reliability centered maintenance

Reliability Centered

Maintenance (RCM)

Evolution of Maintenance

At the very beginning, Maintenance was an appendix to Operations / Production:

It existed only to fix failures, when they happened.These were the days of absolute

Corrective Maintenance

http://www.supertrafego.com/cliparts_graficos_mostra.asp?interno=profissoes_i&numarq=0026&cat=Profiss%C3%B5es&pagina=3&subcat=Profissoes%20I&internopos=Profissoes%20I

As times went by, it was detected that many failures have an almost regular pattern, failing after an

average period. Therefore, one could choose regular intervals to fix the equipment BEFORE the failure:

Preventive MaintenanceAlso know as Time Based Maintenance.


http://www.supertrafego.com/cliparts_graficos_mostra.asp?interno=profissoes_i&numarq=0039&cat=Profiss%C3%B5es&pagina=4&subcat=Profissoes%20I&internopos=Profissoes%20I

However, very often these failures happen in irregular periods. To avoid an unwanted failure, the periods of Preventive Maintenance are shortened. If equipment

conditions were known, the maintenance could be later. Technology development enabled to identify failure

symptoms:Predictive Maintenance

Also know as Condition Based Maintenance.


Many pieces of equipment have sporadic activity (alarms, stand-by equipments, etc.). However, we must be sure that they are ready to run. These are "hidden faults“. Detect and

prevent hidden failure is called:

Detective Maintenance



The different failure modes mean that there’s not one only approach, about Corrective, Preventive or Predictive Maintenance Programs.

The correct balance will give in return better equipment reliability, thus the name:

Reliability Centered Maintenance

Remember, my kid, Prevention is better than

Cure....

Take it easy, grandma, not

always!

Reliability Centered Maintenance (RCM)

John Moubray 1949-2004

After graduating as a mechanical engineer in 1971, John Moubray worked for two years as a maintenance planner in a packaging plant and for one year as a commercial field engineer for a major oil company.

In 1974, he joined a large multi-disciplinary management consulting company. He worked for this company for twelve years, specializing in the development and implementation of manual and computerized maintenance management systems for a wide variety of clients in the mining, manufacturing and electric utility sectors.

He began working on RCM in 1981, and since 1986 was full time dedicated to RCM, founding Aladon LCC, which he led until his premature death in 2004.

John Moubray is today considered a synonym of RCM.

Reliability Centered Maintenance (RCM) Its origins

What about a failure rate of 0.00006/event?

Quite good, no?

This was the average failure rate in commercial flights

takeoffs, in the 50’s. Two thirds of them caused by

equipment failures.

Today, this would mean 2 accidents per day, with

planes with more than 100 passengers!!!

That’s why Reliability Centered Maintenance has begun

in the Aeronautical Engineering. Pretty soon, Nuclear

activities, Military, Oil & Gas industries also began to

use RCM concepts and implement them in their

facilities.

Reliability Centered Maintenance (RCM) Reliability and Availability

Bibliography: Kardec, Alan y Nascif, Julio - Manutenção- Função Estratégica, Editora Qualitymark

Reliability

Reliability is a broad term that focuses on the ability of a product to perform its intended function. Mathematically speaking, reliability can be defined as the probability that an item will continue to perform its intended function without failure for a specified period of time under stated conditions.

Reliability is a performance expectation. It’s usually defined at design.

Availability

Depends upon Operation uptime and Operating cycle.

Availability is a performance result. Equipment history will tell us the availability.



Availability = MTBF

MTBF + MTTR

MTBF = Mean Time Between Failures

MTTR = Mean Time To Repair

A first definition:

Reliability Centered Maintenance (RCM) Availability definitions

Inherent Availability = MTBF

MTBF + MTTR

MTBF = Mean Time Between Failures

MTTR = Mean Time To Repair

MTBM = Mean Time Between Maintenance actions

M = Maintenance Mean Downtime (including preventive and planned corrective downtime)

