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Integrated Reliability

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ReliabilityThe ability of a system to perform its

required function under stated condition fora specic period of time.

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Role of maintenanceIncrease the life of the productIncrease the Reliability of the product

Increase Durability

Lowering the costIncreasing the productivity

Reducing the breakdown

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aintenance!reakdown"reventive

"eriodic

"redictive#orrective

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T"

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Total productive maintenanceIt is a system of maintaining and improving

the integrity of production and qualitysystems through the machines$equipments$ processes and employees.

Denition%a philosophy to permeate throughout an

operating company touching people of alllevelsa collection of techniques and practices

aimed at ma&imi'ing the e(ectiveness )bestpossible return* of business facilities andprocesses

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"illars of T"

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+lements of T"a&imi'e the ,++" for entire life span of equipment

Implemented by various departments

Involves every single employee"romotion of ")preventive maintenance*

through motivation.

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-ctivities of T". -utonomous maintenance/. +quipment improvement

0. 1uality maintenance

2. ")aintenance "revention* system3

building4. +ducation and training

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lossesDown time!reakdown5etup and ad6ustment

5peed lossesIdling and minor stoppagesReduced speed

Defects

"rocess defectsReduced yield

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eaning for total in T"Total e(ectivenessTotal maintenance system 7 " 8 I 39 "

Total participation

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,verall +quipment +(ectiveness

,++ 7 -vailability : "erformance +;ciency :Rate of quality products

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-vailabilityIt is the ratio of operating time to the

loading time.

,perating time 7 loading time < downtime

Loading time 7 working hours

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"erformance +;ciency"+ 7 operating speed rate : net operating

rate

,perating speed rate 7 theoretical cycletime = actual cycle time

>et operating rate 7 actual processingtime = operation time-ctual processing time 7 processed amount

: actual cycle time

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Rate of quality productsIt is the ratio of the no of product produced

without to the total no of productsproduced.

Rate of quality 7 no of products without

defects= no of productsproduced

>o of products without defects 7 total no ofproducts < no ofdefects < no of product reworked

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Down timeTime of equipment when it is not producing

any products )breakdown and failures$accident*

There are two types of downtime can be

occurred$+quipment failures

5etup and ad6ustments

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Example 1

A medium volume manufacturing facility with a capacity of

producing 2 parts/minute actually produced 800 parts in aplanned running 2 shifts of 8 hours each. It had breaks

and scheduled maintenance for 40 minutes and also

faced 40 minutes breakdowns and hour 20 minutes for

changeover and ad!ustment. "umber of re!ects and re#

works were 0 and $ parts respectively. %alculate itsoverall effectiveness

&lanned production time ' 2(8 hrs. ' )$0 minutes

*oading time ' )$0#40 +breaks , scheduled maintenance- ' )20 min.own#time '40 +reakdowns- 80+%hangeover , ad!ustment-' 20

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Example 1 (Contd.) *oading time 1 own time )20 # 20

%Availability' ################################ (00 ' ################(00 ' 87%

*oading time )20

1uantity produced ?@@

%Performance7 33333333333333333333333333333333&@@ 7 3333333333333&@@7 50%

Time run & #apacity=given time )A/@3/@*&/

-mount produced < -mount defects < -mount re3work%Quality7 33333333333333333333333333333333333333333333333333333333333333333333333 &@@

-mount produced ?@@ < @ < B 7 33333333333333333333 & @@ 7 98%

?@@

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!enetsIncreased decision making,perators becomes more valuable

,perators receive additional training

ore cooperative work environmentigher lever of e&pertise

Increased self esteem for betterperformance

ulti skilled operators are in demand-bility to address issue that plagued

operations

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TTEean Time To Eailure )TTE* is the mean time

e&pected until the rst failure of a piece ofequipment.

TTE7 T=>

T 7 total time > 7 no of items under test

TTRean Time To Repair )TTR* is the time needed to

repair a failed hardware module.

T!E

ean Time !etween Eailure )T!E* is a reliabilityterm used to provide the amount of failures permillion hours for a product.

T!E 7 =)sum of all the part failure rates*

Eailure rate 7 no of failure = total time

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+&ample

5uppose @ devices are tested for 4@@ hours.During the test / failures occur.

The estimate of the T!E is%

T!E7 )@F4@@*=/ 7 /$4@@ hours= failure.

Ghereas for TTE

TTE7 )@F4@@*=@ 7 4@@

hours = failure.

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- system has 2@@@ components with afailure rate of @.@/H per @@@ hours.#alculate and T!E.

7 )@.@/ = @@* F ) = @@@* F 2@@@7 ? F @32failures=hour

T!E 7 = )? F @32* 7 /4@ hours