Corrective Maintenance 2t4gf

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Corrective Maintenance INTRODUCTION

Although every effort is made to make engineering systems as reliable as possible through design, preventive maintenance, and so on, from time to time they do fail. Consequently, they are repaired to their operational state. Thus, repair or corrective maintenance is an important component of maintenance activity. Corrective maintenance may be defined as the remedial action carried out due to failure or deficiencies discovered during preventive maintenance, to repair an equipment/item to its opera1–3 tional state. Usually, corrective maintenance is an unscheduled maintenance action, basically composed of unpredictable maintenance needs that cannot be preplanned or programmed on the basis of occurrence at a particular time. The action requires urgent attention that must be added, integrated with, or substituted for previously scheduled work items. This incorporates compliance with “prompt action” field changes, rectification of deficiencies found during equipment/item operation, and performance of repair actions due to incidents or accidents. A substantial part of overall maintenance effort is devoted to corrective maintenance, and over the years many individuals have contributed to the area of corrective maintenance. This chapter presents some important aspects of corrective maintenance.

CORRECTIVE MAINTENANCE TYPES Corrective maintenance may be classified into five major categories as shown in 1,4 Fig. 5.1. These are: fail-repair, salvage, rebuild, overhaul, and servicing. These categories are described below. 1. Fail-repair: The failed item is restored to its operational state. 2. Salvage: This element of corrective maintenance is concerned with disposal of nonrepairable material and use of salvaged material from nonrepairable equipment/item in the repair, overhaul, or rebuild programs. 3. Rebuild: This is concerned with restoring an item to a standard as close as possible to original state in performance, life expectancy, and appearance. This is achieved through complete disassembly, examination of all components, repair and replacement of worn/unserviceable parts as per original specifications and manufacturing tolerances, and reassembly and testing to original production guidelines.

©2002 CRC Press LLC


Salvage

Fail-repair

Overhaul

Corrective maintenance types

Servicing

Rebuild

FIGURE 5.1 Types of corrective maintenance.

4. Overhaul: Restoring an item to its total serviceable state as per maintenance serviceability standards, using the “inspect and repair only as appropriate” approach. 5. Servicing: Servicing may be needed because of the corrective maintenance action, for example, engine repair can lead to crankcase refill, welding on, etc. Another example could be that the replacement of an air bottle may require system recharging.

CORRECTIVE MAINTENANCE STEPS, DOWNTIME COMPONENTS, AND TIME REDUCTION STRATEGIES AT SYSTEM LEVEL Different authors have laid down different sequential steps for performing corrective maintenance. For example, Reference 2 presents nine steps (as applicable): localize, isolate, adjust, disassemble, repair, interchange, reassemble, align, and checkout. Reference 3 presents seven steps (as applicable): localization, isolation, disassembly, interchange, reassemble, alignment, and checkout. For our purpose, we assume that corrective maintenance is composed of five 1 major sequential steps, as shown in Fig. 5.2. These steps are: fault recognition, localization, diagnosis, repair, and checkout. The major corrective maintenance downtime components are active repair time, 1,5 istrative and logistic time, and delay time. The active repair time is made up of the following subcomponents: • • • • • •

Preparation time Fault location time Spare item obtainment time Fault correction time Adjustment and calibration time Checkout time



FIGURE 5.2 Corrective maintenance sequential steps.

Reduction in corrective maintenance time is useful to improve maintenance effectiveness. Some strategies for reducing the system-level corrective maintenance time 6 are as follows: • Efficiency in fault recognition, location, and isolation: Past experience indicates that in electronic equipment, fault isolation and location consume the most time within a corrective maintenance activity. In the case of mechanical items, often the largest contributor is repair time. Factors such as welldesigned fault indicators, good maintenance procedures, well-trained maintenance personnel, and an unambiguous fault isolation capability are helpful in lowering corrective maintenance time. • Effective interchangeability: Good physical and functional interchangeability is useful in removing and replacing parts/items, reducing maintenance downtime, and creating a positive impact on spares and inventory needs. • Redundancy: This is concerned with deg in redundant parts that can be switched in at the moment of need so the equipment/system continues to operate while the faulty part is being repaired. In this case the overall



maintenance workload may not be reduced, but the equipment/system downtime could be impacted significantly. • Effective accessibility: Often a significant amount of time is spent accessing the failed part. Proper attention to accessibility during design can help reduce part accessibility time and, in turn, the corrective maintenance time. • Human factor considerations: Attention paid to human factors during design in areas such as readability of instructions, size, shape, and weight of components, selection and placement of dials and indicators, size and placement of access, gates, and readability, and information processing aids can help reduce corrective maintenance time significantly.

