advanced design for complex systems -...

Advanced Design for Complex systems(2. Problem Formulation)

Advanced Design for Complex systems(2. Problem Formulation)

Associate Professor Hae-Jin ChoiThe School of Mechanical Engineering

Chung-Ang University

n Analysis is ¨Finding behavior of an existing system¨Calculating response of an existing system under

a specified inputn Design is

¨Determining sizes and shapes of various parts of the system to meet performance requirements

¨Selecting the best among multiple candidate to achieve design requirements

Engineering Design vs Analysis

SCHOOL OF MECHANICAL ENG.CHUNG-ANG UNIVERSITY

n Analysis is ¨Finding behavior of an existing system¨Calculating response of an existing system under

a specified inputn Design is

¨Determining sizes and shapes of various parts of the system to meet performance requirements

¨Selecting the best among multiple candidate to achieve design requirements

Engineering Design vs Analysis

Load

Maximum Stress

Height Height


Analysis

Design

HeightH=100 mm

Maximum Stress= ???

Maximum Stress≤ 400 MPa

HeightH=???

Introduction to Problem Formulation

n Correct formulation of a problem takes roughly 50% of the total effort needed to solve

n Important to follow well-defined procedures for formulating design optimization problem

n Designing an engineering system is a complex process, which requires many assumptions, analyses by available methods, verification with experiments, etc.

n Economic considerations is another important aspect; cost-effective systems

n To complete the design of an engineering system, experts from different fields of engineering must usually cooperate : Multidisciplinary Deign Optimization (MDO)


n Correct formulation of a problem takes roughly 50% of the total effort needed to solve

n Important to follow well-defined procedures for formulating design optimization problem

n Designing an engineering system is a complex process, which requires many assumptions, analyses by available methods, verification with experiments, etc.

n Economic considerations is another important aspect; cost-effective systems

n To complete the design of an engineering system, experts from different fields of engineering must usually cooperate : Multidisciplinary Deign Optimization (MDO)

3

Introduction to Problem Formulation

n Design Problem Formulation is transcribing a verbal description of the problem into a well-defined mathematical statement.

n 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

n 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


n Design Problem Formulation is transcribing a verbal description of the problem into a well-defined mathematical statement.

n 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

n 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

4

Problem Formulation Procedure

n STEP 1: Identify and define design variables

n STEP 2: Identify the cost function and develop an expression for it in terms of design variables.

n STEP 3: Identify constraints and develop expressions for them in terms of design variables.


n STEP 1: Identify and define design variables

n STEP 2: Identify the cost function and develop an expression for it in terms of design variables.

n STEP 3: Identify constraints and develop expressions for them in terms of design variables.

5

Example: Design of a Two-bar Structure

n Design a two-member bracket shown in the figure to support a force Wwithout structural failure.

n The force is applied at an angle θwhich is between 0 and 90 degree, his the height and s is the base width for the bracket.

n 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


n Design a two-member bracket shown in the figure to support a force Wwithout structural failure.

n The force is applied at an angle θwhich is between 0 and 90 degree, his the height and s is the base width for the bracket.

n 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

6

STEP1: Identify Design Variables

n Design variables are parameters chosen to describe the design of a system

n 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. n All design variables should be independent of

each other.¨ Example of circular tube (di, do, and t)

n For the two-bar structure, h and s can be treated as design variables in the initial formation

n Other design variables will depend on the cross-sectional shape for members 1 and 2


n Design variables are parameters chosen to describe the design of a system

n 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. n All design variables should be independent of

each other.¨ Example of circular tube (di, do, and t)

n For the two-bar structure, h and s can be treated as design variables in the initial formation

n Other design variables will depend on the cross-sectional shape for members 1 and 2

7

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

n 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.

n 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


¨ 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

n 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.

n 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

8

STEP2: Objective (Cost) Function¨ In case of circular tube,¨ Design variables are

n x1= height h of the trussn x2= span s of the trussn x3 = outer diameter of member 1n x4= inner diameter of member 1n x5 = outer diameter of member 2n x6= inner diameter of member 2

¨ Objective function may ben Total mass of the truss


¨ In case of circular tube,¨ Design variables are

n x1= height h of the trussn x2= span s of the trussn x3 = outer diameter of member 1n x4= inner diameter of member 1n x5 = outer diameter of member 2n x6= inner diameter of member 2

¨ Objective function may ben Total mass of the truss

2 2 1/2 2 2 2 21 2 3 5 4 6(4 ) ( )

8

is the density of the material

Mass x x x x x x

where

pr

r

= + + - -

9

STEP 3: Design Constraintsn Feasible or infeasible designn Implicit constraint is a constraint difficult to express as an explicit

function of design variablesn Linear and Nonlinear constraints

n Equality and Inequality constraints

1 22 21 2 1 2

1 (linear constraints)

