10. optimum design problem formulation -...
TRANSCRIPT
Hae-Jin ChoiSchool of Mechanical Engineering,
Chung-Ang University
10. Optimum Design Problem Formulation
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Correct formulation of a problem takes roughly 50% of the total effort needed
to solve
Important to follow well-defined procedures for formulating design
optimization problem
Designing an engineering system is a complex process, which requires many
assumptions, analyses by available methods, verification with experiments, etc.
Economic considerations is another important aspect; cost-effective systems
To complete the design of an engineering system, experts from different fields
of engineering must usually cooperate : Multidisciplinary Deign Optimization
(MDO)
DOE and Optimization
Introduction to Problem Formulation
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Optimum Design Problem Formulation is transcribing a verbal description of
the problem into a well-defined mathematical statement.
All systems are designed to perform within a given set of constraints
Limitation on resources, material failure, response of the system, member sizes, etc.
The constraints must be functions of the design variables.
If a design satisfies all constraints, we have a feasible (workable) system
If no design exist to satisfy the constraints, we call this as infeasible system
A criterion is needed to judge whether or not a given design is better than
another. This criterion is called the objective function or cost function.
An objective function must be a function of the design variables
DOE and Optimization
Introduction to Problem Formulation
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STEP 1: Identify and define design variables
STEP 2: Identify the cost function and develop an expression for it
in terms of design variables.
STEP 3: Identify constraints and develop expressions for them in
terms of design variables.
DOE and Optimization
Problem Formulation Procedure
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Design a two-member bracket shown in the
figure to support a force W without structural
failure.
The force is applied at an angle θ which is
between 0 and 90 degree, h is the height and s is
the base width for the bracket.
Since the brackets will be produced in large
quantity, the design objective is to minimize its
mass while also satisfying certain fabrication and
space limitation
DOE and Optimization
Example: Design of a Two-bar Structure
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STEP1: Identify Design Variables
Design variables are parameters chosen to describe the design
of a system
At initial stage, designate as many design variables as possible
Which increases design flexibility
Later, we may eliminate (or fix) unimportant design variables
using DOE, RSM, etc.
All design variables should be independent of each other.
Example of circular tube (di, do, and t)
For the two-bar structure, h and s can be treated as design
variables in the initial formation
Other design variables will depend on the cross-sectional
shape for members 1 and 2
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STEP2: Objective (Cost) Function
There can be many feasible design for a system; We need some criterion to compare various
designs
The criterion must be a scalar function whose numerical value can be obtained one a design
is specified.
Objective (or Cost function) is represented by f, or f(x) to emphasize its dependence on
design variable vector x
Maximization of f(x) is converted as minimization of –f(x)
Objective functions may be
minimize cost, maximize profit, minimize weight, minimize energy expenditure, maximize ride
quality of vehicle, etc.
Multiple objectives in cost function is called as multi-objective design optimization problem.
no general rule. We will visit this issue later
Proper formation of objective function is complex problem and needs experience and
expertise of the system
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STEP2: Objective (Cost) Function
In case of circular tube,
Design variables are x1= height h of the truss
x2= span s of the truss
x3 = outer diameter of member 1
x4= inner diameter of member 1
x5 = outer diameter of member 2
x6= inner diameter of member 2
Objective function may be Total mass of the truss
2 2 1/2 2 2 2 2
1 2 3 5 4 6(4 ) ( )8
is the density of the material
Mass x x x x x x
where
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Feasible or infeasible design
Implicit constraint is a constraint difficult to express as an explicit function of
design variables
Linear and Nonlinear constraints
Equality and Inequality constraints
STEP 3: Design Constraints
1 2
2 2
1 2 1 2
1 (linear constraints)
3 0 (nonlinear constraints)
x x
x x x xInfeasible
regionfeasible
region
x1
x2
1 2
1 2
1 (equality constraints)
1 (inequality constraints)
x x
x x
DOE and Optimization
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Principle of static equilibrium
DOE and Optimization
STEP 3: Design Constraints
1 2
1 2
1
2
2 2
sin sin cos
cos cos sin
From the geometry,
sin / 2
cos /
sin 2cos0.5
sin 2cos0.5
(0.5 )
F F W
F F W
s l
h l
F Wlh s
F Wlh s
where
l h s
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Stress in a member is defined as force divided by cross-sectional area.