CORRECTIVE MAINTENANCE MEASURES There are various measures associated with corrective maintenance. This section 1,6–8 presents three such measures.

MEAN CORRECTIVE MAINTENANCE TIME This is defined by ∑ λ j T cm j T mcm = ------------------∑λ j

(5.1)

where Tmcm = mean corrective maintenance time, Tcm j = corrective maintenance time of the jth equipment/system element, λj = failure rate of the jth equipment/system element. Past experience indicates that probability distributions of corrective maintenance times follow exponential, normal, and lognormal. For example, in the case of electronic equipment with a good built-in test capability and a rapid remove and replace maintenance concept, often exponential distribution is assumed. In the case of mechanical or electro-mechanical hardware, usually with a remove and replace maintenance concept, the normal distribution is often applicable. Normally, the lognormal distribution is applicable to electronic equipment that does not possess built-in test capability.

MEDIAN ACTIVE CORRECTIVE MAINTENANCE TIME This normally provides the best average location of the sample data and is the 50th percentile of all values of repair time. It may be said that median corrective maintenance time is a measure of the time within which 50% of all corrective maintenance can be accomplished. The computation of this measure depends on the distribution representing corrective maintenance times. Consequently, the median of the lognormally



6

distributed corrective maintenance times is given by

 ∑ λ j log Tcm j  - T med = antilog  ----------------------------∑λ j  

(5.2)

where Tmed = median active corrective maintenance time.

MAXIMUM ACTIVE CORRECTIVE MAINTENANCE TIME This measures the time needed to accomplish all potential corrective maintenance actions up to a given percentage, frequently the 90th or 95th percentiles. For example, in the case of 90th percentile, the maximum corrective maintenance time is the time within which 90% of all maintenance actions can be accomplished. The distribution of corrective maintenance times dictates the calculation of the maximum corrective maintenance time. In the case of lognormally distributed corrective maintenance 6 times, the maximum active corrective maintenance time is given by: T cmax = antilog ( T mn + z σ cm )

(5.3)

where Tcmax = maximum active corrective maintenance time, Τmn = mean of the logarithms of Tcm j, σcm = standard deviation of the logarithms of the sample corrective maintenance times, z = standard deviation value corresponding to the percentile value specified for Tcmax. The value of σcm can be calculated by using the following equation: 2

M  ∑ ( log T cm j ) –  ∑ log T cm j M j=1 j=1 = ------------------------------------------------------------------------------M–1 M

1/2

2

σ cm

(5.4)

where M = total number of corrective maintenance times.

CORRECTIVE MAINTENANCE MATHEMATICAL MODELS Over the years a vast amount of literature has been published that directly or indirectly concerns corrective maintenance. This section presents a number of mathematical models taken from the published literature. These models take into consideration the item failure and corrective maintenance rates, and can be used to predict item/system



FIGURE 5.3 System transition diagram.

probability of being in failed state (i.e., undergoing repair), availability, mean time to failure, and so on.

MODEL I This mathematical model represents a system that can either be in up (operating) or 8 down (failed) state. Corrective maintenance is performed on the failed system to put it back into its operating state. The system state space diagram is shown in Fig. 5.3. Equations for the model are subject to the following assumptions: • Failure and corrective maintenance rates are constant. • The repaired system is as good as new. • System failures are statistically independent. The following symbols are used to develop equations for the model: i Pi(t) λ µC

= the ith system state, i = 0 (system operating normally), i = 1 (system failed); = probability that the system is in state i at time t; = system failure rate; = system corrective maintenance rate.