Inherent Availability: consider only corrective downtime

Achieved Availability: consider corrective and preventive maintenance

Operational Availability: ratio of the system uptime and total time

Achieved Availability = MTBM

MTBM + M

Operational Availability = Uptime

Operation Cycle


MTBF = (250 + 360 + 200 + 120) / 4 = 232.5 days

MTTR = (9 + 6 + 2) / 3 = 5.67 days

Availability = 232.5 / (232.5 + 5.67) = 97.62 %

250 days 360 days

9 d

200 days

6 2

120 days

Downtime

180 days 400 days

7

120 days

4 3

233 days

Downtime

MTBF = (180 + 400 + 120 + 233) / 4 = 233.25 days

MTTR = (7 + 4 + 3) / 3 = 4.67 days

Availability = 233.25 / (233.25 + 4.67) = 98.04 %

= 947 days

= 947 days



To improve Availability:

Improve MTBM:

•Reduce Preventive Programs to a minimum, or, have Preventive intervals as well defined as possible.

•Using Predictive techniques whenever possible

•Implementing Maintenance Engineering (RCM, TPM...)

Minimize M:

•Implementing Maintenance Engineering (Planning, Logistics...)

•Improving personnel technical skills (training)

•Developing Integrated Planning (Mntce+Ops+HSE+Inspection+...)

Achieved Availability↑ = MTBM↑/ (MTBM+M↓)

Reliability Centered Maintenance (RCM) Improving Productivity


Productivity Improvement Factors:

Detailed work planning

Delivering equipments to Maintenance as clean as possible

Check-list at the end of Maintenance activities

Complete and comprehensive Equipment data available

Supplies available on job site

Skilled personnel

Reliability Centered Maintenance (RCM) Availability benchmark

Reliability Centered Maintenance (RCM) Translating percents to daily routine...

Availability % Downtime per year Downtime per month* Downtime per week

90% 36.5 days 72 hours 16.8 hours

95% 18.25 days 36 hours 8.4 hours

98% 7.30 days 14.4 hours 3.36 hours

99% 3.65 days 7.20 hours 1.68 hours

99.5% 1.83 days 3.60 hours 50.4 min

99.8% 17.52 hours 86.23 min 20.16 min

99.9% ("three nines") 8.76 hours 43.2 min 10.1 min

99.95% 4.38 hours 21.56 min 5.04 min

99.99% ("four nines") 52.6 min 4.32 min 1.01 min

99.999% ("five nines") 5.26 min 25.9 s 6.05 s

99.9999% ("six nines") 31.5 s 2.59 s 0.605 s

Reliability Centered Maintenance (RCM) Maintenance Programs costs

Maintenance Program Cost US$/HP/year

Corrective (unplanned) 17 to 18

Preventive 11 to 13

Predictive / Planned Corrective 7 to 9

NMW Chicago 1998

Reliability Centered Maintenance (RCM) Benchmarking balance between Mtce programs

NMW Chicago 1998

Maintenance activities %

Corrective actions 28

Preventive actions 36

Predictive actions 19

Maintenance studies 17

Reliability Centered Maintenance (RCM) Definitions

Bibliography: Lafraia, João Ricardo - Manual de Confiabilidade, Mantenabilidade e Disponibilidade, Editora Qualitymark

Failure rate (λ)

Failure rate (λ) is defined as the reciprocal of MTBF:

Reliability: R(t)

Let P(t) be the probability of failure between 0 and t; reliability is defined as:

R(t) = 1 – P(t)

MTBFt

1)(

Reliability Centered Maintenance (RCM) Some math...

Considering rate failure (λ) constant, it is proven (check at www.weibull.com), that R(t), meaning the probability of having operated until instant t, is given by:

tetR )(This reinforces the idea that Reliability is function of time, it isn’t a definite number. So, it’s incorrect to affirm: “This equipment has a 0.97 reliability factor...”. We should rather say: “This equipment has 97% reliability for running, let’s say, 240 days...”

http://www.weibull.com/

Reliability Centered Maintenance (RCM) Tricks and tips...

Historically, an equipment has 4 failures per year. Which is the reliability of this equipment for a 100 days run?

