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MAINTENANCE MANAGEMENT
IM –
503
Lecture: Corrective Maintenance
Dr Muhammad Fahad
Associate Professor/Director Product Development Centre
Dept of Industrial & Manufacturing
NED University of Engineering & Technology
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Definition
Maintenance carried out after fault recognition and intended
to put an item into a state in which it can perform a required
function. (BS EN 13306:2010)
Corrective maintenance is an unscheduled maintenance action,
basically composed of unpredictable maintenance needs thatcannot be preplanned or programmed on the basis of
occurrence at a particular time.
Corrective Maintenance
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Operation Types
Corrective Maintenance
Salvage Fail Repair
RebuildServicing
Corrective
Maintenance
Operations
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Operation Types
Fail Repair
The failed item is restored to its original state.
Salvage
This element of corrective maintenance is concerned with disposal of
non-repairable material and use of salvaged material from non-
repairable equipment/item in the repair, overhaul, or rebuild
programs.
Corrective Maintenance
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Operation Types
Rebuild
This is concerned with restoring an item to a standard as close as
possible to a desirable state in performance, life expectancy, andappearance.
Servicing:
Servicing may be needed because of the corrective maintenance
action, for example, engine repair can lead to crankcase refill,
welding on, etc.
Corrective Maintenance
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Downtime
Corrective Maintenance downtime is made up of three components
Corrective Maintenance
CorrectiveMaintenance
Downtime
Administrative/
Logistic Time
Active Repair
Time
Delay
Time
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Downtime
Active repair time is the main ingredient of downtime and is made up
of six components
Corrective Maintenance
PreparationTime
Active Repair
Time
Fault Correction
Time
Fault LocationTime
Spare partsObtainment Time
Adjustment and
Calibration TimeCheck out
Time
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Downtime
Can be reduced by using following strategies
Efficiency in fault recognition, location, and isolation
Effective interchangeability
Redundancy (Parts and/or Machines)
Effective accessibility (instruction related to accessing difficult
locations)
Human factor considerations (readability, weight, lighting etc) Maintenance workers ranking (Education level, Experience,
Certification)
Pre assigning workers to machine (Maintenance brigade system)
Corrective Maintenance
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Downtime
Important Questions
When a machine goes down, what mechanic category (that is
capable of servicing that machine) should be assigned?
Highest category
Lowest category
Corrective Maintenance
(maximum utilization of the highest paid personnel)
(increased probability of machine coverage)
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Downtime
Important Questions
If a mechanic becomes free and there are at least two down
machines among the machine categories that he can service,which should he service?
Calculate Total Expected Cost (TEC) per Piece
Assign the highest TEC machine first
Corrective Maintenance
R
K K t l TEC
))(( 21
l = Total service time (including load/unload), t =Travel time to machine.
K1= Hourly rate of the operator, K2 = Hourly rate of the machine.
R = Rate of production, pieces from machine per hour.
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Downtime
Important Questions
If a mechanic becomes free and all machines that he can
service are attended, should a replacement be undertaken? Calculate Replacement Time
Pre-empt the mechanic of the highest feasible skill level when the
TEC yields a lower value and replacement time savings is at least 0.
Corrective Maintenance
LS FM CM RT time std time std
..
CMstd.time = Standard time to service a machine (current mechanic)
FMstd.time = Standard time to service a machine (free mechanic)R = Time spent by the current mechanic on the machine
L = Loading time for the machine
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Availability
Ability to be in a state to perform as and when required, under
given conditions, assuming that the necessary external resources
are provided (BS EN 13306 –
2010)
The degree to which a system, subsystem, or equipment is in a
specified operable state at the start of an operation (i.e. at an
unknown or random time)
The probability that an item is available for use when
required
Corrective Maintenance
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Availability
Measured as follows:
Corrective Maintenance
) E(Downtime E(Uptime)
E(Uptime) A(t)
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MTTR
Mean Corrective Maintenance Time
Also referred to as Mean Time To Repair (MTTR)
Corrective Maintenance
j
cmj j
mcm λ
T λT
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Model I
This mathematical model represents a system that can either be in up
(operating) or down (failed) state.
Corrective maintenance is performed on the failed system to put it
back into its operating state.
Equations for the model are subject to the following assumptions: Failure (λ) and corrective maintenance (μC) rates are constant.
The repaired system is as good as new.
System failures are statistically independent.