Using the Markov approach presented in Chapter 12, we write the following two 8 equations for the Fig. 5.3 diagram: dP 0 ( t ) ---------------- + λP 0 ( t ) = µ C P 1 ( t ) dt

(5.5)

dP 1 ( t ) ---------------- + µ C P 1 ( t ) = λP 0 ( t ) dt

(5.6)

At time t = 0, P0(0) = 1 and P1(t) = 0. By solving Eqs. (5.5) and (5.6), we get


− ( λ + µ C )t µC λ + ---------------e P 0 ( t ) = --------------λ + µC λ + µC

(5.7)

− ( λ + µ C )t λ µ P 1 ( t ) = --------------- – ---------------e λ + µC λ + µC

(5.8)


The system availability is given by − ( λ + µ C )t µC λ + ---------------e A S ( t ) = P 0 ( t ) = --------------λ + µC λ + µC

(5.9)

where AS (t) = system availability at time t. As t becomes very large, Eq. (5.9) reduces to µC A S = --------------λ + µC

(5.10)

where AS = system steady state availability. Since λ = 1 MTTF and µC = 1 Tmcm, Eq. (5.10) becomes MTTF A S = --------------------------------T mcm + MTTF

(5.11)

where MTTF = mean time to failure. Example 5.1 Assume 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, if the equipment failure and corrective maintenance times are exponentially distributed. Substituting the given values into Eq. (5.11) yields 3000 A S = --------------------- = 0.9983 5 + 3000 There is 99.83% chance that the equipment will be available for service.

MODEL II This mathematical model represents a system that can either be operating normally or failed in two mutually exclusive failure modes (i.e., failure modes I and II). 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 9–10 of the system to put it back into its operational state. The system transition diagram is shown in Fig. 5.4. The following assumptions are associated with this model: • • • •

The system can fail in two mutually exclusive failure modes. The repaired system is as good as new. All system failures are statistically independent. Failure and corrective maintenance rates are constant.




The following symbols are associated with the model: = the 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, for i = 0, 1, 2, λi = system failure rate from state 0 to state i, for i = 1, 2, µCi = system corrective maintenance rate from state i to state 0, for i = 1, 2. i

For Model I, we write the following equations for the Fig. 5.4 diagram: dP 0 ( t ) ---------------- + ( λ 1 + λ 2 )P 0 ( t ) = µ C1 P 1 ( t ) + µ C2 P 2 ( t ) dt

(5.12)

dP 1 ( t ) ---------------- + µ C1 P 1 ( t ) = λ 1 P 0 ( t ) dt

(5.13)

dP 2 ( t ) --------------- + µ C2 P 2 ( t ) = λ 2 P 0 ( t ) dt

(5.14)

At time t = 0, P0 (0) = 1 and P1(0) = P2(0) = 0. By solving Eqs. (5.12)–(5.14), we obtain ( m 1 + µ C1 ) ( m 1 + µ C2 ) m1 t ( m 2 + µ C1 ) ( m 2 + µ C2 ) m2 t µ C1 µ C2 - e – ---------------------------------------------------- e P 0 ( t ) = ---------------- + --------------------------------------------------m1 ( m1 – m2 ) m2 ( m1 – m2 ) m1 m2 (5.15)



λ 1 m 1 + λ 1 µ C2 m1 t ( µ C1 + m 2 )λ 1 m2 t λ 1 µ C2 - e – ------------------------------- e P 1 ( t ) = ------------- + ------------------------------m2 ( m1 – m2 ) m1 ( m1 – m2 ) m1 m2

(5.16)

λ 2 m 1 + λ 2 µ C1 m1 t ( µ C1 + m 2 )λ 2 m2 t λ 2 µ C1 - e – ------------------------------- e P 2 ( t ) = ------------- + ------------------------------m2 ( m1 – m2 ) m1 ( m1 – m2 ) m1 m2

(5.17)

where 2

1/2

– A ± ( A – 4B ) m 1 , m 2 = -------------------------------------------2

(5.18)

A = µC1 + µC2 + λ1 + λ2

(5.19)

B = µC1 µC2 + λ1 µC2 + λ 2 µC1

(5.20)

m1m2 = µC1 µC2 + λ1 µC2 + λ 2 µC1

(5.21)

m 1 + m 2 = – ( µ C1 + µ C2 + λ 1 + λ 2 )

(5.22)

The system availability, AS(t), is given by AS(t) = P0(t)

