Weibull modulus and probabilistic design
0:00 variation in data, histograms of data, average values, standard deviation
9:00 normal distribution, six sigma
10:31 safety factors in engineering design
13:47 the 2-parameter Weibull equation
16:40 defining failure probability, F
18:40 excel example of calculating Weibull modulus
27:00 determining Weibull constants from best fit curves
30:55 using Weibull to determine stress associated with specific failure rate
35:35 using Weibull to determine failure rate at a specific stress
41:48 extending Weibull analysis from test bar to components using effective area ratios
Пікірлер: 6
Great lecture, thanks, helps put things in perspective, thanks
Great explanation of Weibull modulus, Thank you
@TaylorSparks
4 жыл бұрын
Glad it helped!
Great Explanation!! Thanks! Just one comment. I guess people get confused @ 28:15 with the minus sign because it is inside the parenthesis so it is still affected by the exponent m, and you cannot flip it. The equation should be with the minus sign out of the parenthesis and then you can flip it over.
@TaylorSparks
2 жыл бұрын
Oh shoot! Good catch!
Hi Taylor, I am working on fibre matrix composite FE model with Weibull distribution. The idea is each fibre element is assigned with a strength following a Weibull distribution. According to the Weibull function, firstly we have to do a number of single fibre tensile tests to get strengths and probability, so that Weibull parameters can be known . The problem is there are two lengths so called fibre gauge length and fibre length, respectively, in the function and I am quite confused about them. And imagine now we have the right Weibull parameters, we need to calculate the fibre tress (not Weibull scale parameter) with the Weibull function.Here, I think the fibre gauge length is the size of each fibre element, how about the fibre length? F=1-exp(-(l/l_0)*(sigma/sigma_0)^m) l is fibre gauge length l_0 is fibre length. 1. With this equation I will get Weibull parameters. 2. With Weibull parameters from above and this equation, a number of fibre strength (sigma) are obtained for each fibre element. Thank you!