Reliability Eng-Part6.pdf

8/10/2019 Reliability Eng-Part6.pdf

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y = mx + b is the well-known equation for a straight line. Hence, plotted on Weibull (log-log) probability paper, the data will be depicted as a straight line with slope and intercept - ln( ).

Probability Plotting Techniques

Most probability graph papers are based on plots of a variable of interest versus the cumulative percentage probability. To accomplish this, the data needs to be ordered and the cumulative probability calculated. The x- and y-axes of probability plotting papers are designed in such a waythat the true cumulative density function plots as a straight line.

Seven steps of constructing a probability plot of complete data

1. Order the data from the smallest to the largest.

2. Assign a rank to each data point.

3. Calculate probability plotting positions %1005.0 = n

iF i i = 1, 2,...n

4. Plot the data on the appropriate probability paper e.g. Normal, Log-normal, Weibull, Chi-square, etc. probability paper. Plot each data point against its time or cycle on the datascale and against its plotting position on the cumulative probability scale.

5. Draw a straight line through the plotted data as best as you can. The line estimates thecumulative distribution function.

6. Assess the data and assumed probability distribution. If the plotted points follow a straightline, the chosen distribution appears to be valid.

7. Obtain from the graph the desired reliability information.

For suspended or censored data (incomplete data) we must first calculate Adjusted Ranks ( j)using the following formula:

( )( )( )

( )1

1Pr +

++==

Rank InverseNRank AdjustedeviousRank Inverse

Rank Adjusted j

To calculate the probability plotting position we use the Median Rank approximation:

( )( )4.0

3.0+=

N j

Rank Median

Reliability Eng Part 6 2-6 Compiled by: [email protected]


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Example 6.1

A fatigue stress test was performed on an automotive component. Ten units were tested to failure.The failure times in hours were as follows: 1500, 850, 2520, 2380, 1350, 2000, 2400, 1520, 2050,and 1890 hours.a) Rank the data and calculate the plotting position of each data point.

b) Plot the failure times on Weibull probability paper. Is the Weibull model valid?c) Determine the slope and assess the failure mode e.g. early life, random, or wear-out?d) What is the characteristic life?e) What is the B 5 life?f) What percentage of the components is expected to fail before 2000 hours?g) What is the estimated reliability of the components at 3000 hours?

Time to Failure(Hours)

Rank Order(i)

Plotting Position

%1005.0

=

Ni

F i

850 1 51350 2 151500 3 251520 4 351890 5 452000 6 552050 7 652380 8 752400 9 852520 10 95

Answers:

a) See table. b) See graph 6.1. The data tend to follow a straight line. Hence, the assumed Weibull model isvalid.c) To estimate the slope we draw a straight line from the Estimation Point located at the upperleft corner of the probability paper, perpendicular to the Weibull plot line. The point at which thisline intersects with the scale (shown on top of plotting paper) indicates the slope of the Weibull

plot. We find that 3.4 which implies a wear out failure mode.d) The intersection of the dotted estimator line and the Weibull plot line pinpoints thecharacteristic life. We find approximately 2050 hours.e) The B 5 life is approximately 850 hours. It is the age/time at which five percent of thecomponents are expected to fail.f) The graph indicates that 60 percent of the component population can be expected to fail before2000 hours.g) The estimated reliability of the components at 3000 hours is:

R (3000) = 1 F (3000) = 1 0.975 = 0.025 or 2.5 percent.



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Example 6.2A sample of nine electrical devices was tested. Six failed. The test was suspended on three items.The data is shown (ranked low to high) in the table below.

Time tofailure(Hours)

Failureor

Suspension

InverseRank

Adjusted Rank

)( )( ) (( ) 1.

1....+

++=R Inv

NR APR Inv j

Median Rank

( )( )

( )%1004.03.0

+

N j

120 F 9 1 ( )( )

( ) 5.71004.093.01 =

+

400 F 8 2 ( )( ) 1.181004.9

3.02 =

650 F 7 3 ( )( ) 7.281004.9

3.03 =

700 S 6 Suspended Not Plotted


920 F 4 ( )( ) 4.414

1034 =+

+ ( )( ) 6.431004.9

3.04.4 =


2200 F 2 ( )( ) 27.63

104.42 =+ ( )( ) 5.631004.9

3.027.6 =

6400 F 1 ( )( ) 14.82

1027.61 =+ ( )( ) 4.831004.9

3.014.8 =

a) Calculate the adjusted ranks and convert the adjusted ranks into median ranks. b) Plot the failure times and median ranks on Weibull probability paper. Is the model adequate?c) Determine the slope. Assess the failure mode e.g. early life, random, or wear-out?d) What is the characteristic life? What is the B 10 life?e) What is the estimated reliability of the components at 5000 hours?

Answers:a) See table. Note that suspended items affect rank numbers only after a suspension occurs.Earlier failure times have unadjusted rank numbers.

b) See graph 6.2. The data tend to follow a straight line so we may conclude that the Weibullmodel is adequate.c) To estimate the slope we draw a straight line from the Originpoint, located at the top of the

probability paper, parallel to the Weibull plot line. The point at which the parallel line intersectswith the scale (shown on the left side of the paper) indicates the slope of the Weibull plot. Wefind that 0.83 which implies an early life failure mode.d) The intersection of the dotted estimator line (63 rd percentile) and the Weibull plot line marksthe characteristic life. We find approximately 2500 hours. The B 10 life is approximately 160hours. It is the age/time at which ten percent of the devices are expected to fail.e) The estimated reliability of the electrical devices components at 5000 hours is:

R (5000) = 1 F (5000) = 1 0.82 = 0.18 or 18 percent.



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Note: If a straight line does not fit the data well, we can try a three parameter Weibull, or we mayconsider other distribution models such as the Normal or Log-normal distribution.

Graph 6-2


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Reliability Eng-Part6.pdf