(5.23)

As time t becomes very large, from Eqs. (5.15) and (5.23), we get the following expression for the system steady state availability: µ C1 µ C2 µ C1 µ C2 - = --------------------------------------------------------A S = ---------------m1 m2 µ C1 µ C2 + λ 1 µ C2 + λ 2 µ C1

(5.24)

Example 5.2 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 I and II are µC1 = 0.006 repairs per hour and µC2 = 0.009 repairs per hour, respectively. Calculate the system steady state availability. Inserting the specified values into Eq. (5.24) yields 0.006 × 0.009 A S = ------------------------------------------------------------------------------------------------------------------------------( 0.006 × 0.009 ) + ( 0.002 × 0.009 ) + ( 0.005 × 0.006 ) = 0.5294 Thus, the system steady state availability is 0.5294. There is approximately a 53% chance that the system will be available for service when needed.



λ2

λ1 System operating normally

0

λ3 System operating in its degradation mode

µC 1

1

System failed

µC 3

2

µC 2


MODEL III This mathematical model represents a system that can either be operating normally, operating in degradation mode, or failed completely. An example of this type of system could be a power generator, i.e., producing electricity at full capacity, derated capacity, or not at all. Corrective maintenance is initiated from degradation and completely 10 failed modes of the system to repair failed parts. The system state space diagram is shown in Fig. 5.5. The model is subject to the following assumptions: • System complete failure, partial failure, and corrective maintenance rates are constant. • The operating system can either fail fully or partially. The partially operating system can stop operating altogether. • All system failures are statistically independent. • The repaired system is as good as new. The following symbols are associated with the model: = the ith system state, i = 0 (system operating normally), i = 1 (system operating in its degradation mode), i = 2 (system failed), Pi (t) = probability that the system is in state i at time t, for i = 0, 1, 2, λi = system failure rate, i = 1 (from state 0 to state 1), i = 2 (from state 0 to state 2), i = 3 (from state 1 to state 2), µCi = system corrective maintenance rate, i = 1 (from state 1 to state 0), i = 2 (from state 2 to state 0), i = 3 (from state 2 to state 1). i

For Models I and II, we write the following equations for the Fig. 5.5 diagram: dP 0 ( t ) ---------------- + ( λ 1 + λ 2 )P 0 ( t ) = µ C1 P 1 ( t ) + µ C2 P 2 ( t ) dt


(5.25)


dP 1 ( t ) ---------------- + ( µ C1 + λ 3 )P 1 ( t ) = µ C3 P 2 ( t ) + λ 1 P 0 ( t ) dt

(5.26)

dP 2 ( t ) ---------------- + ( µ C2 + µ C3 )P 2 ( t ) = λ 3 P 1 ( t ) + λ 2 P 0 ( t ) dt

(5.27)

At time t = 0, P0 (0) = 1 and P1 (0) = P2 (0) = 0. By solving Eqs. (5.25)–(5.27), we get ( µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 ) P 0 ( t ) = ----------------------------------------------------------------K 1K 2 2

µ C1 K 1 + µ C2 K 2 + µ C3 K 1 + K 1 λ 3 + K 1 + µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 K 1 t - e + -------------------------------------------------------------------------------------------------------------------------------------------------------------Y  µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 +  1 –  -----------------------------------------------------------  K 1K 2  µ C1 K 1 + µ C2 K 1 + µ C3 K 1 + K 1 λ 3 + K 2 + µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3  K 2 t - e – -------------------------------------------------------------------------------------------------------------------------------------------------------------Y  (5.28) 2

where Y = K1(K1 − K2 ). K 1 λ 1 + λ 1 µ C2 + λ 1 µ C3 + λ 2 µ C3 K 1 t λ 1 µ C2 + λ 1 µ C3 + λ 2 µ C3 - + ------------------------------------------------------------------------ e P 1 ( t ) =  -----------------------------------------------------  K 1K 2 Y λ 1 µ C2 + λ 1 µ C3 + λ 2 µ C3 K 1 λ 1 + λ 1 µ C2 + λ 1 µ C3 + λ 2 µ C3 K 2 t - + ------------------------------------------------------------------------- e – -----------------------------------------------------K 1K 2 Y

