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4020-F2011-HW14Solutions

# 4020-F2011-HW14Solutions - MSE 4020 HW14-SOLUTION Problem...

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1 MSE 4020 HW14-SOLUTION 12/04/2011 Problem 1. (a) & (b) ! A n " N = C 0 # n " log ! A + logN = logC 0 # logN = \$ n " log ! A + logC 0 From given picture, we can extract the data listed in the table below, N(cycles) 10 5 10 6 10 7 10 8 5 10 8 σ A (ksi) 43.8 35.7 28.7 23.2 21.0 log(N) 5 6 7 8 8.699 log( σ A ) 1.6415 1.5527 1.4579 1.3655 1.3222 Fitting the data of log(N) vs. log( σ A ), we have following linear equation (and the corresponding plot): y = -11.3 x +23.54 Thus, n = 11.3 Also, logC 0 = 23.54 ! C 0 = 10 23.54

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2 (c) The curves asked are plotted below in different colors:
3 Problem 2. According to the Weibull equation, F x ( ) = 1 ! exp ! x " # \$ % & ' ( k # \$ % % & ' ( ( ) ln ! ln 1 ! F x ( ) ( ) ( ) = k * ln x ( ) ! k * ln " ( ) where F x i ( ) = i ! 0.3 n + 0.4 i is the rankingindex,andnis the total number of experiments (a) For A320 bolts in seawater, we have the following data, i x or N F(x) ln(N) ln(-ln(1-F(x)) 1 47780 0.12963 10.77436 -1.974458694 2 47890 0.31482 10.77666 -0.972686141 3 54360 0.5 10.90338 -0.366512921 4 55730 0.68519 10.92827 0.144767396 5 62780 0.87037 11.04739 0.714455486 Fitting the data of ln(-ln(1-F(x)) vs. ln(N), we have the following linear equation(and the corresponding plot): y = 8.4078 x – 92.019

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4 i.e., k = 8.4078 Also, ! k " ln # ( ) = ! 92.019 \$ # = exp 92.019 8.4078 % & ' ( ) * = 56640 cycles (b) Similarly, for A320 bolts in air, we have the following data, i x or N F(x) ln(N) ln(-ln(1-F(x)) 1 42100 0.15909 10.6478 -1.752894273 2 42830 0.386364 10.66499 -0.716717249 3 44420 0.613636 10.70145 -0.050266149

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