λ =4/365 λ =0.011/day R(100) = e-0.011x100 = e-1.1 = 0.333 = 33.3%

The probability of having no failure until 100 days is 33.3%

Some upgrades have been made, so failure rate now is 2 per year (meaning that MTBF has doubled). Which is the reliability for a 100 days run?

λ =2/365 λ =0.0055/day R(100) = e-0.0055x100 = e-0.55 = 0.577 = 57.7%

The probability of having no failure until 100 days is 57.7%.

As seen, doubling MTBF doesn’t double reliability.

Reliability Centered Maintenance (RCM) Trick and tips...

Historically, an equipment has a MTBF = 200 days. To improve 10% its reliability to operate on a 100 days run, which percent should MTBF be improved?

λ =1/200 λ =0.005/day R(100) =e-0.005x100 = e-0.5 = 0.607 = 60.7%

To improve this reliability in 10%, new reliability should be:

R’(100) = 1.1 x 0.607 = 0.668 = e-λ’x100

Ln 0.668 = -λ’ x 100 -0.403 = -λ’ x 100 λ’= 0.00403

1/MTBF’ = 0.0043 MTBF’ = 232 days

232/200 = 1.16 MTBF should improve 16%

Reliability Centered Maintenance (RCM) Trick and tips...

As per the manufacturer, an equipment has a 90% reliability to run over one year. If you want to have a 95% confidence that it will not fail, how long should it take until the equipment undergo a Preventive maintenance or some predictive technique?

0.9 = e-λx365 ln 0.9 = -λ x 365 -0.1054 = -λ x 365 λ = 2.89 x 10-4/day

0.95 = e-λt ln 0.95 = -λt -0.0513 = - 2.89 x 10-4 x t t = 177.5 days

For practical purposes, this equipment could be in a semester preventive / predictive program.

Reliability Centered Maintenance (RCM) Tricks and Tips...

Reliability and MTBF

0.368 0.368 0.368 0.368 0.368 0.3680.368

0

0.2

0.4

0.6

0.8

1

1.2

1 51 101 151 201 251 301 351

Days

MTBF=50

MTBF=100

MTBF=150

MTBF=200

MTBF=250

MTBF=300

MTBF=365

Reliability Centered Maintenance (RCM) System in series

1 2 3

Let P1=5%, P2=10% and P3=20% be the failure probability of each component of

this system, in a certain period. Which is the reliability of this system, in series?

This system will run, provided that ALL its components run. So, their reliabilities

are multiplied.

R1 = 1 – P1 = 1 – 0.05 = 0.95

R2 = 1 – P2 = 1 – 0.10 = 0.90

R3 = 1 – P3 = 1 – 0.20 = 0.80

R = R1 x R2 x R3 = 0.95 x 0.90 x 0.80 = 0.6840 = 68.4%

System failure probability 31.6%

System failure probability is bigger than each individual component. System

reliability is less than each component.Bibliography: Lafraia, João Ricardo - Manual de Confiabilidade, Mantenabilidade e Disponibilidade, Editora Qualitymark

Reliability Centered Maintenance (RCM) System in parallel

1

2

3

Let P1=5%, P2=10% and P3=20% be the failure probability of each component of this

system, in parallel, in a given period. Which is the reliability of the system, in parallel?

This system will run until ALL components fail. In this case, the failure probabilities

are multiplied.

P = P1 x P2 x P3 = 0.05 x 0.10 x 0.20 = 0.0010

R = 1 – P = 0.999 = 99.9%

System failure probability 0.1%

System failure probability is less than each component. System reliability is bigger

than each component.Bibliography: Lafraia, João Ricardo - Manual de Confiabilidade, Mantenabilidade e Disponibilidade, Editora Qualitymark

Reliability Centered Maintenance (RCM) Mixed systems

1 2 3

4 5

If P1=10%, P2=5%, P3=15%, P4=2% and P5=20%, which is the system reliability?

