engg zc242-l6

BITS PilaniPilani Campus

Anil Jindal

Mechanical Engg Department

BITS Pilani, Pilani Campus

MAINTENANCE & SAFETY

BITS PilaniPilani Campus

Reliability –Centered Maintenance (RCM)

Chapter – 4:Part-1


Reliability

• Reliability refers to the consistency of a measure. A test is

considered reliable if we get the same result repeatedly.

• For example, if a test is designed to measure a trait, then

each time the test is administered to a subject, the results

should be approximately the same.

• Unfortunately, it is impossible to calculate reliability exactly,

but it can be estimated in a number of different ways.

• the probability that no (system) failure will occur in a given

time interval

• A reliable system is one that meets the specifications.


Various aspects of reliabilitycentered maintenance


Practical steps towards achieving reliability–centered maintenance

Step 1: Educate from Top to Bottom on Reliability-

Centered Maintenance

• Shatter the old myths

• Presentation to the staff the better way

• Use multiple formats

• Make the employees understand the importance of

their benefit in following the new techniques

• Planting lots of small seeds



Step2: Benchchmarking the Present position

• Companies will realize that once they are bench-marked,

they will realize that how far they are behind. The realities

will provide the necessary attitude adjustment.

• For safety, the International Standards Organization has

defined to calculate lost time incident rate (LTIR) and

recordable incident rate (RIR).

• It is understood that an RIR of 0.5 and LTIR of 0.05 are

considered to be high. Similar norms are not available for

reliability centred maintenance



Step4: Building up of a Business Case

Some examples of improvement are given below:

A 5 percent increase in availability = 5 percent increase in

revenue for a continuous process plant that can sell all that it

makes.

For example, a plant that produces Rs. 1,000 crore per year

generates another Rs. 50 crore in revenue. Reducing

overtime from 20 to 10 percent moves 10 percent of labour

from overtime rates to straight time rates. If the overtime

multiplier is 1.5 and a plant has a Rs. 10 crore labour budget

towards overtime Rs. 1 crore is saved.



Step5: Conducting a Pilot Program

• It may be necessary to conduct a pilot programme so

as to get the real feel of the benefits of following an

organized maintenance scheme.

• The pilot serves the following critical functions:

• Reduce initial investment.

• The pilot may cost only 10 percent of the full

programme, so the budget approvals are likely to be

fast.

System Reliability“Reliability of the product (made up of a number of components) is

determined by the reliability of each component and also by the

configuration of the system consisting of these components”

–Product design, manufacture, maintenance influence reliability, but

design has a major role

–One common approach for increasing the reliability of the system is

through “redundancy in design”, which is usually achieved by placing

components in parallel.

–As long as one component operates, the system operates

Systems with components in series

–For the system to operate, each component must operate

–It is assumed that the components operate independently of each other

(Failure

of one component has no influence on the failure of any other component)

A CB

Systems with components in

series – contd..

– If there are ‘n’ components in series, then system reliability is given by

Rs = R1 x

R2 x - - - - - - Rn

– System reliability decreases as the number of components in series

increases

– Manufacturing capability and resource limitations restrict the maximum

reliability of any given component

– Product redesign that reduces the no. of components in series is the

viable alternative

– Use of the Exponential Model

• If the system is in chance failure phase, a constant failure rate

could be justified based on which we can calculate failure rate,

mean time to failure and system reliability


series – contd..

• The system reliability is given by

• Thus if each component that fails is replaced immediately with another that has the same failure rate, the mean time tofailure for the system is given by

– Use of the Exponential Model• Suppose the system has ‘n’ components in series

• Each component has exponentially distributed time-to-failure with failure rates given by 1, 2 n

sR e1t X e2t X e3t X ent

n

i

e i1 t

MTTF

i1

•When all components have same failure rate, If

1n

i

1

then nMTTF

Fs 1R11R21Rn(1Ri)

– System reliability can be improved by placing components in parallel as system

will operate as long as at least one of the components operates.

