Exercise 5: Quality management in field data evaluation

Transcription

1 Exercise 5: Quality management in field data evaluation 5 Quality management in field data evaluation 5. Weibull-analysis 5. Median-Rank-Technique 5.3 Extrapolation in the Weibull-diagram Page Contact: Dipl.-Ing. Carsten Scharrenberg Fraunhofer IPT, Room 7 Phone carsten.scharrenberg@ipt.fraunhofer.de References - Pfeifer, T.: Qualitätsmanagement: Strategien, Methoden, Techniken. 3.Auflage; Carl Hanser Verlag; München, Wien, 00 - N.N.: Qualitätskontrolle in der Automobilindustrie, Band 3: Zuverlässigkeits sicherung bei Automobilherstellern und Lieferanten.. Auflage; VDA Verband der Automobilindustrie e.v.; Frankfurt/Main, Pfeifer, T.: Praxishandbuch Qualitätsmanagement.. Auflage; Carl Hanser Verlag, München, Wien, 00

2 Distribution of mileage cumulative frequency H (% figure : distribution of mileage < 6 distance [000km] Page Also known to the manufacturer by comparison with a similar truck is the distribution of mileage for one-year-old trucks. For example. % of all vehicles have travelled 0,000 km or less. 50% of all vehicles have travelled 30,000km or less.

3 Solution ( Determine the cumulative frequency for each category of mileage H S c.f. figure : distribution of travelled distance Determine the individual frequency of each category of mileage H E H E = H S - H S - for mileage category t = km: H E (t = H S (t - H S (t 0 H E (t = 0.035% - 0% = 0.035% for mileage category t = km: H E (t =.7% % =.665% Page 3 Solution The distribution of mileage is a cumulative function. This means that the travelled distances of the respective vehicles are accumulated. The distribution of mileage specifies, which fraction of the vehicles has not exceeded a certain distance. According to the classification of the failure statistics the percentual fraction of the vehicles in the respective distance categories must be determined. In the next step the individual frequencies of the distance categories are determined. For the further solution the allocation of the number of the non defective vehicles to the corresponding distance category t j is required. Therefore, first the percentual fraction of the non defective vehicles in the distance category t j is determined. This percentage is called individual frequency distance category H E. To their calculation the cumulative frequency of the previous distance category t j- is subtracted from the cumulative frequency of the distance category t j. 3

4 Solution ( Determine the number of undamaged parts per category n undamaged : n undamaged = H E * n undamaged n undamaged = n - n damaged = 376 for mileage category t = km: n undamaged (t = H E (t * n undamaged = * 376 for mileage category t = km: n undamaged (t = H E (t * n undamaged = * Page 4 From the distribution of mileage the number of undamaged vehicles n undamaged in each individual category can be calculated. This calculation is based on the whole amount of all vehicles n total. The number of the damaged vehicles in the distance categories must be subtracted from n total. 4

5 Solution (3 Interim result Category of mileage t j [km]. Cumulative frequency per category of mileage H [%]. Individual frequency per category of mileage H E [%] 3. Number of parts not damaged per category of mileage n undamaged table : Number of undamaged parts per category of mileage Page 5 5

6 Solution (4 Calculating the mean ordinal number j : c.f. equations ( and ( for category of mileage t = km: n + j( t N(t = = =.00 + n number of previous parts ( + 0 j(t = j( t + ndamaged( t N( t 0 = * = 5.0 Page 6 Evaluating field failures, it is required to consider the vehicles which still show no damage on account of their very low travelled distance as well. However, these vehicles with low travelled distance can still break down and then would have to be considered in the failure statistics stated above. If for example, a vehicle which travelled a distance of 7,000 km was registered in the distance distribution, this vehicle can absolutely break down, before it has reached the travelled distance of the vehicle with the highest travelled distance. The problem occurs, if the lifespan characteristic feature (here: travelled distance, km does not correspond to the lifetime t. However, the not broke down vehicles contain information which must be considered with the statistic processing of the data. 6

7 Solution (5 For category of mileage t = km: N(t = ( =.07 j(t = 5 + ( *.07 = 7.03 Page 7 7

8 Solution (6 Determining the cumulative frequency H j for the failure distribution: c.f. equation (3 for category of mileage t = km: j( t H (t = = = 0.% n for category of mileage t = km: H (t = = 0.7% Page 8 The cumulative frequency Hj(tj for the failure distribution is calculated using the approximation formula for the Median-Rank-Technique. 8

9 Solution (7 Interim result Category of distance travelled t j [km] Number of damaged parts n damaged Number of undamaged parts per category of distance travelled n not damaged Increase N Mean ordinal number j Cumulative frequency, breakdown H j [%] Table 3: Cumulative frequencies per category of distance travelled for the breakdown distribution Page 9 9

10 Solution (8 Entering the cumulative frequency of the categories of mileage H j in the life expectancy grid (figure lifetime curve composed of two branches. part: failure due to early failure. part: failure due to random failure and wear failure long-run behaviour is affected by wear failure Extrapolation extrapolation of the second straight line intersection point with t = km cumulative frequency H j (0 5 km = 0% n damaged (0 5 km = 0. * 3780 = 378 Page 0 Now the determined cumulative frequencies of the categories of mileage H j can be plotted over the respective upper limits of the categories of mileage. 0

11 Weibull-diagram 99, cumulative frequency H (% 63, figure : Weibull-diagram mileage [000km] Page

12 Extrapolation 99, cumulative frequency H (% 63, figure 3: Extrapolation mileage [000km] Page Extrapolation To estimate the expected failure behaviour, the curve must be extrapolated. However, prerequisite for such prognosis is that thorough experiences with the same or similar products are present. With the extrapolation of the straight lines must be considered that a breaking off of the curve is possible on account of increasing wear failures.