00 fundamentals maintenance management 2006
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THE PROBLEM SOLVING PROCESS
Training course:
Fundamentals Maintenance Management
Author:
Prof. Dr. Attia H. Gomaa
Head of Industrial Eng. Department - Fayoum University
Industrial Engineering Consultant - AUC
Maintenance Engineering Consultant - EMC
attiagomaa@yahoo.com
2006
Fundamentals Maintenance Management
Author:
Dr. Attia H. Gomaa
Head of Industrial Eng. Department - Fayoum University attiagomaa@yahoo.com
Who Should Attend:
Managers, engineers, and other practitioners concerned with maintenance planning and control in government, industrial and services sectors.
Objectives:
To provide the participants with the modern concepts and techniques in maintenance planning and control.
To train the participants on how to use and apply these techniques in practice.
To enhance the participants experience by discussing some maintenance management problems and how to deal with them.
Course Outline:Level I: Traditional Maintenance Management1. Maintenance Management Overview
2. Preventive Maintenance Management
3. Maintenance Control
4. Computer Applications
5. PM Case Studies
6. Machine Failure Analysis
Level II: Advanced Maintenance Management7. Predictive Maintenance Management
8. Risk Based Inspection
9. Reliability Centered Maintenance
10. Total Productive Maintenance
11. Practical cases.
Level I
traditional maintenance management
1. Maintenance Management Overview
What is Maintenance?
BS 3811:1974Maintenance is defined as:
The work under taken in order to keep or restore a facility to an acceptable standard level.
Or
The combination of activities by which a facility is kept in, or restored to, a state in which it can perform its acceptable standard.
Maintenance Policies
To Keep
Planned Maintenance To Restore
Unplanned Maintenance
- Time Based Maintenance
- Condition Based Maintenance
- Risk Based Maintenance- Corrective Maintenance
- Run To Failure
- Emergency Maintenance
- Break down Maintenance
Preventive maintenance - Time-based PM
Pure time )calendar) based: Weekly, monthly, annually, etc.
Used (running) time based: 1000 km, 1000 RH, 3000 RH, etc.
Predictive (Condition-based) Maintenance by monitoring key equipment parameters "Off-line or On-line" Vibration analysis
Oil analysis
Wear analysis
Noise analysis
Temperature analysis
Pressure analysis
Quality analysis
Efficiency analysis, etc.
What are the main factors, which affect the selection of Maintenance Policy?
o Manufacturing maintenance recommendation
o System availability
o Safety factors
o Production process
o Operating conditions
o Information availability
o Resource availability
o Operating & maintenance cost
o Down time cost rate
o Failure and repair characteristics
What is the Maintenance?Example:
1-
How to keep or restore the facility at acceptable standard level in certain operating conditions? System/equipment description
Main parameters
Main items
Functional block diagram
Criticality
Working conditions
2-
How to prevent the failures?
Main failures:
PM:
3-
How to discover the hidden failures?
Main failures:
Policy:
4-
How to detect the early failures?
Main failures:
Policy:
5-
How to minimize the risk of failures?
Main failures:
Risk:
Policy:
According to maintenance information availability:
(1)
Complete Information
Planned PM
70 %(2)
Incomplete information
Planned CM
20%(3)
Without information
Unplanned CM
(or Emergency)
10%
Typical Work (man-hour) distribution in engineering industries
Experience:
Technical
Planning
Analysis
Decision making
Problem solving
Working conditions, etc.
Information:
Catalog
Forms / reports
Data collection
PM levels
Job plans for each PM level
Resources
Cost rates
CM work orders
Failure analysis, etc.
Tools:
Computer programs
International standards
Management tools, etc.
What is the ratio between maintenance cost
& manufacturing costs?
Maintenance costs are a major part of the total operating costs of all manufacturing or production plants.
Depending on the specific industry, maintenance costs can represent between 15% and 40% of the costs of goods produced.
For example in food related industries, the average maintenance cost represents about 15% of the cost of goods produced; while in iron and steel, pulp and paper and other heavy industries maintenance represents up to 40% of the total production costs.
US industry spends more than $200 billion dollars each year on maintenance of plant equipment and facilities,
USA Industries in 1983/ 1984: Maintenance Cost ( $ 35 * 109 Per year
Maintenance Cost: 10 25 % &
Spare parts Cost: 3 10 %
What are the main elements of Maintenance cost?
Direct cost:
Spare parts & supplies cost
Labor cost
Contract cost
Indirect cost:
Overhead cost
Down time cost
Maintenance cost = Direct cost + Overhead cost
Maintenance Costs Elements
Cost to replace or repair
Losses of output
Delayed shipment
Scrap and rework
What is Maintenance Management (MM)?
MM is a powerful systematic methodology to maximize the facility performance and to improve the maintenance resource productivity, through optimizing the maintenance policies for the critical equipment. MM - is the application of knowledge, tools and scientific techniques to identifying and analysis the maintenance activities.
MM - decision-making process to select the best maintenance policies for improving the equipment reliability to an acceptable level.
MM is the art of matching a maintenance's goals, tasks, and resources to accomplish a goal as needed.
MM is do the right things, with the right tools, and in the right way".Through:
1. Define the target and constraints,
2. Information collecting & analysis,
3. Maintenance planning,
4. Maintenance organization,
5. Motivation & direction,
6. Maintenance control,
7. Corrective actions, and
8. Learned lessons.
Maintenance Management History
How do you measure MM success?
Maintenance Planning Concept:
Before you start to maintenance plan, consider...
Who is the ultimate customer?
What are the customer needs?
How long will the maintenance project last?
Where are we now?
Where should we end-up?
What are the cost constraints?
What are the technical challenges?So, Maintenance Planning must determines what, when, where, how, and by whom something is done.
What is to be maintained?
"Description"
Why?
"Target" How?
"Method" By whom?
"Resources" When?
"Schedule"
Where?
"Location"What are the main Types of MM Plans?
1- MM management level plans:
Master plan Top management (10 -15 activity)
Action planControl management (50-100)
Detailed planOperational management (> 500)2- MM Time plans:
Long term
2 to 10 y
Risk 15 to 25%
Medium term
6m to 1 y
Risk 7 to 10%
Short term
1w to 3 mRisk 3 to 5%3- MM risk plans:
Target plan (normal or most likely)
Optimistic plan (best case)
Pessimistic plan (worst case)
4- MM Strategic Plans:
Strategic plan
Tactical plan
Operational plan
Urgent plan
5- MM Planning Level:
Overall plan
Complete information Partial plan
Incomplete information Urgent plan
Without informationWhat is the Maintenance System?
A system is a collection of components (or items) that work together to achieve a certain objective.
Sub-system: Water Pump Unit
Figure - Functional block diagram for a pump
Figure Main Components
Current PM Program:
ItemJob planFrequency
(1)
Motor
(2)
Coupling
(3)
Pump
(4)
Suction line
(5)
Discharge line
(6)
Valves
Root Cause Failure Analysis:
ItemMain FailuresRoot CauseMTBF
(1)
Motor
(2)
Coupling
(3)
Pump
(4)
Suction line
(5)
Discharge line
(6)
Valves
1) Motor:
FailurePMPrDCM
PolicyFreq.PolicyFreq.