(5.29)

K 1 λ 2 + λ 1 λ 3 + λ 2 µ C1 + λ 2 λ 3 K 1 t λ 1 λ 3 + µ C1 λ 2 + λ 2 λ 3 - + ------------------------------------------------------------------ e P 2 ( t ) =  -----------------------------------------------  K 1K 2 Y λ 1 λ 3 + µ C1 λ 2 + λ 2 λ 3 λ 2 K 1 + λ 1 λ 3 + µ C1 λ 2 + λ 2 λ 3 K 2 t - + ------------------------------------------------------------------- e – -----------------------------------------------K 1K 2 Y

(5.30)

where 2

1/2

– D ± ( D – 4F ) K 1 ,K 2 = --------------------------------------------2

(5.31)

D = µ C1 + µ C2 + µ C3 + λ 1 + λ 2 + λ 3

(5.32)

F = K 1 K 2 = µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 + µ C2 λ 1 + λ 1 µ C3 + λ 1 λ 3 + µ C1 λ 2 + λ 2 µ C3 + λ 2 λ 3


(5.33)


The system full/partial availability, A Sf/p ( t ), is given by A Sf/p ( t ) = P 0 ( t ) + P 1 ( t )

(5.34)

As t becomes large, Eq. (5.34) reduces to µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 + λ 1 µ C2 + λ 1 µ C3 + λ 2 µ C3 A Sf/p = ------------------------------------------------------------------------------------------------------------------------K 1K 2

(5.35)

where A Sf/p = system full/partial steady-state availability. Similarly, the system full steady-state availability is µ C1 µ C2 + λ 3 µ C2 + µ C1 µ C3 A Sf = P 0 = -----------------------------------------------------------K 1K 2

(5.36)

Example 5.3 Assume that in Eq. (5.36), we have λ1 = 0.002 failures per hour, λ2 = 0.003 failures per hour, λ3 = 0.001 failures per hour, µC1 = 0.006 repairs per hour, µC2 = 0.004 repairs per hour, and µC3 = 0.008 repairs per hour. Calculate the value of the system full steady-state availability. Inserting the specified data values into Eq. (5.33) yields K 1 K 2 = ( 0.006 × 0.004 ) + ( 0.001 × 0.004 ) + ( 0.006 × 0.008 ) + ( 0.004 × 0.002 ) + ( 0.002 × 0.008 ) + ( 0.002 × 0.001 ) + ( 0.006 × 0.003 ) + ( 0.003 × 0.008 ) + ( 0.003 × 0.001 ) 0.0001 Using the above calculated value and the given data in Eq. (5.36) we get ASf = 0.5170. There is approximately 52% chance that the system will be available for full service.

MODEL IV This mathematical model represents a two identical-unit redundant (parallel) system. At least one unit must operate normally for system success. Corrective maintenance 11 to put it back into its operating state begins as soon as any one of the units fails. The system state space diagram is shown in Fig. 5.6. The following assumptions are associated with the model: • The system is composed of two independent and identical units. • The repaired unit is as good as new. • No corrective maintenance is performed on the failed system (i.e., when both units fail). • Unit failure and corrective maintenance rates are constant.



µCm

Both units operating normally (system up)

2λ

0

One unit failed, other operating normally (system up)

Both units failed (system failed)

λ

1

2

FIGURE 5.6 Two identical unit redundant system transition diagram.

The following symbols pertain to the model: = the ith system state, i = 0 (both units operating normally), i = 1 (one unit failed, other operating), i = 2 (both units failed), Pi (t) = probability that the system is in state i at time t, for i = 0, 1, 2, λ = unit failure rate, µCm = unit corrective maintenance rate. i

For Models I, II, and III we write the following equations for the Fig. 5.6 transition diagram: dP 0 ( t ) ---------------- + 2λP 0 ( t ) = µ Cm P 1 ( t ) dt

(5.37)

dP 1 ( t ) ---------------- + ( µ Cm + λ )P 1 ( t ) = 2λP 0 ( t ) dt

(5.38)

dP 2 ( t ) ---------------- = λP 1 ( t ) dt

(5.39)