– The only time the system fails is when all the parallel components fail

– All components are assumed to operate simultaneously.

– A system having ‘n’ components in parallel, with the reliability of the ith component

denoted by Ri, i=1, 2, ----- n.

– Also assume that the components operate randomly and independently of each

other.

– The probability of failure of each component is given by Fi = 1-Ri.

–System fails only if all the components fail and hence the probability of system

failure is

System with components in parallel

i1

n


parallel – contd..

– Mean time to failure for a system of n components in parallel is given by

1 1 1

MTTF 1/1

– If the time to failure of each component can be modelled by the exponential distribution,

each with a constant failure rate λi, then the system reliability, assuming independence of

component operation is

– Time to failure of the system is not exponentially distributed

– In the special case, where all the components have the same failure rate

the system reliability is Rs = 1- (1-e- t)n

2 3 n

– Reliability of the system is the complement of Fs and

is given by Rs = 1-Fs

– Use of Exponential model


Reliability block diagrams

Once the reliability of the subsystems is determined, the overall

system can be effectively modeled from the reliability perspective.

Once modeled, the weak links usually become evident and can be

addressed with reliability growth measures to eliminate the

deficiencies. Figure illustrates block-diagrammed examples of simple

serial, parallel and combination systems.


Reliability while active and standby


Reliability centered maintenance

Economic optimization of machine reliability relative to organization goal is the primary

objective of RCM.

RCM helps to insure that if we spend on improving reliability, we are getting the full

money back, plus some acceptable return on investment.


• The implementation of RCM follows the law of

diminishing marginal returns.

• The money invested in reliability improvement tends to

yield a higher return on investment than any money

subsequently invested.

• The objective is to reach the point of optimization at

which the benefits of reliability expresses as total

operating cost, are maximized through cost reduction.

Law of diminishing marginal returns


Law of diminishing marginal returns


• The warning time in advance of a functional failure that a monitoring

technique provides is called the P-F interval.

• P refers to the time at which the potential failure occurs.

• F refers to the time at which actual failure occurs.

• Longer the P-F interval, more time one has to make a good decision

and plan action.

Plot between time & condition


• Function oriented: seeks to preserve the system function not just

operability.

• System focused: more concerned about system then components.

• Reliability centered: seeks to define the probability that the system

perform its intended function for a specific operating time interval.

• Accepts design limitations

• Towards safety and economics

• Address failure: failure is defined as loss of function

• Uses logic tree: consistent approach to maintenance

• Needs effective tasks: technically sound and effective

• It should be applicable: must reduce the number and impact of

failure.

Basic guidelines for RCM

• Valuable tool to identify primary functional failure, their related

failure modes, the effect of failure modes on the operation of

system and associated criticality of the failure modes as a function

of impact and criticality.

• Used in advanced maintenance techniques, redesign or

redundancy.

• Example: Safety engineering

Maintainability engineering

Design engineering

23


Failure Modes Effects andCriticality Analysis (FMECA)

• To find the root cause for event.

• Used to find the recurrence, cost to control and control

methods.

Root Cause Failure Analysis (RCFA)

26


It’s to rank failure modes found from FMEA, according to severity

classification and it’s probability of occurrence.

The failure mode criticality number of for each failure mode is calculated

as:

Cm = βα*(λp) t

Where:

Cm = failure mode criticality number

β = conditional probability of failure effect

α = failure mode ratio

λp = part failure rate per million hours

t = duration of relevant mission phase (operation in hours)25


Criticality Analysis/Criticality Matrix.

The next step is to divide the criticality scale into number of sections

according to probability of occurrence.

• LevelA: Frequent. overall probability of failure =>0.2

• Level B: Reasonable. 0.1< overall probability of failure<0.2

• Level C: Occasional. 0.01< overall probability of failure<0.1

• Level D: Remote. 0.001< overall probability of failure<0.0.

