Reliability Centered Maintenance (RCM) Evolution

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


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 Maintenance
Also know as Time Based Maintenance.

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
Take it easy,
Remember, my grandma, not
kid, Prevention always!
is better than
Cure....

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.

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 and Availability

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.

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


MTBF = Mean Time Between Failures
MTTR = Mean Time To Repair
A first definition:

MTBF
Availability =
MTBF + MTTR


Availability definitions
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
MTBF
Inherent Availability =
MTBF + MTTR
MTBM
Achieved Availability =
MTBM + M
Uptime
Operational Availability =
Operation Cycle


250 days 360 days 200 days 120 days = 947 days

Downtime  9d 6 2

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 %

180 days 400 days 120 days 233 days = 947 days

Downtime  7 4 3

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 %


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

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+...)


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


Availability benchmark

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

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

Benchmarking balance between Mtce programs

Maintenance activities %

Corrective actions 28

Preventive actions 36

Predictive actions 19

Maintenance studies 17

NMW Chicago

Definitions

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

1
λ (t ) =
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)

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

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:

− λt
R (t ) = e
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...”

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.

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%

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.

Tricks and Tips...

Reliability and MTBF

1.2

MTBF=50
1 MTBF=100
MTBF=150
MTBF=200
MTBF=250
0.8 MTBF=300
MTBF=365

0.6

0.4 0.368
0.368 0.368 0.368 0.368 0.368 0.368

0.2

0
1 51 101 151 201 251 301 351
Days

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.

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.

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 R1= 1 – 0.10 = 0.90

R2= 1 – 0.05 = 0.95 R123 = 0.9 x 0.95 x 0.85 = 0.7268 P 123= 0.2733
45
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

P123= 0.2733 Psystem = 0.2733 x 0.2160 = 0.0590
System
P45= 0.2160 Rsystem = 1 – 0.0590 = 0.941 = 94.1%

Redundancy

A The pumps A, B y C are feed pumps of a plant. To
operate in full condition, it’s necessary that at least
B 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?
C

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 %

Redundancy

A 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
B 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
C 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%

Systems in series

Systems in series

1

0.9 1 component
2 components
3 components
0.8
4 components
10 components
0.7
1 component
System reliability

0.6

2 components
0.5

3 components
0.4
4 components

0.3

10 components
0.2

0.1

0
0.54

0.64

0.68

0.94
0.5

0.52

0.56

0.58

0.62

0.66

0.72

0.74

0.76

0.78

0.8

0.82

0.84

0.86

0.88

0.9

0.92

0.96

0.98

1
0.6

0.7

Component reliability

Systems in parallel

Systems in parallel

1.2

10 components
1
4 components
3 components
2 components
0.8
System reliability

1 component 1 component
0.6 2 components
3 components
4 components
10 components
0.4

0.2

0
0.52

0.64

0.66
0.5

0.54

0.56

0.58

0.6

0.62

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

System and Component Redundancy

A B A B

A’ B’ A’ B’
Component Redundancy System Redundancy
Which of these systems would have a better overall reliability
(let’s assume all components have the same reliability R)?
AA’ and BB’ subsystems’ reliability: AB and A’B’ subsystems’ reliability:
1 - (1-R)2 =1 – 1 + 2R – R2 = 2R – R2 R2
System reliability: System reliability:
R component redundancy = (2R-R2)2 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

Active and Passive Redundancy

A

B

Active Redundancy: Passive Redundancy:
Both equipment are One equipment is
operating at the same operating, and the other
time, sharing the load. one is at stand-by,
If one fails, the other starting operating after
one will carry the load the failure of the first
alone. one, pending upon a
switch system.

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:


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

2 × 0.00041

R (100) = e − 2×0.00041x100 + 

× e (
− 0.000615 100
×
− e −2×0.00041×100 )
 2 × 0.00041 − 0.000615 
( )
R (100) = e −0.082 + 4 × e −0.0615 − e −0.082
R (100) = 0.9213 + 4 × (0.9404 − 0.9213)
R (100) = 0.9977

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?
R(t) = e-λt (1+Rswλt)
R(t) = e-λt (1+Rswλt),
10−6 ×104 10−2
Switch reliability: Rsw = e =e = e −0.01 = 0.9900

λ = 1/MTBF

−1 ×10000 1
R (10000) = e 28000
× (1 + 0.9900 × ×10000)
28000
R (10000) = e −0.3571 × (1 + 0.3536)
R (10000) = 0.6997 ×1.3536
R (10000) = 0.9471

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

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 important
Useful 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 analyses
Wear-out period:
• symptom of component ageing
• prediction is important for replacement and maintenance
policy

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?

Different bathtub curves

Which of these curves
would be applicable to:
A pump?
An electronic instrument?
A tire?

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...

Preventive maintenance may cause failures earlier....

Failures are likely to happen…
Here begins wear-out period.
Let’s define Preventive
maintenance here…
λ

Time

The failure likelihood is earlier!!!!

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…

Turnarounds

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 Not in excess
during TAR equipments to
be done
during TAR
TAR won’t be
Many able to TAR will carry Good routine
equipments perform all out all services maintenance
left to TAR that has to be needed
done
Many
equipments Unit running
left to well
Routine
Maintenance

Turnarounds

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!

RCM Implementation Flowchart

Will the failure affect No
directly Health, Safety or
Environment?
Will the Failure affect
Yes adversely the Mission, Vision No
and Core Values of the
Company?
Yes Will the failure cause
Yes
major economic losses?
(harm to systems and / or
Is there some Cost- machines)?
No
effective Monitoring
Technology available? No
Yes

Are there regular failure
Deploy Monitoring No
patterns (time
techniques
intervals)?
Yes

Predictive Maintenance Preventive Re-design the system, Run-to-fail?
Maintenance accept failure risk, or
install redundancy

Another RCM Implementation Flowchart

If this thing breaks will it If this thing breaks will it If this thing breaks will it No
Yes No
be noticed? hurt someone or the slow or stop production?
environment?
No Yes
Yes

Can preventing it break Can preventing it break Is it cheaper to prevent it Is it cheaper to prevent
reduce the likelihood of reduce the reduce the breaking than the loss of it breaking than to fix it?
multiple failures? risk to the environment production?
and safety?

Yes No Yes No Yes No Yes No

Prevent it Check to see Prevent it Re-design it Prevent it Let it break Prevent it Let it break
breaking if it is broken breaking breaking breaking

Reliability Centered Maintenance (RCM) Evolution

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