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.

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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,.

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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"

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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.

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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.

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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?

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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#.

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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)

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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;.

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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

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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.

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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#?

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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=

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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?