Transcript PowerPoint - (KAIST) 기계공학과

MAE 430
기계공학에서의 신뢰성 공학
Project 2 : Evaluation of Reliability in S-S Model
항공우주공학 / 20154359 / 김 용 운
항공우주공학 / 20155611 / 홍 승 찬
항공우주공학 / 20155352 / 최 윤 실
1
Summary for Proj. #1
Determination of PDF
Data Set #1 (n=16)
1.00
 Normal Distribution
 = 0.25
 = 0.05
Data Set #3 n=26)
1.00
 Weibull Distribution
 = 0.25
 = 0.05
0.75
0.75
0.50
0.50
0.50
0.25
μ = 263.8737
σ = 210.5263
0.00
F(X)
0.75
F(X)
F(X)
1.00
Data Set #2 n=59)
0.25
0.25
m = 1.3247
ζ = 311.7677
0.00
 Normal Distribution
 = 0.25
 = 0.05
μ = 297.9152
σ = 154.0832
0.00
X
391
130
313
450
89
16
513
30
131
631
401
545
356
108
325
423
49
236
323
454
272
191
254
11
157
134
424
238
398
430
546
498
297
190
515
314
152
151
513
30
209
369
261
470
108
18
547
175
421
488
408
242
365
179
289
347
304
81
364
99
140
93
40
256
322
279
97
601
573
336
441
207
189
319
51
90
240
123
168
281
27
561
136
369
546
139
292
154
460
428
53
163
50
324
384
446
87
463
360
123
325
Normal Dist. / Mean Rank
Weibull Dist. / Mean Rank
Normal Dist. / Rest method
2
Model Determination
Determination of Stress and Strength Model
#1
#2
#3
Stress
Set 1
Set 1
Set 2
Strength
Set 3
Set 2
Set 3
#1
#2
#3
3
Classifying analysis method
Model Determination
Given PDF
∞
𝑅=
Given Data
∞
𝑓𝜎 𝜎
−∞
∞
𝑓𝑠 𝑆 𝑑𝑆 𝑑𝜎
𝑅=
𝜎
−∞
∞
= 1−
∞
𝑓𝜎 𝜎
𝑓𝑠 𝑆 𝑑𝑆 𝑑𝜎
𝜎
∞
𝑓𝜎 𝜎 𝐹𝑠 𝜎 𝑑𝜎
=
0
𝑓𝜎 𝜎 [1 − 𝐹𝑠 𝜎 ]𝑑𝜎
0
Normal Dist. : 𝑓 𝑥 =
𝐹 𝑥 =
𝐻 = 𝐹𝜎 𝜎
1
1 𝑥−𝜇 2
exp
−
2𝜋𝜎
2
𝜎
1
𝑥−𝜇
1 + erf( 𝜎 2 )
2
Weibull Dist. : 𝑓 𝑥 = −
𝐹 𝑥 =1
𝐺 = 1 − 𝐹𝑆 𝜎
𝑚 𝑥 𝑚−1
𝑥 𝑚
exp
−
𝜉 𝜉
𝜉
𝑥 𝑚
− exp − 𝜉
𝐺
1
𝑅
𝑅=
𝐺𝑑𝐻
0
𝐻
4
S-S Model Analysis
Data Set 1 vs Data Set 3
Data Set #3 n=26), Strength
Data Set #1 (n=16), Stress
391
130
131
631
401
545
356
454
272
191
254
11
546
498
297
190
515
369
261
470
108
18
365
179
289
256
322
279
319
51
90
136
369
546
163
50
324
123
325
347
97
304
μ = 263.8737
σ = 210.5263
Normal Dist. vs Normal Dist.