123

45

R1= 1 – 0.10 = 0.90

R2= 1 – 0.05 = 0.95 R123 = 0.9 x 0.95 x 0.85 = 0.7268 P123= 0.2733

R3= 1 - 0.15 = 0.85

R4= 1 – 0.02 = 0.98 R45 = 0.98 x 0.80 = 0.7840 P45= 0.2160

R5= 1 – 0.20 = 0.80

SystemP123= 0.2733 Psystem = 0.2733 x 0.2160 = 0.0590

P45= 0.2160 Rsystem = 1 – 0.0590 = 0.941 = 94.1%

Reliability Centered Maintenance (RCM) Redundancy

A

B

C

The pumps A, B y C are feed pumps of a plant. To operate in full condition, it’s necessary that at least two of these three pumps are running. Failure probability of each one is 10%. Which is the reliability to run this plant at full production?

Failure probability is P= 0.1 (10%), and reliability is R=1-0.1= 0.9 (90%)

Three pumps in parallel, so:

(R + P)3 = R3 + 3R2P + 3RP2 + P3= 0.93 + 3x0.92x0.1 + 3x0.9x0.12 + 0.13

(R + P)3 = 0.729 + 0.243 + 0.027 + 0.001

Three running: 0.729

Two running and one off: 0.243 Reliability = 0.972 = 97.2 %

One running and two off: 0.027

None running: 0.001 No full production = 0.028 = 2.8 %

Reliability Centered Maintenance (RCM) Redundancy

A

B

C

The pumps A, B y C are feed pumps of a plant. Pump A flow rate is 2,000 gpm, pump B flow rate is 1,800 gpm and pump C flow rate is 1,700 gpm. To operate, the plant need at least a feed rate of 3,600 gpm. Reliabilities are: RA=0.95, RB=0.90 and RC=0.85. Which is the plant reliability?

As the plant needs at least 3,600 gpm, to supply this, there will be these cases:

A ∩ B ∩ C 0.95 x 0.90 x 0.85 = 0.72675

A ∩ B ∩ notC 0.95 x 0.90 x (1 – 0.85) = 0.12825

A ∩ notB ∩ C 0.95 x (1 – 0.90) x 0.85 = 0.08075

Plant reliability = 0.93575 93.6%

Reliability Centered Maintenance (RCM) Systems in series

Systems in series

1 component

2 components

3 components

4 components

10 components

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.5

0.52

0.54

0.56

0.58 0.6

0.62

0.64

0.66

0.68 0.7

0.72

0.74

0.76

0.78 0.8

0.82

0.84

0.86

0.88 0.9

0.92

0.94

0.96

0.98 1

Component reliability

Sys

tem

rel

iab

ility

1 component

2 components

3 components

4 components

10 components

Reliability Centered Maintenance (RCM) Systems in parallel

Systems in parallel

1 component

2 components

3 components

4 components

10 components

0

0.2

0.4

0.6

0.8

1

1.2

0.5

0.52

0.54

0.56

0.58 0.6

0.62

0.64

0.66

0.68 0.7

0.72

0.74

0.76

0.78 0.8

0.82

0.84

0.86

0.88 0.9

0.92

0.94

0.96

0.98 1

Component reliability

Sys

tem

rel

iab

ility

1 component

2 components

3 components

4 components

10 components

Reliability Centered Maintenance (RCM) System and Component Redundancy

Component Redundancy

AA’ and BB’ subsystems’ reliability:

1 - (1-R)2 =1 – 1 + 2R – R2 = 2R – R2

System reliability:

R component redundancy = (2R-R2)2

A

A’

B

B’

A

A’

B

B’

System RedundancyWhich of these systems would have a better overall reliability

(let’s assume all components have the same reliability R)?

AB and A’B’ subsystems’ reliability:

R2

System reliability:

R system redundancy = 1 – (1-R2)2

R system redundancy = 1 – 1 + 2R2-R4

R system redundancy = 2R2 - R4

R comp red - R syst red = (2R-R2)2 - (2R2 - R4) = 4R2 – 4R3 + R4 - 2R2 + R4

R comp red - R syst red = 2R4 – 4R3 + 2R2 = 2R2(R2 – 2R + 1) = 2R2(R-1)2≥ 0

R comp red ≥ R syst red

Reliability Centered Maintenance (RCM) Active and Passive Redundancy

A

B

Active Redundancy:

Both equipment are operating at the same time, sharing the load. If one fails, the other one will carry the load alone.