3 0 (nonlinear constraints)

x xx x x x+ £

- + £

x2


n Feasible or infeasible designn Implicit constraint is a constraint difficult to express as an explicit

function of design variablesn Linear and Nonlinear constraints

n Equality and Inequality constraints

1 22 21 2 1 2

1 (linear constraints)

3 0 (nonlinear constraints)

x xx x x x+ £

- + £ Infeasibleregionfeasible

region

x1

1 2

1 2

1 (equality constraints)1 (inequality constraints)

x xx x+ =

+ £

10

STEP 3: Design Constraints

n Principle of static equilibrium 1 2

1 2

1

2

2 2

sin sin coscos cos sin

From the geometry, sin / 2cos /

sin 2cos0.5

sin 2cos0.5

(0.5 )

F F WF F W

s lh l

F Wlh s

F Wlh s

where

l h s

a a qa a q

aa

q q

q q

- + =

- - =

==

é ùê ú=- +ê úë ûé ùê ú=- -ê úë û

= +


1 2

1 2

1

2

2 2

sin sin coscos cos sin

From the geometry, sin / 2cos /

sin 2cos0.5

sin 2cos0.5

(0.5 )

F F WF F W

s lh l

F Wlh s

F Wlh s

where

l h s

a a qa a q

aa

q q

q q

- + =

- - =

==

é ùê ú=- +ê úë ûé ùê ú=- -ê úë û

= +

11

STEP 3: Design Constraintsn Stress in a member is defined as force divided by cross-sectional area. n To avoid over stressing of a member, the calculated stress must be less than

or equal to material allowable stress ( , failure stress).

n Finally, constraints on design variables are written as

which implies the bounds (minimum and maximum) of each design variable space

3 4

5 6

2 sin 2cos( )

2 sin 2cos( )

a

a

Wlx x h s S

Wlx x h s S

sq qp

sq qp

é ùê ú+ £ê ú- ë ûé ùê ú- £ê ú- ë û

as


n Stress in a member is defined as force divided by cross-sectional area. n To avoid over stressing of a member, the calculated stress must be less than

or equal to material allowable stress ( , failure stress).

n Finally, constraints on design variables are written as

which implies the bounds (minimum and maximum) of each design variable space

3 4

5 6

2 sin 2cos( )

2 sin 2cos( )

a

a

Wlx x h s S

Wlx x h s S

sq qp

sq qp

é ùê ú+ £ê ú- ë ûé ùê ú- £ê ú- ë û

_ _ ; =1 to 6i low i i upperx x x i£ £

12

Optimum Design Problem Formulation n Given

¨ Fixed variables, functions, assumptionsn Find

¨ x # design variablesn Satisfy

¨ h(x) = 0 # equality constraints¨ g(x) ≤ 0 # inequality constraints¨ xlow ≤ x ≤ xupper # bound on x

n Minimize¨ f(x) # objective function

1 2 3 4 5 6

13 4

25 6

_ _

7.85, 1.5

( , , , , , )

2 sin 2cos ( ) = 0( )

2 sin 2cos ( ) = 0( )

; = 1 to 6

a

a

i low i i upper

GivenS

Findx x x x x x

SatisfyWlg

x x h s SWlg

x x h s Sx x x i

Mi

r

sq qp

sq qp

= =

=

é ùê ú+ - £ê ú- ë ûé ùê ú- - £ê ú- ë û

£ £

x

x

x

2 2 1/2 2 2 2 21 2 3 5 4 6 ( ) (4 ) ( )

8

nimize

f x x x x x xpr= + + - -x


n Given¨ Fixed variables, functions, assumptions

n Find¨ x # design variables

n Satisfy¨ h(x) = 0 # equality constraints¨ g(x) ≤ 0 # inequality constraints¨ xlow ≤ x ≤ xupper # bound on x

n Minimize¨ f(x) # objective function

1 2 3 4 5 6

13 4

25 6

_ _

7.85, 1.5

( , , , , , )

2 sin 2cos ( ) = 0( )

2 sin 2cos ( ) = 0( )

; = 1 to 6

a

a

i low i i upper

GivenS

Findx x x x x x

SatisfyWlg

x x h s SWlg

x x h s Sx x x i

Mi

r

sq qp

sq qp

= =

=

é ùê ú+ - £ê ú- ë ûé ùê ú- - £ê ú- ë û

£ £

x

x

x

2 2 1/2 2 2 2 21 2 3 5 4 6 ( ) (4 ) ( )

8

nimize

f x x x x x xpr= + + - -x

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

13

Example 1: Design of a Beer Can

n 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.


n 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.

