Reliability in Maintenance

Source : Chapter 8 from Maintenance Engineering and Management by R.C.Mishra

Definition• Reliability: the probability that a

component / system , when operating under given conditions, will perform its intended functions adequately for a specified period of time

• Likelihood that an equipment will not fail during its operation

• Two types– Inherent reliability

• Associated with quality of the material and design of machine parts

– Achievable reliability• Depends on factors such as maintenance and

operation of the equipment

Reliability and Probability

If

R is the probability of reliable function for a specified period of time

F is the probability of failure during the same period of time

Then

R + F = 1

Failure rate

• Defined as the number of failures occurring in a unit time• Denoted as “λ”• Formula

Failure rate λ=

No. of failures recorded

No.of components subjected to operation X no. of hours of operation

Failure rate estimation

Days Mid value Frequency fx

0-10 5 20 100

10-20 15 10 150

20-30 25 8 200

30-40 35 5 175

Using frequency distribution

Mean time to failure = Σ fx

Σf

M = 625 /43 = 14.53

Failure rate = 1 /m = 1 /14.53 = 6.8 x 10-2 failures / day

Failure Pattern of Equipment

The Whole-Life Equipment Failure Profile

• The burn-in phase (known also as infant morality, break-in , debugging):

• During this phase the hazard rate decrease and the failure occur due to causes such as:

Incorrect use procedures Poor test specifications Incomplete final test

Poor quality control Over-stressed parts Wrong handling or packaging

Inadequate materials Incorrect installation or setup

Poor technical representative training

Marginal parts Poor manufacturing processes or tooling

Power surges

The useful life phase:

• During this phase the hazard rate is constant and the failures occur randomly or unpredictably. Some of the causes of the failure include:

A. Insufficient design margins

B. Incorrect use environments

C. Undetectable defects

D. Human error and abuse

E. Unavoidable failures

The wear-out phase (begins when the item passes its useful life phase):• During this phase the hazard rate increases. Some of the

causes of the failure include:A. Wear due to aging. B. Inadequate or improper preventive maintenance C. Limited-life componentsD.Wear-out due to friction, misalignments, corrosion and creep E. Incorrect overhaul practices

Definitions

• Reliability of a system is defined to be the probability that the given system will perform its required function under specified conditions for a specified period of time.

• MTBF (Mean Time Between Failures): Average time a system will run between failures. The MTBF is usually expressed in hours. This metric is more useful to the user than the reliability measure.

Approaches to increase the reliability of a system

Increasing reliability of a system

1. Worst case design

2. Using high quality components

3. Strict quality control procedures

1. Redundancy

2. Typically employed

3. Less expensive

Reliability expressions

• Exponential Failure Law:

• Reliability of a system is often modeled as:

– R(t) = exp(-λt)• where λ is the failure rate expressed as

percentage failures per 1000 hours or as failures per hour.

– When the product “λt” is small,• R(t) = 1 - λt

Relation between MTBF and the Failure rate

• MTBF is the average time a system will run between failures and is given by:

– MTBF = ∫0 R(t) dt = ∫0 exp(-λt) dt = 1 / λ

– In other words, the MTBF of a system is the reciprocal of the failure rate.

– If “λ” is the number of failures per hour, the MTBF is expressed in hours.

∞ ∞

A simple example

• A system has 4000 components with a failure rate of 0.02% per 1000 hours. Calculate λ and MTBF.

• λ = (0.02 / 100) * (1 / 1000) * 4000 = 8 * 10-4 failures/hour

• MTBF = 1 / (8 * 10-4 ) = 1250 hours

Relation between Reliability and MTBF

• R(t) = (1 – λt) = (1 – t / MTBF)• Therefore,

– MTBF = t / (1 – R(t))

Time t

Reliability

R(t)

1.0

0

0.8

0.6

0.4

0.2

1 MTBF 2 MTBF

0.36

An example

• A first generation computer contains 10000 components each with λ = 0.5%/(1000 hours). What is the period of 99% reliability?

• MTBF = t / (1 – R(t)) = t / (1 – 0.99)– t = MTBF * 0.01 = 0.01 / λav

– Where λav is the average failure rate– N = No. of components = 10000– λ = failure rate of a component

• = 0.5% / (1000 hours) = 0.005/1000 = 5 * 10-6 per hour

• Therefore, λav = N λ = 10000 * 5 * 10-6 = 5 * 10-2 per hour

• Therefore, t = 0.01 / (5 * 10-2 ) = 12 minutes

Reliability for different configurations

R R R R R1 2 3 4 N

Overall reliability = Ro = R * R * R…. R = RN

1. Series Configuration

2. Parallel ConfigurationR

R

R

1

2

N

Ro = 1 – (probability that all of the components fail)

Ro = 1 – (1 - R)N

Reliability for different configurations

R R R1 2 N

Overall reliability = Ro = ?

3. Hybrid Configuration

R

R

R

1

2

M

Maintainability

• Maintainability of a system is the probability of isolating and repairing a “fault” in the system within a given time.

• Maintainability is given by:– M(t) = 1 – exp(-µt)– Where µ is the repair rate– And t is the permissible time constraint for the

maintenance action

– µ = 1/(Mean Time To Repair) = 1/MTTR

– M(t) = 1 – exp(-t/MTTR)

20

Availability

• Availability of a system is the probability that the system will be functioning according to expectations at any time during its scheduled working period.

• Availability = System up-time / (System up-time + System down-time)

• System down-time = No. of failures * MTTR

• System down-time = System up-time * λ * MTTR

• Therefore,

– Availability = System up-time / (System up-time + (System up-time * λ * MTTR)

• = 1 / (1 + (λ *MTTR)

– Availability = MTBF / (MTBF + MTTR)