Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07.

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Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07

Transcript of Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07.

Page 1: Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07.

Lin Chen Tom Nierodzinski

Yan Di Lv Zhongyuan

Optimum Sensitivity AnalysisMAE 55012/10/07

Page 2: Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07.

OutlineObjectiveProblem FormulationResults AnalysisConclusion

Page 3: Lin Chen Tom Nierodzinski Yan Di Lv Zhongyuan Optimum Sensitivity Analysis MAE 550 12/10/07.

Objective

Better understand OSA by comparing different parameters for the same design problem

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Problem FormulationParameters Chosen:

P = StressE = Young’s Modulusσ = allowable stressy = deflection

Design Variables:bi and hi i = 1,5

(Total of 10 design variables)

(Total of 21 constraints)

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D.V's Value (cm)

b1 3.13118

b2 2.88012

b3 2.57744

b4 2.20456

b5 1.74969

h1 62.6236

h2 57.6024

h3 51.5488

h4 44.0911

h5 34.9915

Problem Formulation cont.DOT results of optimum

point

Active constraintsg4

g5

g6

g7

g8

g9

g10

g21

Inactive constraints Valuesg1 -0.1275g2 -0.1031g3 -0.6138g11 -2.1311g12 -1.8801g13 -1.5774g14 -1.2046g15 -0.7497g16 -57.624g17 -52.602g18 -46.549g19 -39.091g20 -29.992

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OSA AnalysisLambda values

i Value

4 2855 14136 2097 1928 1729 145

10 10321 20582

* BTB 1BTF(X *)

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OSA AnalysisMatrix dimensions for OSA

A1010 B108

BT 810 088

X101

81

c101

d81

0

)()(22

XgXX

XFXX

A jkiJj

jki

ik

)(XgX

B ji

ij

Jj

)()(22

XgpX

XFpX

c jiJj

ji

i

)(Xgp

d jj

Jj

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Stress(P) = 50,000 N

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Stress(P) = 50,000 NActive to inactive

∆p = 5.6*10^3

Inactive to Active∆p = 3.52*10^3

Minimum %7%

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Young’s Modulus = 2x107 Pa

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Young’s Modulus = 2x107 PaActive to inactive

∆p = 1.56*10^6

Inactive to Active∆p = 7.47*10^5

Minimum %3.7%

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Sigma = 14,000 N/cm2

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σ = 14,000 N/cm2

Active to inactive∆p = 3.64*10^3

Inactive to Active∆p = 345

Minimum %2.5%

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Y(deflection) = 2.5 cm

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Y(deflection) = 2.5 cmActive to inactive

∆p = 0.19

Inactive to Active∆p = 0.0924

Minimum %3.69%

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ConclusionOSA is limited in minimum delta pIn this case inactive constraints are more

sensitive

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Questions?