2) Coupling:
3) Pump:
4) Suction line:
5) Discharge line:
6) Valves:
Developed PM Program:
ItemJob planFrequency
(1)
Motor
(2)
Coupling
(3)
Pump
(4)
Suction line
(5)
Discharge line
(6)
Valves
Modern Maintenance Management Systems:
There are four modern approaches:
1- Optimal system maintenance (OSM),
2- Risk Based Inspection (RBI)
3- Reliability centered maintenance (RCM), and
4- Total productive maintenance (TPM).
Maintenance management methodologies
OSMRBI & RCMTPM
Main objective Improve equipment availabilityPreserve system function & improve system availabilityImprove overall system productivity
ApproachMaintenance information analysis and
Using optimal mathematical modelingImprove the maintenance program
System reliability analysis
Failure mode effect analysis FMEA
Risk analysisSystem overall analysis
Continuous improvement techniques
PolicyApproachGoals
ReactiveRun to failure (fix-it when broke).Minimize maintenance costs for non-critical equipment.
PreventiveUse-based maintenance program.Minimize equipment breakdown.
PredictiveMaintenance decision based on equipment condition.Discover hidden failures and improve reliability for critical equipment.
ProactiveDetection of sources of failures.Minimize the risk of failures for critical systems.
GlobalIntegrated approach.Maximize the system productivity.
PolicyApproachGoals
RCFAIdentification of root causes of failures.Eliminate failures.
FMECAIdentification of criticality of failures.Improve equipment availability.
HAZOPIdentification of hazards and problems associated with operations.Improve HSE effect.
RCMDetermination of best maintenance requirements for critical systems.Preserve system function & improve reliability.
RBIDetermination of an optimum inspection plan for critical systems.Improve system HSE and availability.
PolicyApproachGoals
OSMOptimization approach for the global maintenance system.Maximize reliability measures and minimize maintenance cost rates.
TPMComprehensive productive-maintenance system.Maximize plant effectiveness and resource productivity.
Preventive Maintenance Management
Why Preventive Maintenance should be done? To Prevent FailureTo Detect Early Failure
To Discover a Hidden Failure
Rather, it is better to consider PM only when:
1- High Down time cost rate
2- High Safety level
3- Predictive M. cannot be applied
4- CM cannot be justified
What are the main targets of PM? Improving equipment availability/reliability
Increasing equipment effective life time
Increasing resource utilization
Increasing productivity
Reducing operating cost
Reducing total cost rate
Increasing profitability ratio
PM = Profit
What are the main Elements of Planned Maintenance?
1. Inventory list
2. Layout of facilities
3. Facility register
4. Maintenance program
5. Maintenance job specification
6. Maintenance schedule
7. Job orders
8. Follow up cards
9. Performance evaluation
Note : 1 to 5 Basic data, 6 Scheduling, and 7 to 9 Follow up and evaluation.Maintenance Planning Steps:
1. System criticality analysis
2. Equipment selection
3. Information collection & analysis
4. Target & constraints definitions
5. Requirements & standard levels
6. Main failures determination
7. Root cause failure analysis (RCFA)
8. Best maintenance policy
9. Maintenance policy planning
10. Work orders
11. Measure
12. Analysis
13. Action
14. Performance evaluation & KPI
15. Improvement
Maintenance Planning Steps:StepDescription
1. System criticality analysisHSE - Process Down time Cost
2. Equipment selection Critical equipment
Non-critical equipment
3. Information collection & analysisMaintenance catalog Design information Equipment history- Working conditions- PMs CMs Trouble shooting Reliability information HSE instructions. etc.
4. Target & constraints definitions Targets: Reliability, Availability, Down time, Cost, HSE level, .. etc.
Constraints: Budget, Spare parts, Tools, Manpower, Information,etc.
5. Requirements & standard levels Functional levels: Flow rate, Head, Pressure, Power, .. etc.
HSE levels
6. Main failures determinationFunctional failures - HSE failures Mechanical failures Electrical failures - .. etc.
7. Root Cause Failure Analysis Main failures, Root cause, RRC, Mechanism, Probability, MTBF, MTTR, Remedy.
8. Best maintenance policy Run To Failure (RTF)
Time-based (Preventive) PM
Condition-based (Predictive) PdM
Risk-based (Proactive) PrM
Maintenance Planning Steps:StepDescription
9. Maintenance policy planningFrequency- Levels- Alarm limits- Tools- Job plan- HSE plan- Spare parts- Duration- Manpower- .. etc.
10. Work orders W/O # - W/O type- Dates/time - Responsibility- Level - Alarm limits- Tools- Job plan- HSE plan- Spare parts- Duration- Manpower- Failure - Root cause- .. etc.
Complete Feedback.
11. MeasureRunning hours- Noise- Vibration- Temperature- Oil level- viscosity- Flow rate Head Speed - .. etc.
12. AnalysisNoise analysis- Vibration analysis Temperature analysis - Oil analysis - Flow rate analysis Head analysis Speed analysis - .. etc.
13. Action- Good condition
- Call for service (PM)
- Call for repair (planned CM)
- Breakdown (unplanned CM)
14. Performance evaluation & KPICM/PM- MTBF- MTTR- MTBM- MTTM- Reliability Availability- Maintainability- RAM- Spare parts consumption rates- .. etc.
15. Improvement Information Maintenance levels- Tools Spare parts Manpower skills Time HSE - .. etc.
Approach: FMEA - RCM - RBI- PMIS - .. etc.
What are the main Elements of Maintenance Plan?
1- Equipment name & code,
2- Equipment priority,
3- Maintenance start time,
4- Maintenance down time,
5- Maintenance level and type,
6- Maintenance job description,
7- Maintenance operations time,
8- Maintenance effort (man-hour),
9- Manpower requirements planning,
10- Spare parts and supplies requirement planning,
11- Tools requirements planning,
12- Failure analysis,
13- Maintenance cost estimation,
14- Maintenance budget, and
15- Safety instructions.
Maintenance Work Order
Work order number
Requester Section:
Plant (or department) name / code
Equipment name / code
Equipment priority
Maintenance type & level (PM / Repair / Overhaul)
Job scope & description
Responsibility
Planning Section:
Manpower types & skills
Time estimation
Spare parts
Special tools
Expected equipment down time (from xxx to xxx)
Cost estimation
Safety instructions
Responsibility
Craft Feedback:
Job scope & description
Manpower types & skills
Time estimation
Spare parts
Special tools
Actual equipment down time (from xxx to xxx)
Actual Cost
Responsibility
Coding:
Plant (or department), Equipment
Resources (Manpower, Spare parts, Special tools)
3- Maintenance Control
Total Control Indicators:
1- Work quantity control
Over estimation
Under estimation
2- Time control
Behind schedule (late)
Ahead schedule (early)
3- Cost control
Cost overrun
Cost under-run
4- Quality control
Acceptable level
Non-acceptable level
5- Inventory control
Over estimation
Under estimation
6- Resources control
Over estimation
Under estimation
7- Plant condition control (HSE, etc.)
Acceptable level
Non-acceptable level
Control Steps:
1- What to control?
2- What is the standard (target) performance?
3- What is the actual performance level?
4- Comparison between the actual & target.
5- Detection of variance
6- Identification of causes of variance
7- Corrective actions
8- Learned lessons.
Total Control Levels:
1- Review and data collection.2- Follow-up.
3- Performance evaluation.
4- Productivity analysis.
5- Corrective actions.
6- Learned lessons.