At time t = 0, P0(0) = 1, and P1(0) = P2(0) = 0. Solving Eqs. (5.37)–(5.39), we get λ+µ+C C t λ+µ+C C t P 0 ( t ) = -------------------------1 e 1 – -------------------------1 e 2 C1 – C2 C1 – C2

(5.40)

C t C t 2λ 2λ P 1 ( t ) = ------------------ e 1 – ------------------ e 2 C1 – C2 C1 – C2

(5.41)

C t C t C2 C1 - e 1 – ----------------- e 2 P 2 ( t ) = 1 + ----------------C1 – C2 C1 – C2

(5.42)

where 2

2 1/2

[ – ( 3λ + µ ) ± ( 3λ + µ ) – 8λ ] C 1 , C 2 = -------------------------------------------------------------------------------2


(5.43)


C 1 C 2 = 2λ

2

C 1 + C 2 = – ( 3λ + µ )

(5.44) (5.45)

The system reliability is given by RS ( t ) = P0 ( t ) + P1 ( t )

(5.46)

where RS (t) = redundant system reliability at time t. The system mean time to failure (MTTFS) is given by MTTF S =

∞

∫ RS ( t ) dt 0

3λ + µ Cm = --------------------2 2λ

(5.47)

Since λ = 1MTTFu and µ Cm = 1 MCMT, Eq. (5.47) becomes MTTF MTTF S = --------------------u- ( 3MCMT + MTTF u ) 2MCMT

(5.48)

where MTTFu = unit mean time to failure, MCMT = unit mean corrective maintenance time. Example 5.4 A system is composed of two independent and identical units in parallel. A failed unit is repaired immediately but the failed system is never repaired. The unit times to failure and corrective maintenance times are exponentially distributed. The unit mean time to failure and mean corrective maintenance time are 150 h and 5 h, respectively. Calculate the system mean time to failure with and without the performance of corrective maintenance. Inserting the given values into Eq. (5.48), we get 150 MTTF S = ------------ ( 3 × 5 + 150 ) = 2475 h 2×5 Setting µCm = 0 and substituting the given value into Eq. (5.47) yields 3 MTTF S = ------ = 225 h 2λ This means introduction of corrective maintenance helped increase system mean time to failure from 225 h to 2475 h.



APPROXIMATE EFFECTIVE FAILURE RATE EQUATIONS FOR REDUNDANT SYSTEMS WITH CORRECTIVE MAINTENANCE This section presents approximate effective failure rate equations for two types of redundant systems. The effective failure rate is the reciprocal of the item/system mean time to failure.

EFFECTIVE FAILURE RATE

OF

SYSTEM TYPE I

The system type I is assumed to contain m number of independent and identical active units in parallel and in which k units are allowed to fail without system failure. The corrective maintenance begins as soon as a unit fails. The failed system is never repaired. The unit failure and corrective maintenance rates are constant. Thus, an approximate effective failure rate of a (m − K)-out-of-m system can be calculated 12 using the following equation: K +1

m!λ λ ( m−K )/m = ------------------------------------K( m – K – 1 )!µ

(5.49)

where λ(m−K)/m = system approximate effective failure rate. In this system at least (m − K) units must work normally for the system success, λ = unit failure rate, µ = unit corrective maintenance rate. Example 5.5 A system is composed of three independent and identical units in parallel and at least two units must operate normally for the system success. The unit failure rate is 0.0001 failures per hour. It takes an average of 2 h to repair (exponential distribution) a failed unit to an active state. Calculate the system approximate effective failure rate if the failed system is never repaired. 2

2

6λ 3!λ λ ( 3−1 )/3 = ----------- = -------1!µ µ 2

6 ( 0.0001 ) = -------------------------0.5 = 1.2 × 10

–7

failures per hour −7

The system effective failure rate is 1.2 × 10 failures per hour.