From Eq. 9.1-7
𝑅 = 1−𝐹 −
μ = 297.9152
σ = 154.0832
𝜇𝑠 − 𝜇 𝜎
𝜎𝑠2 + 𝜎𝜎2
𝜇𝑠 − 𝜇𝜎
297.9152 − 263.8737
𝑧=−
=−
= −0.1308
154.08322 + 210.52632
𝜎𝑠2 + 𝜎𝜎2
∴ 𝑅 = 55.17 %
R = 55.19 % (From MATLAB)
5
S-S Model Analysis
Data Set 1 vs Data Set 3
1-FS), Fs(S)
1.0
1-FS)
Fs(S)
interpolated 1-FS)
interpolated Fs(S)
0.5
0.0
0
100
200
300
400
500
600
700
S
0.0
0.5
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4408
R=0.5592
0.8
1.0
high
파손확률-상한
1.0
1-F(S)
0.5
triangle
파손확률-삼각
1.0
1-FS)
1-FS)
low
파손확률-하한
1.0
0.5
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4444
R=0.5556
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
Fs(S)
Pf=0.4479
R=0.5521
6
S-S Model Analysis
Data Set 1 vs Data Set 2
Data Set #1 (n=16) , Stress
Data Set #2 n=59), Strength
391
130
313
450
89
16
513
30
545
356
108
325
423
49
236
323
254
11
157
134
424
238
398
430
190
515
314
152
151
513
30
209
108
18
547
175
421
488
408
242
347
304
81
364
99
140
93
40
601
573
336
441
207
189
90
240
123
168
281
27
561
546
139
292
154
460
428
53
324
384
446
87
463
360
97
μ = 263.8737
σ = 210.5263
m = 1.3247
ζ = 311.7677
Normal Dist. vs Weibull Dist.
R = 50.02 % (From MATLAB)
7
S-S Model Analysis
Data Set 1 vs Data Set 2
1-FS), Fs(S)
1.0
1-FS)
Fs(S)
interpolated 1-FS)
interpolated Fs(S)
0.5
0.0
0
100
200
300
400
500
600
700
S
0.0
0.5
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4754
R=0.5246
0.8
1.0
high
파손확률-상한
1.0
1-FS)
0.5
triangle
파손확률-삼각
1.0
1-FS)
1-FS)
low
파손확률-하한
1.0
0.5
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4783
R=0.5217
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
Fs(S)
Pf=0.4811
R=0.5189
8
S-S Model Analysis
Data Set 2 vs Data Set 3
Data Set #2 n=59) , Stress
Data Set #3 n=26) , Strength
313
450
89
16
513
30
131
631
401
108
325
423
49
236
323
454
272
191
157
134
424
238
398
430
546
498
297
314
152
151
513
30
209
369
261
470
547
175
421
488
408
242
365
179
289
81
364
99
140
93
40
256
322
279
601
573
336
441
207
189
319
51
240
123
168
281
27
561
136
369
139
292
154
460
428
53
163
50
384
446
87
463
360
123
325
m = 1.3247
ζ = 311.7677
μ = 297.9152
σ = 154.0832
Weibull Dist. vs Normal Dist.
R = 56.45 % (From MATLAB)
9
S-S Model Analysis
Data Set 2 vs Data Set 3
1-FS), Fs(S)
1.0
1-FS)
Fs(S)
interpolated Fs(S)
interpolated 1-FS)
0.5
0.0
0
100
200
300
400
500
600
700
Fs(S)
0.5
0.0
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4559
R=0.5441
0.8
1.0
high
파손확률-상한
1.0
1-FS)
0.5
triangle
파손확률-삼각
1.0
1-FS)
1-FS)
low
파손확률-하한
1.0
0.5
0.0
0.0
0.2
0.4
0.6
Fs(S)
Pf=0.4575
R=0.5425
0.8
1.0
0.0
0.2
0.4
0.6
0.8
1.0
Fs(S)
Pf=0.4591
