What are the two types of preventive maintenance. And so today I found out that… Preventive Maintenance Preventive maintenance is a schedule of planned maintenance actions aimed at the prevention of breakdowns and failures. The primary goal of preventive maintenance is to prevent the failure of equipment before it actually occurs. It is designed to preserve and enhance equipment reliability by replacing worn components before they actually fail. Preventive maintenance activities include equipment checks, partial or complete overhauls at specified periods, oil changes, lubrication and so on. In addition, workers can record equipment deterioration so they know to replace or repair worn parts before they cause system failure. Recent technological advances in tools for inspection and diagnosis have enabled even more accurate and effective equipment maintenance. The ideal preventive maintenance program would prevent all equipment failure before it occurs. Value of Preventive Maintenance There are multiple misconceptions about preventive maintenance. One such misconception is that PM is unduly costly. This logic dictates that it would cost more for regularly scheduled downtime and maintenance than it would normally cost to operate equipment until repair is absolutely necessary. This may be true for some components; however, one should compare not only the costs but the long-term benefits and savings associated with preventive maintenance. Without preventive maintenance, for example, costs for lost production time from unscheduled equipment breakdown will be incurred. Also, preventive maintenance will result in savings due to an increase of effective system service life. Long-term benefits of preventive maintenance include: I'mproved system reliability. Decreased cost of replacement. Decreased system downtime. Better spares inventory management. Long-term effects and cost comparisons usually favor preventive maintenance over performing maintenance actions only when the system fails. When Does Preventive Maintenance Make Sense Preventive maintenance is a logical choice if, and only if, the following two conditions are met: Condition #1: The component in question has an increasing failure rate. In other words, the failure rate of the component increases with time, thus implying wear-out. Preventive maintenance of a component that is assumed to have an exponential distribution (which implies a constant failure rate) does not make sense. Condition #2: The overall cost of the preventive maintenance action must be less than the overall cost of a corrective action. (Note: In the overall cost for a corrective action, one should include ancillary tangible and/or intangible costs, such as downtime costs, loss of production costs, lawsuits over the failure of a safety-critical item, loss of goodwill, etc. ) If both of these conditions are met, then preventive maintenance makes sense. Additionally, based on the costs ratios, an optimum time for such action can be easily computed for a single component. This is detailed in later sections. The Fallacy of "Constant Failure Rate" and "Preventive Replacement" Even though we alluded to the fact in the last section of this on-line reference, Availability, it is important to make it explicitly clear that if a component has a constant failure rate (i. E. Defined by an exponential distribution), then preventive maintenance of the component will have no effect on the component's failure occurrences. To illustrate this, consider a component with an MTTF = 100 hours, or = 0. 01, and with preventive replacement all 50 hours. The reliability vs. Time graph for this case is illustrated in Figure 7.

3. In Figure 7. 3, the component is replaced all 50 hours, thus the component's reliability is reset to one. At first glance, it may seem that the preventive maintenance action is actually maintaining the component at a higher reliability. Figure 7. 3: Reliability vs. Time for a single component with an MTTF = 100 hours, or = 0. 01, and with preventive replacement all 50 hours. However, consider the following cases for a single component: Case 1: The component's reliability from 0 to 60 hours: With preventive maintenance, the component was replaced with a new one at 50 hours so the overall reliability is the reliability based on the reliability of the new component for 10 hours, R(t = 10) = 90. 48%, times the reliability of the previous component, R(t = 50) = 60. 65%. The result is R(t = 60) = 54. 88%. Without preventive maintenance, the reliability would be the reliability of the same component operating to 60 hours, or R(t = 60) = 54. 88%. Case 2: The component's reliability from 50 to 60 hours: With preventive maintenance, the component was replaced at 50 hours so this is solely based on the reliability of the new component, for a mission of 10 hours, or R(t = 10) = 90. 48%. Without preventive maintenance, the reliability would be the conditional reliability of the same component operating to 60 hours, having already survived to 50 hours, or . As it can be seen, both cases, with and without preventive maintenance, yield the same results. Determining Preventive Replacement Time As mentioned earlier, if the component has an increasing failure rate, then a carefully designed preventive maintenance program is beneficial to system availability. Otherwise, the costs of preventive maintenance might actually outweigh the benefits. The objective of a good preventive maintenance program is to either minimize the overall costs (or downtime, etc. ) or meet a reliability objective. In order to achieve this, an appropriate interval (time) for scheduled maintenance must be determined. One way to do that is to use the optimum age replacement model, as presented next. The model adheres to the conditions discussed previously, or: The component is exhibiting behavior associated with a wear-out mode. That is, the failure rate of the component is increasing with time. The cost for planned replacements is significantly less than the cost for unplanned replacements. Figure 7. 4: Cost curve for preventive and corrective replacement. Figure 7. 4 shows the Cost Per Unit Time vs. Time plot. In this figure, it can be seen that the corrective replacement costs increase as the replacement interval increases. In other words, the less often you perform a PM action, the higher your corrective costs will be. Obviously, the longer we let a component operate, its failure rate increases to a point that it is more likely to fail, thus requiring more corrective actions. The opposite is true for the preventive replacement costs. The longer you wait to perform a PM, the less the costs; while if you do PM too often, the higher the costs. If we combine both costs, we can see that there is an optimum point that minimizes the costs. In other words, one must strike a balance between the risk (costs) associated with a failure while maximizing the time between PM actions. Optimum Age Replacement Policy To determine the optimum time for such a preventive maintenance action (replacement), we need to mathematically formulate a model that describes the associated costs and risks. In developing the model, it is assumed that if the unit fails before time t, a corrective action will occur and if it does not fail by time t, a preventive action will occur. In other words, the unit is replaced upon failure or after a time of operation, t, whichever occurs first. Thus, the optimum replacement time can be found by minimizing the cost per unit time, CPUT(t). CPUT(t) is given by: (5) Where: R(t) = reliability at time t. CP = cost of planned replacement. CU = cost of unplanned replacement. The optimum replacement time interval, t, is the time that minimizes CPUT(t). This can be found by solving for t such that: (6) Or by solving for a t that satisfies Eqn. (7): (7) Interested readers can refer to Barlow and Hunter [2] for more details on this model. Introduction to Repairable Systems Example 2 The failure distribution of a component is described by a 2-parameter Weibull distribution, with = 2. 5 and = 1000 hours. The cost for a corrective replacement is $5. The cost for a preventive replacement is $1. Estimate the optimum replacement age in order to minimize these costs. Solution to Introduction to Repairable Systems Example 2 Prior to obtaining an optimum replacement interval for this component, the assumptions of Eqn. (5) must be checked. The component has an increasing failure rate, since it follows a Weibull distribution with greater than one. Note that if = 1, then the component has a constant failure rate and if

Have you ever heard a car engine belt squeal? Gates Automotive explains why belt preventative maintenance is so important. The belt performance is critical f. . .

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