Maintenance Control Levels:
- Maintenance Follow-up
-(Actual/Plan)- Maintenance Performance Evaluation- Time Availability
- Reliability
- Mean Time Between Failures (MTBF)- Mean Time To Failures (MTTF)- Mean time to repair (MTTR)- Mean time between repairs (MTBR)- Mean Time Between Maintenance (MTBM)- Preventive Maintenance Rate (PM rate)- Resources Productivity Analysis
Productivity Dimensions
CostQualityQuantityTime
Efficiency
1- Technical Efficiency
2- Operating Efficiency
3- Production Efficiency
4- Economical Efficiency Effectiveness
= Actual output /
Planned output
Maintenance System Effectiveness:
It is related to performance.
It is the degree of accomplishment of objectives.
How well a set of results is accomplished?
Maintenance System Efficiency:
It is related to resource utilization.
It is the degree resources utilization.
How well the resources are utilized to achieve the results.
Productivity:
It is a combination of both effectiveness & efficiency.
Productivity index
= Output obtained / Input expended
= Performance achieved / Resources consumed
Total productivity = Total output / Total input
Partial productivity = Total output / One of the inputs
Measurement of MAINTENANCE EFFECTIVENESS
Equipment Losses Categories
CategoryEquipment lossesIndicator
Down-time losses
(lost availability)Equipment failures
Set-up and adjustmentsEquipment availability
Speed losses
(lost performance)Idling and minor stoppages
Reduced speed operationEquipment performance efficiency
Defect losses
(lost quality)Scrap and rework
Start-up lossesEquipment quality
Rate
Resource lossesCritical resource consumption ratesResource productivity
Cost lossesAll the previous lossesRepair cost
CM/PM cost ratio
Down time cost
Overall equipment effectiveness (OEE)
OEE = Equipment Availability Performance efficiency Quality rate
Total effective equipment productivity (TEEP)
TEEP =Utilization Availability Performance efficiency Quality rate
Net equipment effectiveness (NEE)
NEE = Uptime ratio Performance efficiency Quality rate
Mean unit between assists (MUBA):
MUBA = Total number of units produced / Number of stoppages
What is the effect of Maintenance Policy on the Equipment OEE?
Maintenance PolicyOEE
Operate to failure (RTF) 30 50 %
Good PM Program
Good bonus & incentive system60 80 %
Good PM Program based on RCM
Good bonus & incentive systemMore than 80 %
What are the main factors, which affect the Equipment OEE?
Product quality
Production continuity & rates
Shutdown frequency HSE factors
Equipment availability
Resource availability
Operating & maintenance cost
Down time cost rate
Maintenance Risk levels:
Objective Risk Levels:
Risk %0 - 55 1010 - 1515 - 25> 25
Risk level01234
DescriptionMinorLowMediumHighMajor
Acceptable Risk limits:
Long term
2 to 10 y
Risk 15 to 25%
Medium term
6m to 1 y
Risk 7 to 10%
Short term
1w to 3 mRisk 3 to 5%Maintenance Safety Levels:
LevelSeveritySafety (people)
0NoDoes not apply
1Very low
(Slight)Slight injury
Simple first aid
2Low
(Not Serious)Minor injury
No lost time
No Hospitalize
First aid
3Medium
(Serious)Major injury
Lost time
Hospitalize
Temporary disability
4High
(Very Serious)Fatal injury
Hospitalize
Permanent disability
5Very High
(Catastrophic)Multiple fatalities
Maintenance Performance Evaluation
What are our measures?
What are the units?
What is the time frame? What data is required?
What data is available?
Quality of data
Linking data to measuresHow to measure the performance of PM program?
Four major factors that should control the extent of a PM program:
1- The cost of PM program (PM & repairs costs).
2- Equipment reliability & utilization.
3- HSE (Health, Safety and Environment) level
4- Down time cost.
Availability = A = x 100%
Percentage of downtime = Id = 100% - AMean time between failures = MTBF =
Mean time to repair MTTR =
Where,S = Scheduled production time
d = Downtime
f = Number of failures.
df = Downtime delays from failures.
Example:
Scheduled production time = 31 day
Downtime = 6 day
Number of failures = 3 failure/month
A = x 100% = 80.6 %
Id = 100 - 80.6 = 19.4%
MTBF = = 8.33 days
MTTR= = 2 days
Maintenance Administration Indicators (%):
1- Overtime hours per month
2- Worker activity level
3- Worker productivity
4- Worker utilization
5- Scheduled hours6- Preventive & predictiveMaintenance Effectiveness Indicators (%):
1- Overall effectiveness
2- Gross operating hours
3- Number of failures
4- Breakdown downtime
5- Emergency man-hours
6- Predictive & preventive
Maintenance Cost Indicators (%):
1- Maintenance cost
2- Maintenance cost/unit
3- Maintenance manpower cost
4- Subcontracted cost
5- Cost of maintenance-hour
6- Supervision cost
7- Preventive maintenance cost
8- Cost of spare partsMain Indicators Calculations:
Overtime hours per month = % =
x 100
Worker activity level = % =
x 100
Worker productivity per month = % = x 100
Worker utilization = % =
x 100Scheduled hours versus hours worked = %
= x 100
Preventive and predictive maintenance conducted as scheduled = %=
x 100
Predictive and preventive maintenance coverage% =
x 100
Overall equipment effectiveness (OEE) = A x S x Q
A = Availability indicator
S = Speed indicator
Q = Quality indicator
Availability = A =
Speed = S =
Quality = Q =
Percentage of gross operating hours % =
x 100
Number of failures in the system (NFS) =
Equipment downtime caused by breakdown % =
x 100
Emergency man-hours % =
x 100
Emergency and all other unscheduled man-hours % =
Evaluation of predictive and preventive maintenance % =
x 100
Cost of maintenance to added value of production % =
x 100
Maintenance cost per unit of production = Cost per unit
=
Manpower component in the maintenance cost % =
x 100
Cost of subcontracted maintenance =% =
x 100
Ratio of labor cost to material cost of maintenance =
Cost of maintenance-hour = $ =
Supervision cost as a percentage of total maintenance cost %=
x 100
Progress in cost reduction effects = Index =
Preventive maintenance (PM) cost as related to breakdown maintenance
% = x 100
Inventory turnover rate per year =
Rate =
Cost of spare parts and material to maintenance cost
% = x 100
Ratio of stock value to production equipment value =
4- Computerized CMMS
More than 100 Ready-Made Packages
Most common CMMS: EMPAC
www.plant-maintenance.com
FMMS
www.kdr.com.au
GPS5
www.gps5.com
IMAINT
www.dpsi.com
IMPACT-XP
www.impactxp.com
IMPOWER
www.impower.co.uk
MAINPAC
www.mainpac.com.au MAINPLAN
www.mainplan.com
MAXIMO
www.maximo.com
MP2
www.datastream.net
OEE MANAGERwww.zerofailures.co.uk
OEE SYSTEMSwww.oeesystems.com
OEE TOOLKITwww.oeetoolkit.com
OEE-IMPACTwww.oeeimpact.com
PEMAC
www.pemac.org
PERFORM OEE www.ssw.ie/performoee.asp
RAMS
www.reliability.com.au
RCM Turbo
www.strategic.com
REAL-TPI
www.abb.com
SAP-RLINK
www.osisoft.com
TPM Software
www.tpmsoftware.com
Most MMIS systems can usually:
1. Track components,
2. Provide logistic support (e.g., spares inventory),
3. Store maintenance history,
4. Alarm predetermined maintenance activities,
5. Produce management reports.
A small number of these systems are able to:
6. Analyse maintenance history, and
7. Determine optimal policies for components and sub-systems.For a complex system, MMIS will also have to:
8. Incorporate expert opinion in a knowledge base,
9. Incorporate subjective data from experts,
10. Combine maintenance activities into schedules,11. Update schedules with occurrence of events such as failures etc,
12. Plan resources, and
13. Measure the effectiveness of maintenance activities.
This requires a more quantitative and scientific approach to maintenance management.What is the effect of the Good Computerized Maintenance Package?