To avoid over stressing of a member, the calculated stress must be less
than or equal to material allowable stress ( , failure stress).
Finally, constraints on design variables are written as
which implies the bounds (minimum and maximum) of each design
variable space
STEP 3: Design Constraints
a
_ _ ; =1 to 6i low i i upperx x x i
2 2
3 4
2 2
5 6
2 sin 2cos 0
( )
2 sin 2cos 0
( )
a
a
Wl
x x h s S
Wl
x x h s S
DOE and Optimization
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Given
Fixed variables, functions, assumptions
Find
x # design variables
Satisfy
h(x) = 0 # equality constraints
g(x) ≤ 0 # inequality constraints
xlow ≤ x ≤ xupper # bound on x
Minimize
f(x) # objective function
Optimum Design Problem Formulation
1 2 3 4 5 6
1 2 2
3 4
2 2 2
5 6
_ _
7.85, 1.5
( , , , , , )
2 sin 2cos ( ) = 0
( )
2 sin 2cos ( ) = 0
( )
; = 1 to
a
a
i low i i upper
Given
S
Find
x x x x x x
Satisfy
Wlg
x x h s S
Wlg
x x h s S
x x x i
x
x
x
2 2 1/2 2 2 2 2
1 2 3 5 4 6
6
( ) (4 ) ( )8
Minimize
f x x x x x xx
Verbal statement: With Given fixed parameter ρ and S, Find design variables x1, x2, x3, x4, x5, and x6
Subject to the constraints g1, g2, and bounds to Minimize the objective function
DOE and Optimization
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The cans will be produced in billions, so it is desirable to minimize the cost
of manufacturing them. Since the cost can be related directly to the surface
area of the sheet metal used, it is reasonable to minimize the sheet metal
required to fabricate the can. Fabrication, handling aesthetic, and shipping
considerations impose the following restrictions on the size of the can; (1)
the diameter of the can should be no more than 8 cm; Also, it should not be
less than 3.5 cm; (2) the height of the can should be no more than 18cm and
no less than 8cm; and (3) the can is required to hold at least 400 ml of fluid.
Example 1: Design of a Beer Can
DOE and Optimization
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STEP 1: Identify design variables
Two design variables: D= diameter of the can (cm), H=height of the can (cm)
STEP 2: Identify objective function
to minimize the total surface area of the sheet metal
STEP 3:Identify constraints
volume must be at least 400
bounds on design variables
Design of a Beer Can
D
H
2
Surface Area = 2
DDH
2 4004
Volume D H
3.5 8 and 8 18D H
2
2
Assumption of thin sheet metal
=Diameter of can (cm)
=Height of can (cm)
( , ) 400 04
3.5 8 and 8 18
( , )2
Given
Find
D
H
Satisfy
g D H D H
D H
Minimize
Df D H DH
DOE and Optimization
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A company owns two saw mills and two forests. Table below shows the
capacity of each mill in logs/day and the distances between forests and
mills. Each forest can yield up to 200 logs/day for the duration of the
project and the cost to transport the logs is estimated at 15
cents/km/log. At least 300 logs are needed each day. Formulate the
problem to minimize the cost of transportation of logs each day.