14

Design of a Beer Can

n STEP 1: Identify design variables¨ Two design variables: D= diameter of the can

(cm), H=height of the can (cm) n STEP 2: Identify objective function

¨ to minimize the total surface area of the sheet metal

n STEP 3:Identify constraints¨ volume must be at least 400

¨ bounds on design variables

D

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

FindDH

Satisfy

g D H D H

D HMinimize

Df D H DH

p

pp

= - £

£ £ £ £

= +


n STEP 1: Identify design variables¨ Two design variables: D= diameter of the can

(cm), H=height of the can (cm) n STEP 2: Identify objective function

¨ to minimize the total surface area of the sheet metal

n STEP 3:Identify constraints¨ volume must be at least 400

¨ bounds on design variables

2

Surface Area = 2DDH p

p +

2 4004

Volume D Hp= ³

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

FindDH

Satisfy

g D H D H

D HMinimize

Df D H DH

p

pp

= - £

£ £ £ £

= +

15

Example 2: Saw Mill Operation

n 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.


n 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.

DistanceMill Forest 1 Forest 2 Mill

capacity/dayA 24.0 20.5 240 logsB 17.2 18.0 300 logs

16

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

Findxx=

=

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

xx

Satisfyx xx 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 xx xx x x x

+ £

+ £

+ + + ³

1 2 3 4 1 2 3 4

needed for a day 0; 1, 2,3, 4 # Bounds for realityMinimize ( , , , ) 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

³ =

= + + +


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

Findxx=

=

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

xx

Satisfyx xx 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 xx xx x x x

+ £

+ £

+ + + ³

1 2 3 4 1 2 3 4

needed for a day 0; 1, 2,3, 4 # Bounds for realityMinimize ( , , , ) 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

³ =

= + + +

17

Example 3: Design of Cabinet

A cabinet is assembled from components C1, C2, and C3. Each cabinetrequires eight C1, five C2, and fifteen C3 components. Assembly of C1needs either five bolts or five rivets; C2 six bolts or six rivets; and C3three bolts or three rivets. The cost of putting a bolt, including the cost ofthe bolts, is $0.70 for C1, $1.00 for C2 and $0.60 for C3. Similarly, rivetingcosts are $0.60 for C1, $0.80 for C2, and $1.00 for C3. A total of 100cabinets must be assembled daily. Bolting and riveting capacities per dayare 6000 and 8000, respectively. We wish to determine the number ofcomponents to be bolted and riveted to minimize the cost.

SCHOOL OF MECHANICAL ENG.CHUNG-ANG UNIVERSITY18

A cabinet is assembled from components C1, C2, and C3. Each cabinetrequires eight C1, five C2, and fifteen C3 components. Assembly of C1needs either five bolts or five rivets; C2 six bolts or six rivets; and C3three bolts or three rivets. The cost of putting a bolt, including the cost ofthe bolts, is $0.70 for C1, $1.00 for C2 and $0.60 for C3. Similarly, rivetingcosts are $0.60 for C1, $0.80 for C2, and $1.00 for C3. A total of 100cabinets must be assembled daily. Bolting and riveting capacities per dayare 6000 and 8000, respectively. We wish to determine the number ofcomponents to be bolted and riveted to minimize the cost.

Problem Formulation

cabinet

8xC1 5xC2 15xC3

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

Findxxxxx

=

=

=

=

=

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

xSatify

x xx xx xx x xx x x

x i toMinimize

Cost x x x

=

+ =

+ =

+ =

+ + £

+ + £

³ =

= + +

4 5 6 0.80(6) 0.60(3) 1.00(3)x x x+ + +


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

Findxxxxx

=

=

=

=

=

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

xSatify

x xx xx xx x xx x x

x i toMinimize

Cost x x x

=

+ =

+ =

+ =

+ + £

+ + £

³ =

= + +

4 5 6 0.80(6) 0.60(3) 1.00(3)x x x+ + +

19

Example 4: Tubular Column Design


Problem FormulationGiven

P=10 MN , E=207Gpa, ρ=7833kg/m3, l=5.0, σa=248 Mpa

FindR, t

Satisfy


Minimize

21

Example 5: Beam Design

n A beam of rectangular cross-section is subjected to a bendingmoment of M (Nm) and maximum shear force of V (N). Thebending 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. Theallowable stress in bending and shear are 10 MPa and 2 MParespectively. It is desired to minimize cross-sectional area ofthe beam. For the numerical example, let M=40 kNm andV=150 kN.


n A beam of rectangular cross-section is subjected to a bendingmoment of M (Nm) and maximum shear force of V (N). Thebending 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. Theallowable stress in bending and shear are 10 MPa and 2 MParespectively. It is desired to minimize cross-sectional area ofthe beam. For the numerical example, let M=40 kNm andV=150 kN.

