + (1 – α) y ∈ A
III.5 Convex function
A function f : Rn → R is called convex if:
∀ X∈Rn , ∀ Y∈Rn , ∀ α such that
0 ≤ α ≤ 1, we have :
f(αX + (1 – α) Y) ≤ α f(X) + (1 – α) f(Y)
A Convex
set
A non-convex
set
FigIII.1. : A convex and a non-convex set
FigIII.2. : A convex function

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Chap1: Introduction
• Example 1
Consider 2
4
1
2
1
1
3
2
1
)
(
2
1
2
1
2
1
+











−
+


















=
x
x
x
x
x
x
f
T
T
x
minimum
at (1.2, -2.6):
f
x1
x2
FigIII.3. : A convex function

R. Zakhama 45
Chap1: Introduction
• Example 2
Consider 2
1
2
2
2
2
1
2
2
2
2
1 5
5
10
)
10
(
5
.
0
10
)
10
(
4
)
( x
x
x
x
x
x
x
f −
−





 −
+
+
+





 −
−
+
=
x2
x1
f(x)
x1
x2
FigIII.4. : A non convex function

R. Zakhama 46
Chap1: Introduction
III.6 Convex optimisation problem
If all inequality constraint functions gj(x) are convex and the
equality constraint functions hk(x) are linear, then the resulting
feasible domain is convex, and if the objective function is
convex then the optimisation problem is convex and has only
one local = global minimum.
(reverse not true in general)







=
=
=
≤
1
,
0
)
(
1
,
0
)
(
s.t.
)
(
min
e
k
g
j
x
..n
k
x
h
..n
j
x
g
x
f

R. Zakhama 47
Chap1: Introduction
IV. Problems
IV.1 Problem 1
In this problem, we want to optimise a truss structure with 2 bars, where we want to
minimise the weight under some constraints (figure IV.1.). The bars of the structure
are made with a material defined by a young modulus E and a density ρ. The load
F0 and α=30°. We want to find the optimal section areas X1 and X2 that minimise
the weight structure Z. The stresses of the truss structure members should not
exceed the elasticity limit in tension σ0 and in compression -σ0, respectively. The
displacement δy of node A should not exceed the limit δ0.
1- Give the expression of the objective function Z with
respect to the design variables.
2- Express the stresses in the bars with respect to the
design variables.
3- Find the vertical displacement expression δy with
respect to X1 and X2.
4- Formulate the truss optimisation problem. FigIV.1. : Problem 1 structure

R. Zakhama 48
Chap1: Introduction
IV. Problems
IV.1 Problem 2
Consider the three truss problem (see figure IV.2.) of minimum weight design under
stress constraints.
We adopt the following formulation where we
consider only the significant constraints:
s.t. :
where x1 and x2 are the bar sections as shown in figure IV.2.
FigIV.2. : Problem 2 structure
2
1
3
Min
2
1
x
x
Z
,x
x
+
=

R. Zakhama 49
Chap1: Introduction
IV. Problems
IV.1 Problem 2
1- Sketch the remaining constraints in the two-dimensional design in figure IV.3.
Identify the feasible domain D.
2- Plot the contours of the objective function. Draw the gradient of the objective
function, and indicate the minimising direction of this function by an arrow.
3- Locate the optimum solution on the graph. Determine the optimal sections x1
* and
x2
*, and the corresponding objective value Z*.

R. Zakhama 50
Chap1: Introduction
IV. Problems
IV.1 Problem 2
FigIV.3. : Problem 2 plot

R. Zakhama 51
Chap1: Introduction
Sources
Presentation based on material from:
• Haftka Gürdal (1994): Elements of Structural Optimization
• van Keulen: TUD Optimization course
• Etman (2006): EM course Engineering Optimization

Chapter 1 (1).pdf

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