1- Increase labor utilization by 5 25 %
2- Increase equipment utilization by 5-15%
3- Decrease spare parts inventory by 10-20%
4- Decrease down time cost by 5-15%
CMMS Block diagram:InputsToolOutputs
1- Reference data
2- Equipment list
3- Equipment priority
4- PM information
5- Resource list
6- Working conditions7- CM information8- Cost rates
9- Other data
10-Actual performanceExcel1- Maintenance labor force.2- Average system availability.
3- Annual downtime cost losses.
4- Annual maintenance cost.
5- Annual PM plan.
6- Maintenance resources
7- Monthly PM plans.
8- Maintenance work order9- Other reports
10- Maintenance Control
CMMS main Steps:
5- PM Case StudiesCase (1):
How to construct the coding & criticality systems:
Equipment CodingLocationEquipment TypeEquipment Tag #
123478
Propose a coding system and priority rules for the following equipment:
Plant SystemsLocationEquipment TypeNumber of Machines
Productive systemsMachining shopTurning
Milling
Drilling
Grinding
Press4
2
2
2
1
Foundry shopInduction furnaces
Molding machines2
5
Welding shopArc Welding1
Supportive systemsMaterial handlingFork lift
4
Air roomCompressor2
Water roomPump 50 HP
Pump 100 HP2
2
Power roomDiesel generator2
Equipment Coding Structure:
LocationEquipment TypeEquipment Tag #
123478
LocationEquipment Type
01 Machining shop01 Turning
02 Milling
03 Drilling
04 Grinding
05 Press
02 Foundry shop10 Induction furnaces
11 Molding machines
03 Welding shop20 Arc Welding
04 Material handling30 Fork lift
05 Air room40 Compressor
06 Water room51 Pump 50 HP
52 Pump 100 HP
07 Power room06 Diesel generator
Example: 010202
010202
Machining shopMilling# 2
Example: 065201
065201
Water roomPump 100 HP# 1
EIGHT LEVEL DECOMPOSITION:
LevelCharacterization
0System
1Sub-System
2Major Assembly
3Assembly
4Sub-Assembly
5Component
6Part
7Material
Equipment PriorityFailure effect:
- Effect on HSE
- Effect on Production
- Effect on Cost
Failure Probability:
- Failure Frequency
Example:
Factors%Levels
1- Production 30V- Very Important
I- Important
N- Normal
2- HSE30V- Very Important
I- Important
N- Normal
3- Stand by15WO- Without
WS- With Standby
4- Value5H- High Value
M- Medium
L- Low
Priority LevelDescription
AGroup A: Equipment with 100% duty factor, whose failure involves production losses and potential safety hazards.
BGroup B: Equipment with a ratio duty factor, i.e., having some standby, whose failure involves production losses and potential safety hazards.
CGroup C: Equipment with standby, whose failure involves either production losses or potential safety hazards.
DGroup D: Equipment with standby, whose failure involves neither production losses nor safety hazards.
Equipment PrioritiesLocationEquipment TypePriority Level
Machining shopTurning
Milling
Drilling
Grinding
PressB
B
B
B
D
Foundry shopInduction furnaces
Molding machinesA
B
Welding shopArc WeldingA
Material handlingFork liftC
Air roomCompressorC
Water roomPump 50 HP
Pump 100 HPC
C
Power roomDiesel generatorA
Case (2):
How to select the best maintenance policy?
Number of Engine 2000
Capital maintenance policy for engine is as follows:
Four Policies:
Replacement after first failure (after 36 month)
Repair (010) after first failure & Replacement after second failure (after 30 month)
Repair (020) after second failure & Replacement after third failure (after 24 month)
Repair (030) after third failure & Replacement after fourth failure (after 15 month)Cost rate:
Replacement $ 10,000& Repair $ 3,500
Required:
Select the best maintenance policy
Estimate the annual budget for the best policy
Target maintenance plan
Case (3):The yearly maintenance information for ten gas generators (GG) in a site are as follows:
1- Working conditions for each GG:
Average working hours 7000 hour/year2- PM Levels for each GG:
Check oil level every 150 R.H. (about 2 liter) Change oil every 750 R.H. (about 20 liter) Change oil filter every 1500 R.H.3- CM for each GG:
1. Average oil quantity is 100 liter/year/G.G.
4- Cost rates:
2. Oil cost 5 $/liter
3. Filter cost 50 $/unitRequired:
1. Annual materials (oil and filters) requirements Planning.
2. Annual materials cost
3. Annual PM plans
4. Materials profile (histogram)
5. Maintenance work order for each PM level Case (4):The yearly PM programs information for six similar gas turbines in a power station are as follows:
1- PM information:
Maintenance levels per gas turbine
PM TypeFrequencyDurationNo. of
WorkersSpare parts Cost
$1000
Y Level 1Yearly15 days2010
S Level 26 Monthly10 days208
3M Level 33 Monthly5 days155
M Level 4Monthly2 days102
2- Working conditions:
Gas turbine operating conditions: 24 hour/day
Workers operating conditions: 300 day/year & 8 hour/day
3- CM information:
Average effort of CM = 380 man-day per gas turbine
Average annual spare parts CM = $ 12000 per gas turbine
Average CM downtime = 15 days/year per gas turbine
Average downtime cost rate = $ 1000 per day
4- Cost rates:
Average labor cost rate = $ 10 per man-day
Overhead cost = 25 % direct cost (spare parts & labor)
Required:
1) The size of maintenance labor force.2) Average system availability.
3) Annual downtime cost losses.
4) Annual maintenance cost.
5) Annual PM plan.
6) Maintenance resource profiles.
7) Monthly PM plans.
8) Maintenance work orderThe size of maintenance labor forcePM TypeAnnual
FrequencyDuration
(day)No. of WorkerMan-day
per PM type
Y11520300 * 1= 300
S11020200 * 1 = 200
3M251575 * 2 = 150
M821020 * 8 = 160
Annual PM man-day per gas turbine810
Total PM annual man-day Required810 * 6 = 4860
The size of PM labor force = 4860/300 =16.2 = 17 workers
The size of CM labor force = 380 * 6 / 300 = 8 workers
Total labor force = 17 + 8 = 25 workers
Crew check is ok (25 more than 20).
The average down time per year
PM TypeAnnual
FrequencyDuration
(day)PM Downtime
(day)
Y11515 * 1= 15
S11010 * 1 = 10
3M255 * 2 = 10
M822 * 8 = 16
PM downtime per gas turbine51
Average down time = 51 + 15 = 66 day/year per gas turbine
Annual downtime cost losses = 66 * 6 * 1000 = $ 396000
Average equipment availability =
Active operating time / Total time
= (364 66) / 364 = 82 %
System Reliability:
Series or chain structure: Rs = R1 * R2 * R3 * etc.
Parallel structure: Rs = 1 (1-R1)* (1-R2)* (1-R3) * .etc.
