By Dr. Robert B. Abernethy SuperSMITH Software WinSMITH Weibull Student Version Step-by-Step Tutorial using Case Studies By Dr. Robert B. Abernethy Copyrighted 2006

Problem#1:Plotting a Weibull with suspensions Produce a Weibull plot based on the following data: Note the differences between the two plots. Use Median Ranked Regression (MRR), 2. then Maximum Likelihood Estimates (MLE).

Step by Step Open WinSMITH Weibull Student Version (WSWS): double click the icon on the desktop or go to “Start--> Programs-->Supersmith Weibull Input the data, using a negative sign for suspensions. Or you can cut and paste from EXCEL.

Inputting data to SuperSMITH Then click on the “paste” bottle graphic.

to put your titles on the plot After a few seconds, SuperSMITH automatically produces your Weibull plot Now, click on “Labels” to put your titles on the plot Notice that the suspensions that were put in as a negative numbers show up as a >1750 and >5000

Customize your plot with titles Finally, click “Exit” Click here to put a label on the plot Type in your title and click the green check Do the same for the y-axis title, and put your initials in

Getting your plot into PowerPoint In order to save this plot to a PowerPoint slide click on the “printer” graphic.

Getting your plot into Powerpoint(continued) Now click on the “clipboard” graphic; Then go to a blank page in Powerpoint and “paste.”

Final Median Ranked Regression Weibull

To do an Maximum Likelihood (MLE) Weibull Click on this symbol

To do an MLE Weibull(continued) Click on this symbol

To do an MLE Weibull(continued) This is an MLE Weibull, and note the poor fit

Problem #2 The following data represents the life of precision grinder wheels measured in number of pieces produced. Fit a Weibull to this data.. Is a t0 needed?

Do a Ranked Regression as in Problem 1: Note the curvature in the Weibull plot

To do a t0 correction in WSWS Click on t0 – 3 parameter Weibull button

To do a t0 correction in WSWS (continued) Click on this button, then click On the green “check”

To do a t0 correction in WSWS (continued) . Note the improvement in r2.. from 0.764 before to 0.98 now

Problem #3 The following data represents the shear strength of brass and steel brake rivets. Do a Weibull of each… is there a significant difference between the two?

First, the brass rivet Weibull Cut and paste the data from EXCEL™, or punch in directly, then … Click here to put confidence bounds on the plot

Put confidence bounds on this Weibull Click on this button for 2-sided Confidence bounds……………...

Continuing to put Confidence bounds on a Weibull line Accept 90% confidence…….

You now have a Weibull Plot for Shear strength of Brass rivets , with 90% confidence bounds 18 110 Note that the confidence bound at 10% failure or 90% reliability is (8.7-104.7) This may be read from the plot or the “report.”

How to bring up the “Report” To see the “Report” Click on the right Tab

The Report Shows Exact Readings From The Plot Here are the exact 90% Confidence B10 Bounds which are also 90% Reliability Bounds. Other B lives may be added using the report icon. Also shown are confidence bounds for eta and beta.

Now repeat this procedure for the steel rivets 90% confidence bounds at 10% reliability are (152.9-513.1)

Declare a Significant Difference Confidence bounds for Brass rivets at 90% reliability are (8.7-104.7). Confidence bounds for steel rivets at 90% reliability are (152.9-513.1). Conclusion: Since these bounds do not overlap at the B10 level, there is a significant difference in the strength of the brass and steel rivets with 90% confidence.

Problem #4 The following data represents the pull strength of spot welds from the lab: Fit a Weibull to this data.. Anything surprising … do you believe the value of the calculated b?

Doing a Ranked Regression Weibull as before b seems high, try Lognormal and Normal.

Now try a Lognormal Click on this button

Now try a Lognormal, continued Click on this button (if you wait a second or two The name of the option will appear below it in a yellow box).

Now let’s try a Normal Click on this button

Now let’s try a Normal (continued) Click on this button

Now let’s try a Normal (continued) Click on this button, then Then “NO” on “Lower bias?”

So far, where are we? Hmmm, looks like log-normal or Normal.. But wait , look at the original Weibull… It’s a little hard to tell, but it seems like There may be curvature .. Let’s try a T0 correction.

Doing a t0 correction Click on this button

Doing a t0 correction (continued) Click on this button, and the “No” will change to “Yes”,then click the Green check

Doing a t0: The fit is better, R2=.995, so, 3-parameter Weibull is your best choice????

To Find the Best Distribution r2 is a good measure of fit but (r2 – CCC2) is more accurate as explained in Chapter 3. Remember the “Report” we used in Problem 3? It contains (r2 – CCC2) and this allows us to do an accurate distribution analysis. If we click on the tab above the plot for each distribution, the results are: Weibull 2-parameter (r2 – CCC2) = 0.1487 Weibull 3-parameter = 0.0739 Log Normal = 0.1511 Therefore from a statistical view the Log Normal best fits our data set. However, the physics of failure and prior experience may provide more information, at least equally important, as discussed in Chapter 3.

Remember the Pump Problem in Chapter 4? -1000 2000 -2000 3000 -3000 -4000 Let’s see if we can do the Abernethy Risk failure forecast to predict the number of failures on the 18 remaining pumps in the next year. To make the Weibull plot, Figure 4.1, enter the data shown in Table 4-1 and repeated here.

Here is the plot. R squared is 1. 0 because we only have two failures Here is the plot. R squared is 1.0 because we only have two failures. The expected usage for 1 year is 1000 hours for each pump or 83.3 hours per month. Select the Abernethy Risk icon.

To make a failure forecast click on the Abernethy Risk Icon which is here.

Enter Usage = 83.3 hours per month and click on the Green Check

What does this tell us. Expect 2. 3 failures in the next year What does this tell us? Expect 2.3 failures in the next year. Expect the next failure in four months. The Now Risk is 2.5, close to the observed number 2, so a batch problem is not indicated. In five years expect 14 failures.

Summary We have illustrated how to input data, failures and suspensions, obtain MRR and MLE plots, add confidence bounds, do a distribution analysis and a failure forecast. We hope we have helped introduce you to Weibull Analysis and we would be pleased if you would send us your questions and/or comments. Bob Abernethy …..weibull@att.net