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Probability of Failure

Submitted by David A Bennett on

I think this is the appropriate place for this. What I am trying to determine is a "probability of failure" based upon a von Mises stress and/or displacement. I am currently breaking the ultimate tensile strength into about 20 divisions.  I divided the ultimate tensile strength by the safety factor. From this I used the Normal Distribution, and determined where 50, 55, 60, etc 'percent of failure' occured at. I used the 'percent of failure' values to determine what 'x' on the normal distribution this occured at. Then, I created standard deviations based upon the 20 divisions. The table I created can be viewed below. 

 

  Percent 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
Max str (MPa) 0.00 7.05 14.11 21.16 28.21 35.26 42.32 49.37 56.42 63.48 70.53 77.58 84.63 91.69 98.74 105.79 112.85 119.90 126.95 134.00 141.06
Normal x pos -10.00 -1.645 -1.282 -1.036 -0.842 -0.674 -0.524 -0.385 -0.253 -0.126 0.00 0.126 0.253 0.385 0.524 0.674 0.842 1.036 1.282 1.645 10.00
Standard Deviation -5.000 -4.500 -4.000 -3.500 -3.000 -2.500 -2.000 -1.500 -1.000 -0.500 0.00 0.500 1.000 1.500 2.000 2.500 3.000 3.500 4.000 4.500 5.000
0.239 0.262 0.285 0.310 0.335 0.361 0.388 0.416 0.444 0.472 0.500 0.528 0.556 0.584 0.612 0.639 0.665 0.690 0.715 0.738  

 

 

the max stres values were created by dividing max tensile stress by 20.

x pos was created by used the normsdist command in excel

standard deviation was created by taking that Max stress value and subtracting it by max stress @ 50% and dividing by the [(max tensile divided by safety factor) or resulting stress]

the bottom row was created by using the normdist comman in excel.

the bottom row values were used to calculate the probability of failure. if the stress fell between the max stress values then it used the bottom row to calculate the probability of failure. i would like to know if i have any errors in my logic or if you know of a better way to do this, please let me know.