10. optimum design problem formulation -...

Hae-Jin ChoiSchool of Mechanical Engineering,

Chung-Ang University

10. Optimum Design Problem Formulation

SCHOOL OF MECHANICAL ENG.

CHUNG-ANG UNIVERSITY-1-

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



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


Introduction to Problem Formulation



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.


Problem Formulation Procedure



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


Example: Design of a Two-bar Structure


CHUNG-ANG UNIVERSITY-5-DOE and Optimization

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



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



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



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




Principle of static equilibrium



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



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


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




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




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




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




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.


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



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



Example 3: Design of Cabinet



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 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


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



Example 4: Tubular Column Design



Problem FormulationGiven

P=10 MN , E=207Gpa, ρ=7833kg/m3,

l=5.0, σa=248 Mpa

Find

R, t

Satisfy

Minimize



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.


Example 5: Beam Design



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



Graphical Solution

Example for Graphical Solution



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)



Design Problem with Multiple Solutions Design Problem with Unbound Solutions



Graphical Solution for Tabular ColumnGiven

P=10 MN , E=207Gpa, ρ=7833kg/m3,

l=5.0, σa=248 Mpa

Find

R, t

Satisfy

Minimize



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

10. optimum design problem formulation -...

Documents