8 maintenance optimization

8/17/2019 8 Maintenance Optimization

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Approach)!o types of app"oaches&

• *ualitatively based maintenance "ules

• *uantitatively based maintenance "ules

)he qualitatively based maintenance "ules !ill give an initialinsight in vital components high failu"e consequences(components !ith high "epai" costs( etc.#.

)hese dominant components !ill mainly dictate themaintenance concept. )he othe" components a"econsequential( and do not "equi"e fu"the" effo"t.



Types of costs)h"ee types of costs in maintenance planning&

1. +osts of "epai" o" "eplacement

2. +osts of inspection

$. %is,-costs costs of failu"e multiplied !ith failu"e p"obability#

)a"get& minimum ife +ycle +osts( so !e have to balance these costson time basis( the /et P"esent 0alue app"oach#.

)he longe" the component !ill fulfil its function( the lo!e" the cost pe"unit a"e( but the highe" the "is, of failu"e may be o" the mo"einspections a"e needed to lo!e" that "is,.



Simplified Cost formulasailu"e based co""ective maintenance&

• c# 3 +" 4 +f #')c

5se-based p"eventive maintenance&

• c# 3 +" 4 Pf t"#⋅+f #')c

+ondition-based p"eventive maintenance&

• c# 3 +" 4 n⋅+i4 Pf n(g#⋅+f #')c

in !hich&

c# 3 e6pected total cost pe" unit time7 +" 3 cost of "epai" o" "eplacement7 +i 3 cost of

inspection

+f 3 cost of failu"e7 Pf 3 p"obability of failu"e7 )c 3 mean lifetime7 n 3 numbe" of

inspections

g 3 bounda"y of "e8ection to "epai"



Use and limitations of cost formulas

)he simplified cost fo"mulas p"esent a ve"y "ough

app"o6imation( and a"e to simple to be used in p"actice.)hey do not ta,e into account&

•

)he time value of money no discounting#

• )he"e is no deg"adation of the st"uctu"e included

9o !e need anothe" app"oach. :o!eve"( fo" ;sc"eening;pu"poses this simplified app"oach might be used.



Degradation<n hyd"aulic enginee"ing( a component# of a

st"uctu"e can be in a numbe" of diffe"ent ;states;(depending on its deg"ading "esistance.

A failu"e is defined as the event in !hich- due to dete"io"ation - the "esistance d"ops belo!the design st"ess o" the failu"e level.

o" e6ample& the minimum !ate" to be !ithstood(o" the c"est-level of a dy,e.



Example

9uppose the c"est level of an emban,ment is 2mete"s( and the failu"e level is 1(=> mete"s.9uppose the preventive maintenance level is

1(> mete"s. )he settlement each yea" is >(>2 mno unce"tainty#.

What is the p"eventive maintenance inte"val?



Preventive maintenance levels

<n the e6ample the maintenance level is given.

@ut ho! can !e dete"mine this level?

Ans!e"& ma,e calculations fo" diffe"ent levels o"

"elated maintenance inte"vals#( using all types of

costs "is, costs( "epai" costs( inspection cost#.



optimum

minimum m a i n t e n a n c e c o s t s p e r y e a r

maintenance interval

repair cost

cost of failuretotal cost

OPTIMAL MAINTENANCE INTERVAL(PREVENTIVE MAINTENANCE)



A preventive maintenance strategy

One of the simplest p"eventive maintenance

st"ategies is the !ell-,no!n age replacement

strategy .

A component is "eplaced upon failu"e o" upon

"eaching a p"e-dete"mined age( the so-called;age-"eplacement inte"val;.



010 20 30 40 50

0

1

2

3

4

5

6

7

number of

replacements

replacement

preventive replacement

corrective replacement

unused remaining life

preventive age

replacement interval

time [year]

AGE REPLACEMENT WITH AN AGE REPLACEMENTINTERVAL OF 10 YEARS



Uncertain deterioration and time of failure

<n mechanical and elect"ical enginee"ing( the life time is often"ep"esented by ;lifetime dist"ibutions; de"ived f"om statistics#.

<n :yd"aulic nginee"ing !e often have p"eventivemaintenance( so !e cannot have failu"e data to obtain;lifetime dist"ibutions;.

A st"uctu"e can be in a va"iety of states( depending on itsdeg"ading "esistance.

<f the"e is no unce"tainty in the deto"io"ation( it is "athe" easyto calculate the p"eventive maintenance inte"val.

<t becomes mo"e difficult if the deto"io"ation is unce"tain.



Statistical calculation of lifetime9uppose that the ave"age deto"io"ation pe" yea" is µ( and the standa"ddeviation is σ. We assume linea" dete"io"ation.

)han afte" n yea"s !e have as dete"io"ation& n.µ( and the standa"ddeviation& √n.σ2#.

<f !e assume a no"mal dist"ibution !e can calculate in each yea" thefailu"e p"obability.

Example

9uppose !e have fo" a pa"ticula" emban,ment ave"age settlement of>(>2 m pe" yea"( !ith a standa"d deviation of >(>>2. What is thedete"io"ation afte" 1> yea"s ave"age and standa"d deviation#?



6ampleWe have an emban,ment( and !e !ant to ,no! the p"eventive maintenanceinte"val.

)he c"est level decline is >(>1 m pe" yea"( and the standa"d deviation is >(>2 mno"mal dist"ibution#. )he c"est level is no! 1(= m( and the failu"e level is 1($>m. 9uppose that an acceptable safety level is a failu"e p"obability of >(>1 pe"yea".

)he failu"e p"obability after 5 years& the mean dec"ease of the emban,ment

is =6>(>13>(>= and standa"d deviation3√=6>(>23>(>BP6C1($>#3PuC1($>-1(=#'>(>B#3PuC-$($=# 3 >(>>> After 10 years:P6C1($>#3PuC1($>-1(>#'>(>D$2# 3 PuC-1(=# 3 >(>=B1 After 15 years:P6C1($>#3PuC1($>-1($=#'>(>BB=# 3 PuC->(D# 3 >(2D11

After 20 years:P6C1($>#3PuC1($>-1($>#'>(>E# 3 PuC># 3 >(= After 25 years:P6C1($>#3PuC1($>-1(2=#'>(1# 3 PuC>(=# 3 >(DE1=



6e"cise

We have an emban,ment( and !e !ant to ,no! the p"eventive

maintenance inte"val.)he p"obability of failu"e Pf is the follo!ing&

Fea" 1& >(>>=7 yea" 2& >(>27 yea" $& >(>7 yea" & >(>

Fea" =& >(1D7 yea" D& >($27 yea" B& >(D+ost fo"mula is& +"4P0Pf +f #'A>(>=(n !ith n the maintenance inte"val

ailu"e +ost +f 3 1>.>>> eu"o( %epai" +ost +" 3 1>>> eu"o. )he

inte"est "ate is =G. Please ma,e the cost figu"e. What is the optimalmaintenance inte"val?

8 maintenance optimization

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