Passive Redundancy:

One equipment is operating, and the other

one is at stand-by, starting operating after

the failure of the first one, pending upon a

switch system.

Reliability Centered Maintenance (RCM) Getting closer to real world...

In systems with active redundancy all redundant components are in operation and are sharing the load with the main component. Upon failure of one component, the surviving components carry the load, and as a result, the failure rate of the surviving components may be

increased.

The reliability of an active, shared load, parallel system can be calculated as follows:

where: λ1 is the failure rate for each unit when both are working and λ2 is the failure rate of the surviving unit when the other one has

failed.

If 2λ1 = λ2, then:


9977.0)100(

)9213.09404.0(49213.0)100(

4)100(

000615.000041.02

00041.02)100(

082.00615.0082.0

10000041.02100000615.010000041.02

R

R

eeeR

eeeR x

In a system with active redundancy, reliability of each of the two components for 100 days is R=0.96, when sharing the load. If one compontents fails, the

surviving one will have a 50% increase in its failure rate. Which is it the system reliability for 100 days?

R(100) = 0.96 = e-λx100 ln 0.96 = -100λ λ1 = 0.00041

λ2 = 1.5 x λ1 = 0.000615

If there were no increase in failure rate, system reliability would be 0.9984. Look like nothing, but this means a 30.5% decrease in system MTBF!!!


The redundant or back-up components in passive or standby systems start operating only when one or more fail. The back-up components remain dormant until needed.

For two identical components (primary and back-up) the formula is:

R(t) = e-λt (1+λt), considering a perfect switch

If the reliability of the switch is less than one, the reliability of the system is affected by the switching mechanism and is reduced accordingly:

R(t) = e-λt (1+Rswλt), Rsw switch reliability

The reliability of a standby system consisting of one primary component with constant failure rate λ1 and a backup component with constant failure rate λ2 is given by:


Two feed pumps in a nuclear power plant are connected in a stand-by mode. One is active and one is on standby. The power plant will have to shut down if both feed pumps fail. If the time between failures of each pump has an exponential distribution with MTBF = 28,000 hours, and the failure rate of the switching mechanism λsw is 10-6 what is the probability that the power plant will not have to shut down due to a pump failure in 10,000 hours?

9900.001.0101010 246

eeeRsw

R(t) = e-λt (1+Rswλt),

Switch reliability:

λ = 1/MTBF

9471.0)10000(

3536.16997.0)10000(

)3536.01()10000(

)1000028000

19900.01()10000(

3571.0

10000280001

R

R

eR

eR

R(t) = e-λt (1+Rswλt)

Reliability Centered Maintenance (RCM) Bathtub Curve

Early Life (Burn-in, infant mortality)• large number of new component failures which decreases with time

Useful Life• small number of apparently random failures during working life (λ constant)

Wear-out• increasing number of failures with time as components wear out

Reliability Centered Maintenance (RCM) Bathtub Curve

Early Life: • sub-standard materials• often caused by poor / variable manufacturing and poor quality control• prevented by effective quality control, burn-in, and run-in, de-bugging techniques• weak components eventually replaced by good ones• probabilistic treatment less importantUseful Life:• random or chance failures• may be caused by unpredictable sudden stress accumulations outside and inside of the components beyond the design strength• over sufficiently long periods frequency of occurrence (λ) is approximately constant• failure rate used extensively in Safety & Reliability analysesWear-out period: • symptom of component ageing• prediction is important for replacement and maintenance policy

Reliability Centered Maintenance (RCM) Different bathtub curves

These statistics are from aeronautical industry. In a

process plant, like a refinery, do you think the

percent of each one would be about the

same?

Reliability Centered Maintenance (RCM) Different bathtub curves

Which of these curves would be applicable to:

A pump?An electronic instrument?

A tire?

Reliability Centered Maintenance (RCM) Failure modes

Common sense tells that the best way to optimize the availability of plants is to implement some Preventive maintenance.