• Level E: Unlikely. 0.001> overall probability of failure

26


Criticality Analysis/Criticality Matrix

Severity Classification:

• Category I:Catastrophic - death, weapon system loss, eg

aircraft, tank, missile etc.

• Category II: Critical – severe injury, property damage, mission

loss

• Category III: Marginal – Minor injury, minor property damage,

delay of mission

• Category IV: Minor – unscheduled maintenance or repairs

27



• Benefits of Criticality Analysis

• Quickly identifies risk and high exposure.

• Ranks functional areas and equipment based levels on exposure.

• Sums exposure levels for user defined areas and entire facility.

• Reduces red zone exposure with engineering follow up and action plans.

• Prioritize programs, initiatives and maintenance on critical ranking.

• Establish guidelines for determining maintenance work order priorities.


RCM:

• To find the root cause of event.

• Use criticality matrix.

• Driven by preventive maintenance strategy.

• Can find symptoms of the event.

29


ENGG ZC242, Maintenance & Safety,

02/009/14, Lecture-7

RCM vs RCA (Root Cause Analysis)

RCA:

30


• To find the underlying reason and to find necessary step to eliminate that event

• Use logic tree method

• Driven by maintenance prevention strategy

• To find and correct the cause

RCM vs RCA

31


Reliability prediction modelThe following diagram show that how field reliability of a system is predicted

and factors associated with it.

32


Software in RCMComputerized RCM tool is depicted below in the figure


An amplifier has an exponential time to failure distribution

with a failure rate of 8% per 1000h. What is the reliability

of the amplifier at 5000 h? Find the mean time to failure.

Example


Solution:


What is the highest failure rate for a product if it is to have a

probability of survival (that is, successful operation) of

95% at 4000h? Assume that the time to failure follows an

exponential distribution.

Example


Solution:


A module of a satellite monitoring system has 500

components in series. The reliability of each component

IS 0.999. Find the reliability of the module. If the number

of components in series is reduced to 200. What is the

reliability of the module?

Example


Solution:


The automatic focus unit of a television camera has 10

components in series. Each component has an

exponential time –to-failure distribution with z constant

failure rate of 0.05per 4000h. What is the reliability of

each component after 200h of operation? Find the

reliability of the automation focus unit for 200h operation.

What is it mean time-to-failure?

Example


Solution:


In continuation with previous example, the automatic focus

unit of television camera, which has 10 similar

components in series. It desired for the focus unit to

have a reliability of 0.95 after 2000h of operation. What

would be the mean time to failure of individual

components?

Example


Solution:


Find the reliability of the system shown in Fig

below with three components (A,B and C) in

parallel. The reliability of A,B and C are

0.95,0.92 and 0.90, respectively.

Example

A

C

B


Solution:


Example

For the system shown in fig, determine the system reliability for

2000 h of operation, and find the mean time to failure. Assume

that all three components have an identical time-to failure

distribution that is exponential, with a constant failure rate of

0.0005/h. what is the mean time to failure of each component?

If it desired for the system to have a mean time to failure of

4000h, what should the mean time to failure be for each

component?


Solution:


Find the reliability of the eight component system shown in

fig below some components are in series and some are

in parallel. The reliabilities of the components are as

follows: RA1=0.92, RA2=0.90 RA3=0.88 RA4=0.96

RB1=0.95 RB2=0.90 RB3=0.92 and RC1=0.93.

Example


Solution:


Find the system failure rate and the mean time to failure for

the eight-component system shown in fig the failure rate

( number of units per hour ) for the components are as

follows: λA1=0.0006, λA2=0.0045, λA3=0.0035,

λA4=0.0016, λB1=0.0060, λB2=0.0060, λB3=0.0060,

λC1=0.0050

Example


Solution:

engg zc242-l6

Documents

pilani campuspractical

overtime rs

percent increase

percent of labour

overtime rates

pilot programme

crore labour budget

overtime multiplier