EFFECTIVE FAILURE RATE

OF

SYSTEM TYPE II

System type II is composed of two independent and nonidentical units in parallel. Corrective maintenance begins as soon as either unit fails. The failed system is never repaired. Unit failure and corrective maintenance rates are constant. 12 An approximate formula to obtain system effective failure rate is as follows: λ1 λ2 [ ( µ1 + µ2 ) + ( λ1 + λ2 ) ] λ se = -----------------------------------------------------------------µ1 µ2 + ( µ1 + µ2 ) ( λ1 + λ2 )

(5.50)

where λse = two unit parallel system effective failure rate, λi = unit i failure rate, for i = 1, 2, µi = unit i corrective maintenance rate, for i = 1, 2. Example 5.6 A system is composed of two independent and nonidentical units in parallel. Unit 1 failure and corrective maintenance rates are 0.004 failures per hour and 0.005 repairs per hour, respectively. Similarly, the unit 2 failure and corrective maintenance rates are 0.002 failures per hour and 0.003 repairs per hour, respectively. The failed system is never repaired. Calculate the system approximate effective failure rate. Substituting the given data into Eq. (5.50), we get ( 0.004 × 0.002 ) [ ( 0.005 + 0.003 ) + ( 0.004 + 0.002 ) ] λ se = -----------------------------------------------------------------------------------------------------------------------------( 0.005 × 0.003 ) + ( 0.005 + 0.003 ) ( 0.004 + 0.002 ) = 0.0018 failures per hour The system effective failure rate is 0.0018 failures per hour.

PROBLEMS 1. Define corrective maintenance. 2. Describe the following types of corrective maintenance: • Overhaul • Rebuild • Servicing 3. Discuss sequential steps associated with corrective maintenance. 4. Define main components of active repair time. 5. Discuss at least four strategies for reducing the system-level corrective maintenance time. 6. Define median corrective maintenance time. 7. Assume that exponential mean time to failure and mean corrective maintenance time of a system are 2500 h and 4 h, respectively. Calculate the system steady-state availability.



8. A system can fail in two mutually exclusive failure modes. Failure mode I constant failure and corrective maintenance rates are 0.005 failures per hour and 0.02 repairs per hour, respectively. Similarly, failure mode II constant failure and corrective maintenance rates are 0.001 failures per hour and 0.03 repairs per hour, respectively. Calculate the system steadystate availability. 9. A system is composed of two independent and identical units in parallel. Although a failed unit is repaired immediately, the failed system is never repaired. The unit times to failure and corrective maintenance times are exponentially distributed. Thus, the unit mean time to failure and mean corrective maintenance time are 200 h and 2 h, respectively. Calculate the system mean time to failure. 10. Assume that a system is composed of two independent and identical units in parallel and at least one unit must operate normally for system success. The unit failure and repair rates are 0.002 failures per hour and 0.01 repairs per hour, respectively. The failed system is never repaired. Calculate the value of the system approximate effective failure rate.

REFERENCES 1. AM 706-132, Engineering Design Handbook: Maintenance Engineering Techniques, Department of Defense, Washington, D.C., 1975. 2. Omdahl, T.P., Reliability, Availability, and Maintainability (RAM) Dictionary, ASQC Quality Press, Milwaukee, Wisconsin, 1988. 3. McKenna, T. and Oliverson, R., Glossary of Reliability and Maintenance , Gulf Publishing Company, Houston, Texas, 1997. 4. MICOM 750-8, Maintenance of Supplies and Equipment, Department of Defense, Washington, D.C., March 1972. 5. NAVORD OD 39223, Maintainability Engineering Handbook, Department of Defense, Washington, D.C., June 1969. 6. Blanchard, B.S., Verma, D., and Peterson, E.L., Maintainability, John Wiley & Sons, New York, 1995. 7. AM-766-133, Engineering Design Handbook: Maintainability Engineering Theory and Practice, Department of Defense, Washington, D.C., 1976. 8. Dhillon, B.S., Design Reliability: Fundamentals and Application, CRC Press, Boca Raton, Florida, 1999. 9. Dhillon, B.S. and Singh, C., Engineering Reliability: New Techniques and Applications, John Wiley & Sons, New York, 1981. 10. Dhillon, B.S., Reliability Engineering in Systems Design and Operation, Van Nostrand Reinhold Co., New York, 1983. 11. Shooman, M.L., Probabilistic Reliability: An Engineering Approach, McGraw-Hill, New York, 1968. 12. RADC Reliability Engineer’s Toolkit, prepared by the Systems Reliability and Engineering Division, Rome Air Development Center, Griffiss Air Force Base, Rome, New York, July 1988.


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