System time availability =
Parallel structure: As = 1 (1-A1)**6
= 1 (1-0.82)**6
= 1 (0.18)**6 = 99%
Annual maintenance costPM TypeAnnual
FrequencyCost
$1000Spare parts PM Cost $1000
Y11010 * 1= 10
S188 * 1 = 8
3M255 * 2 = 10
M822 * 8 = 16
Annual spare parts PM per gas turbine =44
Total annual spare parts PM cost =44 * 6 = 264
The average annual spare parts CM cost =
$ 12000 * 6 = $ 72,000
Annual spare parts maintenance cost =
264000 + 72000 = $ 336,000
Annual labor cost =
25 workers * 300 day/year * $ 10 per man-day= $ 75,000
Annual direct maintenance cost = $ 336000 + $ 75000
= $ 411000
Overhead cost = 25 % direct cost
Annual maintenance cost = $ 411000 * 1.25 = $ 513750
Annual maintenance cost = $ 513750
Annual downtime cost losses = $ 396000
Basic Annual PM Plan
Eq.
codeMonth #
123456789101112
G01YMM3MMMSMM3MMM
G02MMYMM3MMMSMM3M
G03M3MMMYMM3MMMSM
G04SMM3MMMYMM3MMM
G05MMSMM3MMMYMM3M
G06M3MMMSMM3MMMYM
Resource analysis:
Man-day580230580230580230580230580230580230
Day/
month242424242424242424242424
Workers241024102410241024102410
SP cost261826182618261826182618
DT331833183318331833183318
Y= 300S= 2003M= 75 M= 20 man-day
Y= 10S= 8
3M= 5 M= 2 $1000
Y= 15S= 103M= 5 M= 2 day
Target Annual PM Plan # 1
Eq.
codeMonth #
123456789101112
G01YMM3MMMSMM3MMM
G02MMYMM3MMMSMM3M
G03M3MMMYMM3MMMSM
G04MSMM3MMMYMM3MM
G053MMMSMM3MMMYMM
G06MM3MMMSMM3MMMY
Resource analysis:
Man-day455355455355455355355455355455355455
Workers191519151915151915191519
SP cost232123212321212321232123
DT282328232823232823282328
Y= 300S= 2003M= 75 M= 20 man-day
Y= 10S= 8
3M= 5 M= 2 $1000
Y= 15S= 103M= 5 M= 2 day
Target Annual PM Plan # 2
Eq.
codeMonth #
123456789101112
G01YMM3MMMMSM3MMM
G02MMYMM3MMMMSM3M
G03M3MMMYMM3MMMMS
G04MSM3MMMYMM3MMM
G05MMMSM3MMMYMM3M
G06M3MMMMSM3MMMYM
Resource analysis:
Man-day400410400410400410400410400410400410
Workers171717171717171717171717
SP cost202420242024202420242024
DT252625262526252625262526
Y= 300S= 2003M= 75 M= 20 man-day
Y= 10S= 8
3M= 5 M= 2 $1000
Y= 15S= 103M= 5 M= 2 day
Monthly Maintenance Plan: Month # 1
DayG01G02G03G04G05G06PM worker
1. Y20
2. Y20
3. Y20
4. Y20
5. Y20
6. Y20
7. Y20
8. Y20
9. Y20
10. Y20
11. Y20
12. Y20
13. Y20
14. Y20
15. Y20
16. SB-
17. M10
18. M10
19. SB-
20. M10
21. M10
22. SB-
23. M10
24. M10
25. SB-
26. M10
27. M10
28. SB-
29. M10
30. M10
31. SB-
Maintenance Work Order
010120
Requester Section:
Power Station PS03 - Gas Turbine G01 - Priority: A
Maintenance type/level: Annual PM1- Check .
2- Clean ..
3- Replace ..
4- Adjust
5- Repair ..
Eng. Attia Gomaa
Planning Section:
Labor: 4 Mech. 2 Helper 5 days
5 Elec. 4 Helper 10 days
Spare parts: 2 valve xx1, 4 air filter yy3, .. etc.
Special tools: xxx, yyyy, etc,
Expected down time (from 01/01 to 15/01/2004)
Cost estimation ($ 10,000)
Safety instructions:
- Check Eng. Aly Ahmed
Craft Feedback:
1- Check .
2- Clean ..
3- Replace ..
4- Adjust
5- Repair ..
Labor: 3 Mech. 2 Helper 5 days
6 Elec. 3 Helper 11 days
1 Vib. 1 Helper 2 days
Spare parts: 2 valve xx1, 4 air filter yy3, .. etc.
Special tools: Vibrometer, etc,
Down time (01/01 to 17/01/2004) Actual Cost ($ 12,000)
Eng. Omer AlyCoding:
Case (5):
The yearly maintenance information for three generators in a site are as follows:
1- Working conditions:
Two gas generators (GG01 and GG02), one operating and the other standby
Diesel generator for emergency
Site operating hours 24/day * 365 day
2- PM Levels (Catalog information):
Check oil level every 150 R.H. (about 2 liter)
Change oil every 750 R.H. (about 20 liter)
Change oil filter every 1500 R.H.
Check cooling level every 150 R.H.
Clean/ drain cooling system every 1500 R.H.
Check and clean batteries every 1500 R.H.
Lubricate bearing every 750 R.H. (about 1 liter)
Change bearing every 3000 R.H.
Replace thermostat every 3000 R.H.
3- CM for each GG:
Average oil quantity is 100 liter/year/G.G.
4- Cost rates:
Oil cost 3 $/liter
Filter cost 10 $/unit
Bearing oil cost 5 $/liter
Bearing cost 30 $/each
Thermostat cost 30 $/each
Required:
1. Maintenance work order for each PM level 2. Annual materials requirements Planning & materials cost
3. Annual PM plans
4. Cost & materials profiles (histogram)
Case (6):Maintenance spare parts cost ($):Year
1999Year
2000Year
2001Year
2002Exp.
2003Forecasting limits
2003
1450130012001000??
X12345
Y1450130012001000?
XY1450260036004000
n = 4
Sum X = 10
Sum X2 = 30
Sum Y = 4950
Sum XY = 11650
Sum Y = n . a + b Sum X , Sum XY = a Sum X + b Sum X2
4950 = 4 a + 10 b
11650 = 10 a + 30 b
14850 = 12 a + 30 b
a = 1600
b = - 145
Y = 1600 145 X
Y5 = 1600 145 (5) = 875
X12345
A1450130012001000-
F1445131011651020875
(A-F)5103520
(A-F)2251001225400
MSE = 1750 / (4 -1) = 583
S = 24
Z = 2
CLs = 0 Z S = 0 48Case (7):
Uncertain spare parts cost
Spare parts cost
$ 100,000Probability
%
920
1050
1120
127
133
Estimate the spare parts budget based on the following:
1- Average method ($ 1,100,000)
2- Probability method ($ 1,023,000)
3- PERT method ($ 1,033,000)
Solution
Maintenance shutdown Planning
Using CPM
Case (8): The monthly PM programs information for a machining shop are as follows:
Machine CodeT01D01M01T02M02
Machine DescriptionTurningDrillingMillingTurningMilling
Predecessors---T01M01
Duration (day)856820
Worker/day58755
Spare Parts cost $ 1000543612
Maximum worker limit is 12 worker/dayRequired:
1. Monthly maintenance plan.
2. Calculate the monthly spare parts cost.
3. Construct the monthly spare parts cost profile.
Case (9): Monthly Maintenance Plan for Wire Production Line
Project Name : MMPW
Project start: 1 Jan. 2004Planning unit : Day
6 DAYS /WEEK
1- Activity List
ActivityIDDuration (day)PredecessorsRelations
(SS, FS, FF, and SF)
1PreparationPRP2--
2Mech. maintenance # 01MM17PRP-
3Elec. maintenance # 01EM19MM1SS 3
4Mech. maintenance # 02MM26PRP-
5Elec. maintenance # 02EM28MM2SS 2
6Mech. maintenance # 03MM35PRP-
7Elec. maintenance # 03EM37MM3SS 2
8SetupSTP1EM1
EM2
EM3-
2- Resource List
Resource
CodeResources descriptionUnitLimits/dayPrice
LE/unit
Norm.Max.
