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Published byAdam McBride Modified over 9 years ago
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Project II Team 9 Philippe Delelis Florian Brouet 이성혁
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DATA 1DATA 2DATA 3 µ1= 285.095µ2= 297.962µ3= 262.552 σ = 188.796σ = 148.135σ = 178.233 Stress Strentgh Because DATA 2 > DATA 1 > DATA 3
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Data 1&2 : Project 1 Results Data set 1 (N = 21) Normal Distribution µ = 288.696 σ = 217.391 Data set 2 (N = 26) Weibull Distribution m = 1.9055 ξ = 340.52 Weibull Normal
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Data 1&2 : Project 2 Analysis Use of the equation 9.1.1 Strength : Weibull Distribution Stress : Normal Distribution
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x=[0:1:1000]; Fsig=0.5*(1+erf((x-288.696)/(217.391*sqrt(2)))); Fs=1-exp(-(x/340.52).^1.9055); fsig=diff(Fsig); fs=diff(Fs); plot([fs, fsig],'r‘) Data 1&2 : Using Matlab Stress Strength R= 0.5091
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Data 1&2 :
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Data 1&2 : Original Graph Pf = 0.4153 R = 0.5847
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Data 1&2 : Triangle Method Pf = 0.4486 R = 0.5514
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Data 1&2 : Upper Limit Pf = 0.4548 R = 0.5452
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Data 1&2 : Lower Limit Pf = 0.4424 R = 0.5576
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Conclusion Data 1&2 Matlab values Reliability Calculation Method TriangleUpperLower Probability of Failure (Pf) 49.09%44.86%45.48%44.24% Reliability (R)50.91%55.14%54.52%55.76% R = 0.5091
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Data 2&3 : Project 1 Results Data set 2 (N = 26) Weibull Distribution m = 1.9055 ξ = 340.52 Data set 3 (N = 29) Normal Distribution µ = 262.78 σ = 180.17 Weibull Normal
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x=[0:1:1000]; Fsig=0.5*(1+erf((x-262.78)/(180.17*sqrt(2)))); Fs=1-exp(-(x/340.52).^1.9055); fsig=diff(Fsig); fs=diff(Fs); plot([fs, fsig],'r‘) Data 2&3 : Using Matlab Stress Strength R= 0.5497
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Data 2&3 :
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Data 2&3 : Original Graph… Pf = 0.4298 R = 0.5702
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Data 2&3 : Triangle Method Pf = 0.4586 R = 0.5414
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Data 2&3 : Upper Limit Pf = 0.4611 R = 0.5389
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Data 2&3 : Lower Limit Pf = 0.4561 R = 0.5439
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Conclusion Data 2&3 Matlab values Reliability Calculation Method TriangleUpperLower Probability of Failure (Pf) 45.03%42.98%46.11%45.61% Reliability (R)54.97%57.02%53.89%54.39% R = 0.5389
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