Preventive maintenance means fixing or replacing some pieces of equipments and/or components in fixed intervals. Useful lifespan of equipments may be calculated with Failure Statistical Analysis, enabling Maintenance Department to implement Preventive Programs.

This is true for some simple pieces of equipment and components, which may have a prevailing failure mode. Many components in contact with process fluids have a regular lifespan, as well as cyclic equipment, due to fatigue and corrosion.

But, for many pieces of equipment there’s no connection between reliability and time. Furthermore, as seen in Reliability curves, defining the optimum interval for Preventive maintenance may be a hard task. Besides, fixing or even replacing the equipment may bring you back to Infant Mortality period...

Reliability Centered Maintenance (RCM) Preventive maintenance may cause failures earlier....

Time

λ

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ere

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Reliability Centered Maintenance (RCM) Turnarounds

Turnarounds are often seen by Operations as an unique opportunity to have all problems solved, all equipment fixed…

Meanwhile, for Maintenance, a Turnaround is a huge event, time & resources & costs consuming, in which ONLY should be done whatever CANNOT be done on the run, during normal operation.

Frequently, Maintenance is asked to perform General Maintenance in ALL rotating equipment of a Unit, during its Turnaround. Matter of fact, if these equipment have spares, this General Maintenance should be done out of the TAR.

Why do Operations want everything to be done during the TAR?

1) Because Ops don’t have enough confidence that it will be done during routine maintenance.

2) Because they don’t feel comfortable running with an equipment momentarily without spare… the same way when we have a flat tire, we just drive with the spare tire enough to hit the tire repair shop…


1) Ops don’t have enough confidence that it will be done during routine maintenance.

To improve TAR results, reversing the vicious cycle below, Maintenance management has to improve Routine Maintenance!

To much to be done

during TAR

TAR won’t be able to

perform all that has to be

doneMany

equipments left to

Routine Maintenance

Many equipments left to TAR

Not in excess equipments to

be done during TAR

TAR will carry out all services

needed

Unit running well

Good routine maintenance


2) Because they don’t feel comfortable running with an equipment momentarily without spare… the same way when we have a flat tire, we just drive with the spare tire enough to hit the tire repair shop…

Consider these two pumps in a Passive Redundancy (one will be as stand-by). Assume that during the first 100 h after a General Maintenance such a pump will have a 70% reliability, and after this, for an one year period, it would run with 97% reliability (which are reasonable assumptions!!!).

If General Maintenance is performed in a Preventive or Predictive Program, during normal operations, during repair time the unit will be running pending upon a unique pump, with a 97% reliability.

If during TAR both pumps will be under General Maintenance, during the first 100 hours the system reliability (considering a perfect switch) would be 94.5% (using the R(t) = e-λt(1+λt) formula) . So, the unit would run for a period of time with two available pumps, but with an overall reliability below if it would be running with only one pump!

Reliability Centered Maintenance (RCM)RCM Implementation Flowchart

Will the failure affect directly Health, Safety or

Environment?

Will the Failure affect adversely the Mission, Vision

and Core Values of the Company?

Will the failure cause major economic losses?

(harm to systems and / or machines)?Is there some Cost-

effective Monitoring Technology available?

Deploy Monitoring techniques

Predictive Maintenance Preventive Maintenance

Run-to-fail?Re-design the system, accept failure risk, or

install redundancy

Are there regular failure patterns (time

intervals)?

Yes

No

Yes

Yes

Yes

Yes

No

No

No

No

Reliability Centered Maintenance (RCM)Another RCM Implementation Flowchart

If this thing breaks will it be noticed?

If this thing breaks will it hurt someone or the

environment?

If this thing breaks will it slow or stop production?

Can preventing it break reduce the likelihood of

multiple failures?

Is it cheaper to prevent it breaking than the loss of

production?

Is it cheaper to prevent it breaking than to fix it?

No

Yes

YesYes

No

Yes No

No

Can preventing it break reduce the reduce the risk to the environment

and safety?

Yes Yes YesNo No No

Prevent it breaking

Prevent it breaking

Prevent it breaking

Prevent it breaking

Check to see if it is broken

Re-design it Let it breakLet it break