L01Mechanical workermd3640
L02Electrical worker md4860
SPSSpare parts & suppliescost--1000
3- Resource Allocation
ActivityIDResource
L01/
dayL02/
daySPS
(Total)
1PreparationPRP211
2Mech. maintenance # 01MM14-3
3Elec. maintenance # 01EM1-54
4Mech. maintenance # 02MM23-2
5Elec. maintenance # 02EM2-43
6Mech. maintenance # 03MM32-2
7Elec. maintenance # 03EM3-33
8SetupSTP2
21
4- Base Calendar (Working periods)
SaturdaySundayMondayTuesdayWednesdayThursdayFriday
XXXXXX
1/01/04
Holidays:20 to 21 Jan. 2004
Required:
1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?
3. Construct the corresponding smoothed worker loading?4. Construct the corresponding worker leveling?
5. Construct the target action plan?.
6. Construct the cost profile & S-curve?7. Construct the target master plan?Case (10): Annual Maintenance Plan for AUC-IT Labs.
Project Name : AMIT
Project start: 1 Jan. 2004Planning unit : Day
6 DAYS /WEEK
1- Activity List
ActivityIDDuration (day)PredecessorsRelations
1PreparationPRP1--
2Server maintenanceSRM3PRP-
3Hardware maintenance Lab #01HM14SRM-
4Software maintenance Lab #01SM15HM1
5Hardware maintenance Lab #02HM23SRM-
6Software maintenance Lab #02SM24HM2SS 1
7Hardware maintenance Lab #03HM33SRM-
8Software maintenance Lab #03SM34HM3SS 1
9SetupSTP1SM1
SM2
SM3-
2- Resource List
ResourceCodeResources descriptionUnitLimits/dayPrice
LE/unit
Norm.Max.
L01Hardware Engineermd36120
L02Software Engineermd48100
SPSSpare parts & suppliescost--1000
3- Resource Allocation
ActivityIDResource
L01/
dayL02/
daySPS
(Total)
1PreparationPRP211
2Server maintenanceSRM111
3Hardware maintenance Lab #01HM14-2
4Software maintenance Lab #01SM1-53
5Hardware maintenance Lab #02HM23-1
6Software maintenance Lab #02SM2-42
7Hardware maintenance Lab #03HM32-1
8Software maintenance Lab #03SM3-32
9SetupSTP2
21
4- Base Calendar (Working periods)
SaturdaySundayMondayTuesdayWednesdayThursdayFriday
XXXXXX 1/01/04
Holidays:20 to 21 Jan. 2004
Required:
1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?
3. Construct the corresponding smoothed worker loading?4. Construct the corresponding worker leveling?
5. Construct the target action plan?.
6. Construct the cost profile & S-curve?7. Construct the target master plan?Materials Requirements Planning (MRP) for Maintenance
Case (11): A monthly maintenance plan for 50 similar equipment to replace the gear box for these equipment. The gear box structure is shown below.
ComponentABCDEFG
Lead time (week)1211232
On-Hand101520101050
Required:
1. Time-phased for the gear box structure
2. Gross requirements plan for 50 gear box
3. Net material requirements plan for 50 gear box.Case (12): The monthly plan and the actual maintenance spare parts in ABC Company are as follows:
Spare part #Plan (Jan. 2001)Actual (Jan. 2001)
Planned quantity
(unit)Standard cost (L.E./unit)Actual quantity
(unit)Actual cost
(L.E./unit)
A11401000401100
A12301200201200
A1350900401000
A142085010800
A152095020900
Based on these data, determine the different performance indicators.
Total Maintenance Control
Case (13):Monthly production information on Foundry Shop FS510 was as follows:
ItemJan.
2004Feb.
2004
Working days3128
Standard production rate (ton/hr)88
Average daily time (hr/day)2424
Average down time (hr/day)64
Average standby (hr/day)33
Average target quantity (ton/day)120136
Average actual quantity (ton/day)80105
Average sound quantity (ton/day)7098
Average defect quantity (ton/day)107
Average energy consumption
(1000 kwh/day)4967
Material cost (1000 L.E/day)100130
Based on these data, determine the different PE indicators for the productive system.
Basic data
ItemJan 04Feb 04Feb. / Jan.
Production rate (ton/hr)88100 %
Total time (hr/day)2424100 %
Average down time (hr/day)6467 %
Average available time (hr)1820111 %
Average standby (hr/day)33100 %
Average used time (hr/day)1517113 %
Average target quantity (ton/day)120136113 %
Average actual quantity (ton/day)80105125 %
Average sound quantity (ton/day)7098129 %
Average defect quantity (ton/day)10 (14%)7
(7%)64 %
Energy productivity (kwh/ton)70068498 %
Material productivity (1000 L.E/ton)1429132692 %
Performance Evaluation
IndicatorJanuary
2004February
2004Feb. / Jan.
Availability
18/24= 75 %20/24= 83 %111 %
Performance
efficiency80/120= 67 %105/136= 77 %115 %
Quality rate
70/80= 88 %98/105= 93 %106 %
Utilization ratio
15/18= 83 %17/20= 85 %102 %
Uptime (hr/day)
70/8= 8.7598/8= 12.25140 %
Uptime ratio
8.75/15= 49%12.25/17=72 %147 %
OEE
44 %60 %136 %
TEEP
37 %51 %138 %
NEE
29 %52 %179 %
Energy productivity (kwh/ton)70068498 %
Material productivity (1000 L.E/ton)1429132692 %
Case (14):
The six-monthly maintenance costs ($1000) for a productive system are as follows:
Target Costs:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost:
Spar parts
Labor100
50100
50100
50100
50100
50100
50100
50
CM Cost:
Spar parts
Labor200
150200
150200
150200
150200
150200
150200
150
DT Cost300300300300300300300
Actual Costs:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost:
Spar parts
Labor23
3238
6549
9656
9468
9465
9054
72
CM Cost:
Spar parts
Labor231
503213
370181
293185
164199
201196
193157
142
DT Cost407397320290330320362
Based on these data, determine the different performance evaluation indicators for the maintenance system.