22

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

GivenM kN m V kN

Finddb

Satisfy

g b dbd

g b dbd

g b d d bb d

Minimize

= =

´= - £

´= - £

= - £

³ ³

g

( , )f b d bd=


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

GivenM kN m V kN

Finddb

Satisfy

g b dbd

g b dbd

g b d d bb d

Minimize

= =

´= - £

´= - £

= - £

³ ³

g

( , )f b d bd=

23

Example 6: Bio-sensor Designn A high performance and cost effective

biosensor is designed using a radial contour-mode disk resonator (RCDR). This sensormeasures tiny biological mass attached on adisk vibrating at a high frequency, producinghigh quality of output signal. A series ofanalysis and simulation models is developedto predict the mass sensitivity, dynamicstability, and motional resistance of theRCDR biosensor with given geometry andsignal input. In order to decrease motionalresistance while keeping the fabrication costlow, a layer of dielectric material is depositedwithin the capacitor gap

Two-port bias and excitation scheme (Clark et. al., 2005)


n A high performance and cost effectivebiosensor is designed using a radial contour-mode disk resonator (RCDR). This sensormeasures tiny biological mass attached on adisk vibrating at a high frequency, producinghigh quality of output signal. A series ofanalysis and simulation models is developedto predict the mass sensitivity, dynamicstability, and motional resistance of theRCDR biosensor with given geometry andsignal input. In order to decrease motionalresistance while keeping the fabrication costlow, a layer of dielectric material is depositedwithin the capacitor gap

24

Example 6: Bio-sensor Design

n




Goal Programming Approach (discussed later)

n Sandwich Structure Panel

Front face sheet

Example 7: Blast Resistant Panels (BRPs)


Thompson, S. C., H. Muchnick, H.-J. Choi, D. L. McDowell, J. K. Allen and F. Mistree, 2006, "Robust Materials Design of Blast Resistant Panels,” 11th AIAA/ISSMO Multidisciplinary Analysis and Optimization ConferencePortsmouth, Virginia. AIAA-2006-7005.

Front face sheet

Core

Back face sheet

Example 7: Blast Resistant Panels (BRPs)

Fleck, N.A. and V.S. Deshpande, 2004, "The Resistance of Clamped Sandwich Beams to Shock Loading". Journal of Applied Mechanics, Vol. 71, p. 386-401.

TopFacesheet (RHA)TopInterlayer(PU)Core(metalfoam)BottomInterlayer(PU)BottomFacesheet(RHA)(Lim Y.W. et al., 2012)


n Designed to dissipate blast energy through allowable plastic deformation (crushing) of the core layer

n Three layer metal sandwich panel with a square honeycomb core or metal foam core

n Design objective: minimize deflection of the back face sheet while meeting mass and failure constraints

n Robustness : minimize variance of deflection of the back face sheet due to variation in noise factors

Design Problem Formulation

(BRP with Metal Foam Core and Interlayers) TopFacesheet(RHA)TopInterlayer(PU)Core(metalfoam)BottomInterlayer(PU)BottomFacesheet(RHA)


BottomInterlayer(PU)BottomFacesheet(RHA)(Lim Y.W. et al., 2012)

Design Problem Formulation

A feasible alternative: 3 layer sandwich core panel with square honeycomb core

Assumptions:• air blast impulse pressure of exponential form• blast occurs on the order of 10-4 seconds• no strain hardening

Parameters: area of panel, geometric constants for analysis, mean and variance of noise factors

Constraints:• mass/area less than 100 kg/m2

• deflection less than 10% of span, relative density greater than 0.07 to avoid buckling

• constraints for front face sheet shear-off

Bounds: bounds on design variables and deviation variables

Given Satisfy


A feasible alternative: 3 layer sandwich core panel with square honeycomb core

Assumptions:• air blast impulse pressure of exponential form• blast occurs on the order of 10-4 seconds• no strain hardening

Parameters: area of panel, geometric constants for analysis, mean and variance of noise factors

Design variables:• material density• yield strength• layer thicknesses• cell wall thickness• cell spacing

• minimize deflection of back face sheet • minimize variance of deflection of back

face sheet

Find Minimize

Discussion

Why are your doing yourexperiments in your project?simulation in your project?modeling in your project?

How about your projects???


What is the design problem in your project?

Why are your doing yourexperiments in your project?simulation in your project?modeling in your project?

What is your final goal of your project?

Assignment

n Explain your project (at least 1 page of A1)¨ Introduction, approaches, your tasks, goals

n Identify and define a design problem in your project (at least 1 page of A1)¨ Design problem description, design problem formulation

¨ Due on 6th Oct (make-up class at 10am-1pm on the next Saturday)


n Explain your project (at least 1 page of A1)¨ Introduction, approaches, your tasks, goals

n Identify and define a design problem in your project (at least 1 page of A1)¨ Design problem description, design problem formulation

¨ Due on 6th Oct (make-up class at 10am-1pm on the next Saturday)

33

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