Post on 22-Oct-2014
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)