Corrective Maintenance
System Operating
Normally (0)
System
Failed (1)
λ
μC
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Model I
The following symbols are used to develop equations for the model:
i = ith system state, i = 0 (system operating normally), i = 1 (system failed)
Pi (t) = probability that the system is in state i at time t
λ = system failure rate μC = system corrective maintenance rate
Corrective Maintenance
System Operating
Normally (0)
System
Failed (1)
λ
μC
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Model I
Corrective Maintenance
)t μ( λ-
C C
1
c
e μ λ
λ
μ λ
λ
(t) P
)t μ( λ-
C C
C
0
ce μ λ
λ
μ λ
μ(t) P
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Model I
Availability
Corrective Maintenance
) E(Downtime E(Uptime)
E(Uptime) A(t)
)t μ( λ-
C C
C
0
ce μ λ
λ
μ λ
μ(t) P A(t)
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Model I
Since: λ = system failure rate =
And μC = system corrective maintenance rate =
Therefore;
Corrective Maintenance
)t μ( λ-
C C
C
0
ce
μ λ
λ
μ λ
μ(t) P A(t)
MTTF
1
MTTR
1
C
C
0 μ λ
μ(t) P A(t)
For large values of t
MTTR MTTF
MTTF
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Example
Assume that the MTTF of a piece of equipment is 3000 h
and its mean corrective maintenance time is 5 h. Calculate
the equipment steady-state availability.
Corrective Maintenance
MTTR MTTF
MTTF A
%.or. 839999830
53000
3000 A
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Model II
This mathematical model represents a system that can either be in up
(operating) or failed in two mutually exclusive failure modes.
A typical example of this type of system or device is a fluid flow
valve (i.e., open and close failure modes).
Corrective maintenance is performed from either failure mode of the
system to put it back into its operational state.
Corrective Maintenance
System Operating
Normally (0)
System
Failed (1)
λ1
μC1
SystemFailed (2)λ2
μC2
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Model II
Equations for the model are subject to the following assumptions:
The system can fail in two mutually exclusive failure modes (1 and 2).
Failure and corrective maintenance rates are constant.
The repaired system is as good as new.
System failures are statistically independent.
Corrective Maintenance
System Operating
Normally (0)
System
Failed (1)
λ1
μC1
SystemFailed (2)λ2
μC2
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Model II
The following symbols are used to develop equations for the model:
i = ith system state, i = 0 (system operating normally), i = 1 (system failed
in failure mode type I), i = 2 (system failed in failure mode type II)
Pi (t) = probability that the system is in state i at time t (i = 0, 1, 2)
λi = system failure rate from state 0 to state i
μCi = system corrective maintenance rate from state I to 0
Corrective Maintenance
System Operating
Normally (0)
System
Failed (1)
λ
μC
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Model II
Corrective Maintenance
t m
211
C21C11
21
C2C1
0
1
]e )m(mm
) μ )(m μ(m
[ mm
μ μ
(t)P
t m
212
C22C12 2 ]e
)m(mm
) μ )(m μ(m[
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Model II
Corrective Maintenance
t m
212
1C12t m
211
C2111
21
C21
1
21
]e )m(mm
) λ μ(m
[ ]e )m(mm
μ λm λ
[ mm
μ λ
(t)P
t m
212
2C12t m
211
C1212
21
C122
21 ]e )m(mm
) λ μ
(m[ ]e
)m(mm μ λ
m λ[
mm μ λ(t)P
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Model II
Corrective Maintenance
2
4B A Am ,m
2
21
21C2C1 λ λ μ μ A
C12C21C2C1 μ λ μ λ μ μ B
C12C21C2C12 μ λ μ λ μ μmm
1
) λ λ μ( μmm21C2C121
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Model II
Corrective Maintenance
For large values of t
t m
211
C21C11
21
C2C1
0
1
]e )m(mm
) μ )(m μ(m
[ mm
μ μ
(t)P
t m
212
C22C12 2 ]e
)m(mm
) μ )(m μ(m[
21
C2C1
0mm
μ μ(t) P
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Example
An engineering system can fail in two mutually exclusive failure
modes. Failure modes I and II constant failure rates are λ1 =
0.002 failures per hour and λ2 = 0.005 failures per hour,respectively. The constant corrective maintenance rates from
failure modes 1 and 2 are μC1 = 0.006 repairs per hour and
μC2 = 0.009 repairs per hour, respectively. Calculate the system
steady state availability.
Corrective Maintenance
0.006) x(0.0050.009) x(0.0020.009) x(0.006
0.009 x0.006 A
53%0 5294