Target:
Cost itemMonth #
JanFebMarAprMayJunJlyTotal
PM Cost1501501501501501501501050
CM Cost3503503503503503503502450
TM Cost8008008008008008008005600
DT Cost3003003003003003003002100
TM+DT11001100110011001100110011007700
PM/TM0.140.140.140.140.140.140.140.955
CM/PM2.332.332.332.332.332.332.3316.33
Actual:
Cost itemMonth #
JanFebMarAprMayJunJlyTotal
PM Cost55103145150162155126896
CM Cost7345834743494003692993208
TM Cost119610839397898928647876550
DT Cost4073973202903303203622426
TM+DT16031480125910791222118411498976
PM/TM0.050.100.150.190.180.180.161.007
CM/PM13.355.663.272.332.472.382.3731.82
Change %:
Cost itemMonth #
JanFebMarAprMayJunJlyTotal
PM Cost
CM Cost
TM Cost
DT Cost
TM+DT
PM/TM
CM/PM
Case (15):The yearly PM programs information for six similar gas turbines in a power station are as follows:
Target work performed:
ItemPMCMTotal
Total labor force (worker)18725
Annual spare parts cost ($1000)26472336
Annual labor cost ($1000)----75
Overhead cost ($1000)----514
Average down time
(day/year per gas turbine)511566
Average downtime cost rate = $ 1000 per day
Actual work performed:
ItemPMCMTotal
Total labor force (worker)201030
Annual spare parts cost ($1000)300100400
Annual labor cost ($1000)----80
Overhead cost ($1000)----520
Average down time
(day/year per gas turbine)45550
Based on these data, determine the different performance evaluation indicators for the maintenance system.
Performance Evaluation Sheet:
ItemTargetActualChange %
Total labor force (worker)2530+ 20
Annual s. parts cost ($1000)336400+ 19
Annual labor cost ($1000)7580+ 6.6
Overhead cost ($1000)514520+ 1.2
Total m. cost ($1000)9251000+ 8.1
Average down time6650- 24.3
Down time cost ($1000)6650- 24.3
TMC + DTC9911050+ 6.0
Availability %81.986.3+ 5.3
CM/PM % (labor force)7/18 =
38.910/20 =
50+ 28.5
CM/PM % (Spare parts)72/264 =
27.3100/300 =
33.3+ 22.0
Overhead %514/411=
1.25520/480=
1.08- 13.6
Labor productivity %
(worker/gas turbine)25/6=
4.1730/6=
5.00- 16.6
Case (16):The six-monthly maintenance costs ($1000) for a productive system are as follows:
Target Costs:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost:
Spar parts
Labor100
50100
50100
50100
50100
50100
50100
50
CM Cost:
Spar parts
Labor200
150200
150200
150200
150200
150200
150200
150
DT Cost300300300300300300300
Actual Costs:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost:
Spar parts
Labor23
3238
6549
9656
9468
9465
9054
72
CM Cost:
Spar parts
Labor231
503213
370181
293185
164199
201196
193157
142
DT Cost407397320290330320362
Based on these data, determine the different performance evaluation indicators for the maintenance system.
Target:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost150150150150150150150
CM Cost350350350350350350350
DT Cost300300300300300300300
TM Cost800800800800800800800
Actual:
Cost itemMonth #
JanFebMarAprMayJunJly
PM Cost 55 103145150162155126
CM Cost 734 583474349400369299
DT Cost 407 397320290330320362
TM Cost11961083939789892864787
7- Machine Failure Analysis
Parameters used for detection of machine faults
Type of faultParameters
VibrationTemp.Oil
Out of balancexxx--
Misalignment / bent shaftxxxx-
Damage of rolling bearingxxxxxx
Damage of journal bearingxxxxxx
Damage of gear boxxxxxxx
Belt problemsxx--
Motor problemsxxx-
Mechanical loosenessxxxxx
Resonancexxx--
xxx
High, easy and soft to measure.
xx
Medium to measure.
x
Low to measure.
-
Non.
Parameters used for detection of pump faults
Type of faultParameters
VibrationTemp.Oil
Out of balancexxx--
Misalignment / bent shaftxxxx-
Damage of rolling bearingxxxxxx
Damage of journal bearingxxxxxx
Damage of gear boxxxxxxx
Belt problemsxx--
Motor problemsxxx-
Mechanical loosenessxxxxx
Resonancexxx--
Minimum flow / Cavitationsxxxxx-
xxx
High, easy and soft to measure.
xx
Medium to measure.
x
Low to measure.
-
Non.
Bearing Failure Analysis
Bearing Failure: Causes and Cures
Excessive Loads:
Excessive loads usually cause premature fatigue. Tight fits, brinelling and improper preloading can also bring about early fatigue failure.
The solution is to reduce the load or redesign using a bearing with greater capacity.
Overheating: Symptoms are discoloration of the rings, balls, and cages from gold to blue.
Temperature in excess of 400F can anneal the ring and ball materials.
The resulting loss in hardness reduces the bearing capacity causing early failure.
In extreme cases, balls and rings will deform. The temperature rise can also degrade or destroy lubricant.
True Brinelling:
Brinelling occurs when loads exceed the elastic limit of the ring material.
Brinell marks show as indentations in the raceways which increase bearing vibration (noise).
Any static overload or severe impact can cause brinelling.
False Brinelling:
False brinelling - elliptical wear marks in an axial direction at each ball position with a bright finish and sharp demarcation, often surrounded by a ring of brown debris indicates excessive external vibration.
Correct by isolating bearings from external vibration, and using greases containing antiwear additives.
Normal Fatigue Failure:
Fatigue failure - usually referred to as spalling - is a fracture of the running surfaces and subsequent removal of small discrete particles of material.
Spalling can occur on the inner ring, outer ring, or balls.
This type of failure is progressive and once initiated will spread as a result of further operation. It will always be accompanied by a marked increase in vibration.
The remedy is to replace the bearing or consider redesigning to use a bearing having a greater calculated fatigue life.
Reverse Loading:
Angular contact bearings are designed to accept an axial load in one direction only.
When loaded in the opposite direction, the elliptical contact area on the outer ring is truncated by the low shoulder on that side of the outer ring.
The result is excessive stress and an increase in temperature, followed by increased vibration and early failure.
Corrective action is to simply install the bearing correctly.
Contamination:
Contamination is one of the leading causes of bearing failure.
Contamination symptoms are denting of the bearing raceways and balls resulting in high vibration and wear.
Clean work areas, tools, fixtures, and hands help reduce contamination failures.
Keep grinding operations away from bearing assembly areas and keep bearings in their original packaging until you are ready to install them.
Lubricant Failure:
Discolored (blue/brown) ball tracks and balls are symptoms of lubricant failure. Excessive wear of balls, ring, and cages will follow, resulting in overheating and subsequent catastrophic failure.
Ball bearings depend on the continuous presence of a very thin -millionths of an inch - film of lubricant between balls and races, and between the cage, bearing rings, and balls.
Failures are typically caused by restricted lubricant flow or excessive temperatures that degrade the lubricants properties.
Corrosion:
Red/brown areas on balls, race-way, cages, or bands of ball bearings are symptoms of corrosion.
This condition results from exposing bearings to corrosive fluids or a corrosive atmosphere.
In extreme cases, corrosion can initiate early fatigue failures.
Correct by diverting corrosive fluids away from bearing areas and use integrally sealed bearings whenever possible.
Misalignment:
Misalignment can be detected on the raceway of the nonrotating ring by a ball wear path that is not parallel to the raceways edges.
If misalignment exceeds 0.001 in./in you can expect an abnormal temperature rise in the bearing and/or housing and heavy wear in the cage ball-pockets.
Appropriate corrective action includes: inspecting shafts and housings for runout of shoulders and bearing seats; use of single point-turned or ground threads on non hardened shafts and ground threads only on hardened shafts; and using precision grade locknuts.
Loose Fits:
Loose fits can cause relative motion between mating parts. If the relative motion between mating parts is slight but continuous, fretting occurs.
