Lecture - Corrective Maintenance

8/11/2019 Lecture - Corrective Maintenance

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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



Slide

1-2

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



Slide

1-3

Operation Types


Salvage Fail Repair

RebuildServicing

Corrective

Maintenance

Operations



Slide

1-4

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.




Slide

1-5

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.




Slide

1-7

Downtime

Corrective Maintenance downtime is made up of three components


CorrectiveMaintenance

Downtime

Administrative/

Logistic Time

Active Repair

Time

Delay

Time



Slide

1-8

Downtime

Active repair time is the main ingredient of downtime and is made up

of six components


PreparationTime

Active Repair

Time

Fault Correction

Time

Fault LocationTime

Spare partsObtainment Time

Adjustment and

Calibration TimeCheck out

Time



Slide

1-9

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)




Slide

1-10

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


(maximum utilization of the highest paid personnel)

(increased probability of machine coverage)



Slide

1-11

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


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.



Slide

1-12

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.


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



Slide

1-14

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




Slide

1-15

Availability

Measured as follows:


) E(Downtime E(Uptime)

E(Uptime) A(t)



Slide

1-16

MTTR

Mean Corrective Maintenance Time

Also referred to as Mean Time To Repair (MTTR)


j

cmj j

mcm λ

T λT



Slide

1-17

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.


System Operating

Normally (0)

System

Failed (1)

λ

μC



Slide

1-18

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


System Operating

Normally (0)

System

Failed (1)

λ

μC



Slide

1-19

Model I


)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

Slid



Slide

1-20

Model I

Availability


) 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)

Slid



Slide

1-21

Model I

Since: λ = system failure rate =

And μC = system corrective maintenance rate =

Therefore;


)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

Slid



Slide

1-22

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.


MTTR MTTF

MTTF A

%.or. 839999830

53000

3000 A

Slide



Slide

1-23

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.


System Operating

Normally (0)

System

Failed (1)

λ1

μC1

SystemFailed (2)λ2

μC2

Slide



Slide

1-24

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.


System Operating

Normally (0)

System

Failed (1)

λ1

μC1

SystemFailed (2)λ2

μC2

Slide



Slide

1-25

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


System Operating

Normally (0)

System

Failed (1)

λ

μC

Slide



Slide

1-26

Model II


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[

Slide



Slide

1-27

Model II


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

Slide



Slide

1-28

Model II


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

Slide



Slide

1-29

Model II


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

Slide



Slide

1-30

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.


0.006) x(0.0050.009) x(0.0020.009) x(0.006

0.009 x0.006 A

53%0 5294

Lecture - Corrective Maintenance

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