DOE and Optimization
Example 2: Saw Mill Operation
Distance
Mill Forest 1 Forest 2 Mill capacity/day
A 24.0 20.5 240 logs
B 17.2 18.0 300 logs
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Problem Formulation for Saw Mill Operation
1
2
Cost to tranport = 15 cents/km/log
Table of data for saw mill operation
number of logs shipped from Forest 1 to Mill A
number of logs shipped from Forest 2 to M
Given
Find
x
x
3
4
1 2
3 4
ill A
number of logs shipped from Forest 1 to Mill B
number of logs shipped from Forest 2 to Mill B
240 # capacity constraint of Mill A
300
x
x
Satisfy
x x
x x
1 3
2 4
1 2 3 4
# capacity constraint of Mill B
200 # yield constraint of Forest 1
200 # yield constraint of Forest 2
300 # Constraints on the number of logs
x x
x x
x x x x
1 2 3 4 1 2 3 4
needed for a day
0; 1,2,3,4 # Bounds for reality
Minimize
( , , , ) cost=24(0.15) 20.5(0.15) 17.2(0.15) 18(0.15)
ix i
f x x x x x x x x
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Example 3: Design of Cabinet
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Problem Formulation
cabinet
8xC1 5xC2 15xC3
5
bolts
5
rivets
6
bolts
6
rivets
3
bolts
3
rivets
x1 x2 x3 x4 x5 x6
1 1
2 1
3 2
4 2
5 3
Prblem Definition
number of C to be bolted
number of C to be riveted
number of C to be bolted
number of C to be riveted
number of C to be bol
Given
Find
x
x
x
x
x
6 3
1 2
3 4
5 6
1 3 5
2 4 6
1 2 3
ted
number of C to be bolted
800
500
1500
5 6 3 6000
5 6 3 8000
0; 1 6
0.70(5) 0.60(5) 1.00(6)
i
x
Satify
x x
x x
x x
x x x
x x x
x i to
Minimize
Cost x x x
4 5 6 0.80(6) 0.60(3) 1.00(3)x x x
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Example 4: Tubular Column Design
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Problem FormulationGiven
P=10 MN , E=207Gpa, ρ=7833kg/m3,
l=5.0, σa=248 Mpa
Find
R, t
Satisfy
Minimize
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A beam of rectangular cross-section is subjected to a bending moment
of M (Nm) and maximum shear force of V (N). The bending stress in
the beam is calculated as σ=6M/bd2 (Pa) and the average shear stress is
calculated as τ=3V/2bd (Pa), where b is the width and d is the depth of
the beam. The allowable stress in bending and shear are 10 MPa and 2
MPa respectively. It is desired to minimize cross-sectional area of the
beam. For the numerical example, let M=40 kNm and V=150 kN.
DOE and Optimization
Example 5: Beam Design
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Problem Formulation
8
1 2
5
2
3
40 , 150
= depth of the beam (mm)
= width of the beam (mm)
2.4 10 ( , ) 10 0
2.25 10 ( , ) 2 0
( , ) 2 0
0, 0
Given
M kN m V kN
Find
d
b
Satisfy
g b dbd
g b dbd
g b d d b
b d
Minimize
( , )f b d bd
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Graphical Solution
Example for Graphical Solution
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Graphical Solution
Problem Formulation
1
2
1 2
Problem Description
= number of A machines manufactured each day
= number of B machines manufactured each day
16 # shipping and handling constraint
Given
Find
x
x
Satisfy
x x
x1 2
1 2
1 2 1 2
1 # manufacturing constraint28 14
1 # sales constraint14 24
( , ) (400 600 )
x
x x
Minimize
f x x x x
Linear Programming -> Simplex Method (Numerical Approach)
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Design Problem with Multiple Solutions Design Problem with Unbound Solutions
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Graphical Solution for Tabular ColumnGiven
P=10 MN , E=207Gpa, ρ=7833kg/m3,
l=5.0, σa=248 Mpa
Find
R, t
Satisfy
Minimize
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Graphical Solution for Beam Design
8
1 2
5
2
3
40 , 150
= depth of the beam (mm)
= width of the beam (mm)
2.4 10 ( , ) 10 0
2.25 10 ( , ) 2 0
( , ) 2 0
0, 0
Given
M kN m V kN
Find
d
b
Satisfy
g b dbd
g b dbd
g b d d b
b d
Minimize
( , )f b d bd