performance analysis and fleet maintenance … · performance analysis and fleet pagina 14...

PERFORMANCE ANALYSIS AND FLEET

MAINTENANCE DEVELOPMENT FOR ITALIAN

PUBLIC TRANSPORT COMPANY (FS-BUSITALIA)

Student: Naraharisetti Prabhu Rayudu

Matricola: 1794230

Relatore: Prof. Lorenzo Fedele

Content

Historical data

Reliability modelling Weibull distribution

Output highlighting

MTBF

Equipment trend recognition

Maintenance decision

27/03/2020 Pagina 2PERFORMANCE ANALYSIS AND FLEET


PUBLIC TRANSPORT COMPANY (FS-

BUSITALIA)

Why I choose only the 3 components

After analyzing the past failures

System and equipment reliability prioritization was done by taking the below in to consideration

1.Operational Cost (A)

2.Throughput Process (B)

3.Safety (C)

Here the safety is taken as the major aspect to be taken into consideration

PERFORMANCE ANALYSIS AND FLEET



BUSITALIA)

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𝑆𝑦𝑠𝑡𝑒𝑚 𝑎𝑛𝑑 𝑒𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 𝑟𝑒𝑙𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝐷

=𝐴2 + 𝐵2 + 𝐶2

12

4

Other aspects

• Operations criticality analysis

• Asset criticality ranking

• Asset failure probability factor (AFPF)

After taking the above aspects in to consideration the main components that are to be analyzed

are the below:

1. Engine

2. Brake

3. Suspension

Pagina 4PERFORMANCE ANALYSIS AND FLEET



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27/03/2020

ENGINEMercedes Benz Model OM 936 engine

OUTPUT [KW] TORQUE [Nm]




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Engine • Problem statement

Here we can see that the defects occur just before and after of

the maintenance.

0 39

37

39

90

70

35

45

53

7

55

62

5

10

28

59

15

48

78

15

74

87

16

27

13

16

42

08 18

57

03

18

57

03

23

49

63

0100002000030000400005000060000700008000090000

100000110000120000130000140000150000160000170000180000190000200000210000220000230000240000250000

ST

AR

T D

AT

E

25 A

pr

201

6

02 M

ay 2

01

6

14 M

ay 2

01

6

21 N

ov 2

01

6

22 A

ug

20

16

29 M

ay 2

01

7

31 J

ul 2

01

8

31 J

ul 2

01

8

30 A

ug

20

18

18 S

ep

20

18

28 D

ec 2

01

8

28 D

ec 2

01

8

06 M

ar

20

19

The brown

lines are PM

scheduled at

(60000KMS

interval)




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Percentile Report:Percentage Percentile

(Time)

0.1 0.189591

0.135 0.244983

0.5 0.750237

1 1.359

5 5.462

10 10.09671209

25 23.798

50 50.413

75 91.097

90 140.48

95 175.86

99 253.84

99.5 286.11

99.865 345.47

99.9 358.82

-3,46

-2,96

-2,46

-1,96

-1,46

-0,96

-0,46

0,04

0,54W

EIB

UL

L S

CO

RE

Ln (Time-to-Fail)

Weibull Analysis: Time-to-Fail

Distribution

Characteristics:Estimate

Mean (MTTF) 65.258




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Preliminary analysis




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Fig.6 Pressure vs Crank angle

47.5849.5

Main analysis




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Some corrections still

needed

SUSPENSION




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BRAKE




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y = -0,0002x + 27,667R² = 0,9061

0

5

10

15

20

25

30

0 10000 20000 30000 40000 50000 60000 70000 80000

MILEAGE VS THICKNESS

Mileage Thickness

1000 27

2000 28

3000 26

4000 26

5000 27

6000 28

7000 26

8000 28

U = Total time under use

t = time at which wear is measured

ΔD= degree of wear

R= brake pad thickness remaining

• X-mileage

• Y-thickness

• C- coefficient of intercept

• M- coefficient of slope

𝑌 = 𝑀𝑋 + 𝐶

𝑊𝑒𝑎𝑟 𝑟𝑎𝑡𝑒 = 𝑊𝑒𝑎𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 / 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑎𝑓𝑡𝑒𝑟 𝑏𝑟𝑒𝑎𝑘

COST

BENEFIT

ANALYSIS




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Cost benefit analysis-One time maintenance

X = total cost for one-time maintenance

• C(pm) = cost for the preventive maintenance

• C(cm) = cost of corrective maintenance

• R(B) = reliability at time B

• B= time interval for preventive maintenance

• C= mean time to perform preventive maintenance

• D= mean time required to conduct corrective maintenance

• t= MTTF

• E= mean time for corrective and preventive maintenance. (MTTR)

FAILURE

REPLACEME

NT

PREVENTIVE

MAINTENANCEFAILURE

REPLACEME

NT

A B

C DE

CYCLE 1 CYCLE 2

t




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𝑋 =𝐶 𝑝𝑚 ∗ 𝑅 𝐵 + 𝐶 𝑐𝑚 ∗ 1 − 𝑅 𝐵

𝐵 + 𝐶 ∗ 𝑅 𝐵 + 0𝐵𝑡𝑓 𝑡 𝑑𝑡 + 𝐸 ∗ 1 − 𝑅 𝐵

If no maintenance is performed




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I (n) = expected time between breakdowns

I (n) = ∑ nx probability of breakdown,

where

n = time period when a breakdown may occur,

❑ x = probability of a breakdown in period n

= Number of machine breakdowns in period n / Total number of breakdowns in total

❑ Expected number of breakdowns = 1

𝐼 𝑛

❑ Expected cost of letting Y identical machines breakdown, CT is:

a= cost of repairing a break down machine

E= mean time for corrective and preventive maintenance. (MTTR)

[So finally, the cost of letting Y machines breakdown = cost, if no maintenance is

performed]

𝑪𝑻 = (𝒀 ∗ 𝒂) / (𝑬 (𝒏))

Final formula




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[𝑵𝒆𝒘 𝒑𝒓𝒆𝒅𝒊𝒄𝒕𝒊𝒗𝒆𝒎𝒂𝒊𝒏𝒕𝒆𝒏𝒂𝒏𝒄𝒆 𝒄𝒐𝒔𝒕

= 𝑪𝒐𝒓𝒓𝒆𝒄𝒕𝒍𝒚 𝒑𝒓𝒆𝒅𝒊𝒄𝒕𝒆𝒅

× 𝑵𝒆𝒘 𝒓𝒆𝒑𝒂𝒊𝒓 𝒄𝒐𝒔𝒕 – 𝑾𝒓𝒐𝒏𝒈𝒍𝒚 𝒑𝒓𝒆𝒅𝒊𝒄𝒕𝒆𝒅

× 𝑪𝒐𝒔𝒕 𝒐𝒇 𝒓𝒆𝒑𝒂𝒊𝒓 − 𝑨𝒎𝒐𝒖𝒏𝒕 𝒊𝒏𝒗𝒆𝒔𝒕𝒆𝒅]

Each correct prediction saves the additional cost and each

wrong prediction adds additional cost to the budget.

performance analysis and fleet maintenance … · performance analysis and fleet pagina 14...

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