R=0.5409
10
Summary
Summary
Set 1 (stress), Set 3 (Strength)
하한
Reliability, R
[%]
(by MATLAB)
삼각
상한
Set 1 (stress), Set 2 (Strength)
하한
55.19
삼각
상한
Set 2 (stress), Set 3 (Strength)
하한
50.02
삼각
상한
56.45
Reliability, R
[%]
55.92
55.56
55.21
52.46
52.17
51.89
54.41
54.25
54.09
Failure, Fs [%]
44.08
44.44
44.79
47.54
47.83
48.11
45.59
45.75
45.91
모델2 (Set 1 vs Set 2) 및 모델3(Set 2 vs Set 3)의 신뢰도 계산 결과, 도식적 계산결과와 PDF를 이용한
이론적 계산결과에 큰 차이를 보임 (2,44 %, -2.36 %)
: Set 2에 대한 PDF 결정에 오류가 있다고 판단
Set 2의 PDF를 Weibull Dist. 에서 Normal Dist.로 변경
1.00
 Weibull Distribution
 = 0.25
 = 0.05
0.75
0.75
0.50
0.50
F(X)
F(X)
1.00
0.25
m = 1.3247
𝜉 = 311.7677
0.00
X
0.25
0.00
 Normal Distribution
 = 0.25
 = 0.05
μ = 273.7274
σ = 180.5054
11
Summary
Summary
Set 1 (stress), Set 3 (Strength)
하한
Reliability, R
[%]
(by MATLAB)
삼각
상한
Set 1 (stress), Set 2 (Strength)
하한
55.19
삼각
상한
Set 2 (stress), Set 3 (Strength)
하한
51.42
삼각
상한
54.06
Reliability, R
[%]
55.92
55.56
55.21
52.46
52.17
51.89
54.41
54.25
54.09
Failure, Fs [%]
44.08
44.44
44.79
47.54
47.83
48.11
45.59
45.75
45.91
Set 2의 PDF를 Weibull Dist. 에서 Normal Dist.로 변경 결과
 모델2 (Set 1 vs Set 2) 및 모델3(Set 2 vs Set 3)의 신뢰도 계산 결과
 도식적 계산결과와 PDF를 이용한 이론적 계산결과 차이 대폭 감소 (1.04 %, 0.35 %)
12
Appendix
Reliability calculation with MATLAB
Set1 (Normal Dist.) Set2(Weibull Dist.) Set3(Normal Dist.)
clear all; clc;
sigma1 = 210.5263; sigma3 = 154.0832; mu1 = 263.8737; mu3 = 297.9152; m = 1.3247; xi = 311.7677;
Fs1 = @(x) 1/2*(1+erf((x-mu1)/(sigma1*sqrt(2))));
Fs2 = @(x) 1-exp(-(x/xi).^m);
Fs3 = @(x) 1/2*(1+erf((x-mu3)/(sigma3*sqrt(2))));
%% 1vs3
funfs = @(sigma) 1/(sqrt(2*pi)*sigma1)*exp(-1/2*((sigma-mu1)/sigma1).^2).*Fs3(sigma);
%% 1vs2
funfs = @(sigma) 1/(sqrt(2*pi)*sigma1)*exp(-1/2*((sigma-mu1)/sigma1).^2).*Fs2(sigma);
%% 2vs3
funfs = @(sigma) m/xi.*(sigma/xi).^(m-1).*exp(-(sigma/xi).^m).*Fs3(sigma);
R = 1-integral(funfs, 0, inf) % or 1-integral(funfs, -inf, inf); %for case1
Set1 (Normal Dist.) Set2(Normal Dist.) Set3(Normal Dist.)
clear all; clc
sigma1 = 210.5263; sigma3 = 154.0832; mu1 = 263.8737; mu3 = 297.9152; mu2 = 273.7274; sigma2 = 180.5054;
Fs1 = @(x) 1/2*(1+erf((x-mu1)/(sigma1*sqrt(2))));
Fs2 = @(x) 1/2*(1+erf((x-mu2)/(sigma2*sqrt(2))));
Fs3 = @(x) 1/2*(1+erf((x-mu3)/(sigma3*sqrt(2))));
%% 1vs3
funr = @(sigma) 1/(sqrt(2*pi)*sigma1)*exp(-1/2*((sigma-mu1)/sigma1).^2).*Fs3(sigma);
%% 1vs2
funr = @(sigma) 1/(sqrt(2*pi)*sigma1)*exp(-1/2*((sigma-mu1)/sigma1).^2).*Fs2(sigma);
%% 2vs3
funr = @(sigma) 1/(sqrt(2*pi)*sigma2)*exp(-1/2*((sigma-mu2)/sigma2).^2).*Fs3(sigma);
R = 1-integral(funr, -inf, inf)
13