Fretting is the generation of fine metal particles which oxidize, leaving a distinctive brown color. This material is abrasive and will aggravate the looseness. If the looseness is enough to allow considerable movement of the inner or outer ring, the mounting surfaces (bore, outer diameters, faces) will wear and heat, causing noise and runout problems.
Tight Fits:
A heavy ball wear path in the bottom of the raceway around the entire circumference of the inner ring and outer ring indicates a tight fit.
Where interference fits exceed the radial clearance at operating temperature, the balls will become excessively loaded. This will result in a rapid temperature rise accompanied by high torque.
Continued operation can lead to rapid wear and fatigue.
Corrective action includes a decrease in total interference.
Case (17): Pump Failure Analysis
Pump Station: PS01
8 Centrifugal pump
Failure Type: Bearing failure
Part code: xxxxx
PM every 1600 R.H. (change oil , filter and bearing)
Bearing failures for centrifugal pumps
(Year 2004)
# of failureEquipment codeRun time
(hr)Repair time (hr)Failure
Mechanism
1100712508Corrosion
2100814506Corrosion
31001100010Temperature
4100415007Corrosion
5100610004Oil
6100212507Corrosion
710037009Oil
810076008Temperature
910085008Temperature
10100612509Corrosion
111001100010Oil
12100214508Corrosion
1310057008Temperature
141004125011Corrosion
15100510009Corrosion
1610037006Oil
1710086009Temperature
18100110008Oil
Based on these data, Determine the different PE indicators for this system. Construct how to analyze and eliminate the bearing failure.Failure Analysis:
Pump Station: 8 Centrifugat pumpCode: 1000
Failure Type: Bearing failure
Part code: xxxxx
(Year 2004)
# of failureEquipment codeRun time
(hr)Repair time (hr)Failure
Mechanism
1100712508Corrosion
2100814506Corrosion
31001100010Temperature
4100415007Corrosion
5100610004Oil
6100212507Corrosion
710037009Oil
810076008Temperature
910085008Temperature
10100612509Corrosion
111001100010Oil
12100214508Corrosion
1310057008Temperature
141004125011Corrosion
15100510009Corrosion
1610037006Oil
1710086009Temperature
18100110008Oil
Total18200145
MTBF = 18200/18 = 1011 hr ( = 0.989 * 10-5 failure/hr
MTTR =145 /18 = 8 hr
A =1011/(1011+8) =99.21%
MTBF at which less than 20 % of the pumps are assumed to fail
Run time
hrFrequencyCumulative FrequencyC.F.
%
1500115.56
14502316.67
12504738.89
100051266.67
70031583.33
60021794.44
500118100
100066.67
?
80.00
MTBF = 760 hr
70083.33
Max. running time =1650 hr. Min. running time= 300 hr
Run time
hrFrequencyMid pointC.F.C.F.
%
1650-1400 31525316.67
1400-115041275738.89
1150-900510251266.67
900-65037751583.33
650-300347518100
Freq
6
5
4
3
2
1
300
650650
900900
11501150
14001400
1650
MTBF
Equipment Level:
Equipment codeMTBF
(hr)MTTR (hr)A
%Failure
Mechanism
10011000
1000
1000
100010
10
8
9.3399.00Temperature
Oil
Oil
10021250
1450
13507
8
7.599.44Corrosion
Corrosion
1003700
700
7009
6
7.598.94Oil
Oil
10041500
1250
13257
11
999.33Corrosion
Corrosion
1005700
1000
8508
9
8.599.01Temperature
Corrosion
10061000
1250
11254
9
6.599.43Oil
Corrosion
10071250
600
9258
8
899.14Corrosion
Temperature
10081450
500
600
8506
8
9
7.6699.10Corrosion
Temperature
Temperature
Average1011899.21Corrosion 8
Oil 5
Temperature 5
Failure Mechanism Level:
Failure
MechanismMTBF
(hr)MTTR (hr)RangesEquipment code
Corrosion1500
1450
1450
1250
1250
1250
1250
10007
8
6
8
7
9
11
9MTBF
1000 -
1500
MTTR
6 111004
1002
1008
1007
1002
1006
1004
1005
Oil1000
1000
1000
700
7004
10
8
9
6MTBF
700 -
1000
MTTR
4-101006
1001
1001
1003
1003
Temperature1000
700
600
600
50010
8
8
9
8MTBF
500 -
1000
MTTR
8-101001
1005
1007
1008
1008
Average1011899.21Corrosion 8
Oil 5
Temperature 5
Remedy:
Maintenance Policy
Condition BasedTime Based
Every 300 hours
Oil analysis
Temperature analysis
Vibration analysis(1) Change oil every 600 hour
(2) Change bearing & oil every
1200 hour
Down time: (1) 1 hr & (2) 8 hr
Cost Analysis:
Cost elements:
Spare parts cost = 1000 L.E./failure
PM impact = 2000 L.E./failure CM impact = 4000 L.E./failure
ParameterCurrentProposed
PM frequency (failure/year)-18
CM frequency (failure/year)181
Spare parts cost (1000 L.E. / year)1819
PM impact (1000 L.E. / year)-36
CM impact (1000 L.E. / year)724
PM & CM impact (1000 L.E. / year)7240
Total cost (1000 L.E. / year)9059
Cost ratio %10065.5
Cost saving %-34.5
Maintenance Policy:
I- Vibration analysis:
1- Frequency: Every 300 Running Hours
2- Tool:
Vibration Equipment: accelerometers, charge amplifier and analyser.
Computer program for trend analysis and prediction.3- International Standard: CDA/MS/NVSH107
4- Method:
1. Record the vibration spectrum, specify the peaks corresponds to the bearing components
2. Record each component peak and frequency.
3. By using the soft ware and the standard limits, determine the trend of each peak.
4. Determine the bearing state(good need service need change)
5- Limits: According to CDA/MS/NVSH107
1. Pre-failure: vibration level5.6 m/s
2. Failure: vibration level 5.610 m/s
3. Near catastrophic failure: vibration level >10 m/s
6- Actions:
1. Bearing is Good
2. Call for bearing change
3. Bearing must be changed immediately
II- Temperature analysis:
1- Frequency: Every 300 Running Hours
2- Tool:
Temperature measuring equipments as thermocouple or infrared camera.
Computer program for trend analysis and prediction.3- International Standard: SKF
4- Method:
Measure the temperature of the bearing on line and take the average value every day.
By using the software analyze the data, determine the max. & average temperature values.
According to the allowable range specified in SKF standard, determine the bearing state.
5- Limits: According to CDA/MS/NVSH10
1. Pre-failure: ( 100 C)
2. Faiulre: (100 125 C)3. Near catastrophic failure: (>125 C)
6- Actions:
1. Bearing is good
2. Call for bearing change
3. Bearing must be changed immediately
III- Oil analysis:
1- Frequency: Every 300 Running Hours
Viscosity change
Acidic content
Wear rate
2- Tool:
Viscometer, PH meter, and particle counter
Computer program for trend analysis and prediction.3- International Standard: ASTM-445 & 664 & 3984- Method:
Take a suitable oil sample volume, to be used in analyses After each 300 hours.
Put it in a closed container, isolated from air, heat and contamination exposes.
Measure the previous mentioned properties then enter the obtained data to the software to be trended.
5- Limits:Viscosity: according to ASTMD-445:
1. Pre-failure: Viscosity change
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