student tutorial.ppt

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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.

1. Use Median Ranked Regression (MRR),2. then Maximum Likelihood Estimates (MLE).

Failure Time

(cycles) Status

1500 Failure

1750 Suspension

2250 Failure

4000 Failure

4300 Failure

5000 Suspension

7000 Failure



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 pastefrom EXCEL.



Inputting data to SuperSMITH

Then click on the “paste”

bottle graphic.



Notice that the suspensions that were put in as a negativenumbers show up as a >1750 and >5000

After a few seconds, SuperSMITH automatically

produces your Weibull plot

Now, click on “Labels”

to put your titles on the plot



In order to save this plot to a PowerPoint slide click on the

“printer” graphic.

Getting your plot into PowerPoint



Now click on the “clipboard” graphic;

Then go to a blank page in Powerpoint and “paste.”

Getting your plot into Powerpoint(continued)



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



To do a t0 correction in WSWS

Click on t0 – 3 parameter Weibull button



.

Note the

improvementin r2.. from 0.764

before to 0.98 now

To do a t0 correction in WSWS (continued)



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…….



To see the “Report” Click on the right Tab

How to bring up the “Report”



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% reliabilityare (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.



b seems high, try Lognormal and

Normal.

Doing a Ranked Regression Weibull as before



Now try a Lognormal

Click on this button



Now let’s try a Normal





Now let’s try a Normal (continued)



Doing a t correction



Doing a t0 correction


Doing a t correction (continued)



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,

R 2=.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 fitsour 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

-1000

2000

-2000

-2000

-2000

-2000

3000

-3000

-3000

-3000

-3000

-4000

-4000

-4000

-4000

-4000

Let’s see if we cando the Abernethy

Risk failure forecast

to predict the

number of failureson 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. 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

Wh t d thi t ll ? E t 2 3 f il i th t



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 WeibullAnalysis and we would be pleased if you would send us

your questions and/or comments.

• Bob Abernethy …[email protected]

student tutorial.ppt

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