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MAE430 Reliability Engineering in ME Term Project II Jae Hyung Cho 20101103 Andreas Beckmann 20156476.

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Presentation on theme: "MAE430 Reliability Engineering in ME Term Project II Jae Hyung Cho 20101103 Andreas Beckmann 20156476."— Presentation transcript:

1 MAE430 Reliability Engineering in ME Term Project II Jae Hyung Cho 20101103 Andreas Beckmann 20156476

2 Contents Project I results summary Results using theoretical probability distribution Results using the graphical procedure Conclusion 2

3 Contents Project I results summary Results using theoretical probability distribution Results using the graphical procedure Conclusion 3

4 Project I results summary Jae Hyung’s Data Set (n = 63) –Best fitting distribution: Biexponential Distribution –Best CDF estimation method: Median Rank Andreas’s Data Set (n = 59) –Best fitting distribution: Weibull Distribution –Best CDF estimation method: Symmetric S. C. D. 4

5 Contents Project I results summary Results using theoretical probability distribution Results using the graphical procedure Conclusion 5

6 Strength and Stress 6 42207371440 60225380449 96232386449 96233390461 114235393468 130239393473 132239399484 134259400490 150262405499 150268410506 159299412514 186305413527 187306414544 188315415546 194340422606 205345435  Strength 16139281428 27140292430 30151313441 30152314446 40154323450 49157325460 53168336463 81175360488 87189364513 89207384513 93209398547 99236408561 108238421573 123240423601 134242424  Stress

7 Calculation of PDF Using Wolfram Alpha 7 Strength: Biexponential CDF calculation: Median Rank ξ = 118.9061 X0 = 391.3413 Strength: Weibull CDF calculation: Symmetric S.C.D. m = 1.41297 ξ = 308.2052

8 Theoretical probability distribution 8 Strength Stress

9 Numerical Integration Using Matlab 9 >> f = @(x)(1-exp(-exp((x-391.3413)/118.9061))).*(0.000430008.*exp(-0.000304329.*( x.^1.41297)).*(x.^0.41297)); >> P_f = integral(f, 0, Inf) P_f = 0.3792 >> g = @(x)(1-exp(-(x/308.2052).^(1.41297))).*(0.000312936*exp(-0.0372099.*exp(0.0 0841*x)+0.00841*x)); >> R = integral(g, 0, Inf) R = 0.6208 Stress-basedStrength-based The two formulas yield the same result !

10 10 % f_stress_smaller0 = 0; % F_stress_smaller0 = 0; % f_stress_larger0 = 0.000430008.*exp(-0.000304329.*(x.^1.41297))*(x.^0.41297); % F_stress_larger0 = 1-exp(-(x./308.2052).^(1.41297)); % f_strength = 0.000312936.*exp(-0.0372099.*exp(0.00841.*x)+0.00841.*x); % F_strength = 1-exp(-exp((x-391.3413)/118.9061)); % term_for_R_smaller0 = (f_strength * F_stress_smaller0); % term_for_R_larger0 = (f_strength * F_stress_larger0); integrand_R_smaller0 = @(x) 0.000312936.*exp(-0.0372099.*exp(0.00841.*x)+0.00841.* x).* 0; integrand_R_larger0 = @(x) (0.000312936.*exp(-0.0372099.*exp(0.00841.*x)+0.00841.* x)).* (1-exp(-(x./308.2052).^(1.41297))); integrand_Pf_smaller0 = @(x) 0.* ( 1-exp(-exp((x-391.3413)./118.9061))); integrand_Pf_larger0 = @(x) (0.000430008.*exp(-0.000304329.*(x.^1.41297)).*(x.^0.4 1297)).* (1-exp(-exp((x-391.3413)./118.9061))); R = integral(integrand_R_smaller0, -inf, 0) + integral(integrand_R_larger0, 0, i nf) Pf = integral(integrand_Pf_smaller0, -inf, 0) + integral(integrand_Pf_larger0, 0, inf) R = 0.620773031855217 Pf = 0.379229034575878 0.6208 + 0.3792 = 1

11 Contents Project I results summary Results using theoretical probability distribution Results using the graphical procedure Conclusion 11

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20 20 Theoretical LowerUpperTriangle Theoretical LowerUpperTriangle Most conservative values

21 Contents Project I results summary Results using theoretical probability distribution Results using the graphical procedure Conclusion 21

22 Conclusion 22


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