ENGM03 Advanced Maintenance Practices
Added on 2021-05-31
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UNIVERSITY OF SUNDERLAND FACULTY OF Engineering and Advanced Manufacturing ENGM03 ADVANCED MAINTENANCE PRACTICES Individual Assessment 2
INTRODUCTION The report hereinafter details the major aspects of effective equipment maintenance. From the given scenario, it is clear that the company is facing serious maintenance problems courtesy of the weak maintenance strategy. Notably, reactive maintenance, especially for a complex production system, is never a sure means of maintaining the system. It has got a number of challenges such as: longer downtime adding onto cost of production, frequent machine and equipment breakdown prompting the maintenance team to work on the system in a haphazard fashion hence contributing to low quality of workmanship. Therefore, this report aims at highlighting some of the challenges that the existing strategy possesses and consequently, a new maintenance strategy is proposed thereafter. Preventive Maintenance strategy is therefore the better option. Notably, the governing parameter is OEE (overall equipment effectiveness)OVERALL EQUIPMENT EFFECTIVENESSThe OEE is a parameter that is used to estimate the level of effectiveness of a plant. In this case, three factors are used toarrive at the OEE value, namely: performance, quality and availability. Therefore OEE is a function of these factors. Factors required to calculate OEE(i)Performance In this case, speed loss is considered such that it is given by the formula: P = ideal cycle time/(operating time x total pieces ). Performance analysis is essential as it indicates the level of production outcome in terms of number of pieces produced so far. (ii)Quality – This considers the number of parts that have been produced and are non-defective. To arrive at this factor, we will use the number of defective parts. (iii)Availability
This the portion of time in which the equipment was running without any breakdown. It is also referred to as uptime. It normally given by the formula: A= operating time/planned timeTherefore, OEE = availability x performance x qualityTable 1: OEE calculated Day TotalUnitsProduced Discard FaultsOccurring Stoppage(hours) CorrectiveActions PercentdefectivePERFROMANCE QUALITY AVAILABILITY OEEE1900400Shaft alignment 5Realigned. 44.00%0.135637.52.64621200450Rollers chipped (SCRAP) 1Check Oil. 3750.00%0.1062.587.55.2084383180050Damaged by heat Temp, measuring system needed 277.00%0.0697.230.0617330.014185000.00%0.06005185000.00%0.0600
6800150Shaft alignment 3Realigned. 1875.00%0.1481.2562.57.2515637185000.00%0.060081000550Rollers chipped (SCRAP) 1Check Oil. 5500.00%0.114587.54.4966259185000.00%0.0600101000850Shaft alignment 4Realigned. Vibration system required 8500.00%0.1115500.856511185000.00%0.070012185000.00%0.0700131200550Rollers chipped (SCRAP) 1Check Oil. Oil analysesneeded 4583.00%0.1054.1787.54.51236114185000.00%0.0700151000850Shaft alignment 3Realigned. 8500.00%0.111562.51.07062516185000.00%0.060017185000.00%0.060018185000.00%0.060019185000.00%0.060020185000.00%0.0600Total 31100385018 (hrs) 26.11384Table 2: Faults and fault mode tableFAULTSFAULT MODE (CORRECTIVE ACTION)
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