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POLAR 1998.
The MAINTENANCE(CONTENTS) of WORK:
1. Introduction. Page 3
2. The task for course work of page 4
3. Calculation of parameters of distribution of resources of a detail automobile
Engines three methods of page 5
4. Definition of confidential borders of measurement of structural parameter
And operating time to the first resource diagnosing page 15
5. Forecasting a residual resource of a detail
Groups of the automobile engine on the basis of results of diagnosing of page 17
6. Conclusions. Page 21
7. The literature. Page 22
1. Introduction. By results of numerous researches annual production rate of automobiles by the end of term their service is reduced in 1,5 - 2 times in comparison with initial, safety of a design of automobiles is reduced. For service life of the automobile charges on his(its) maintenance service and repair surpass initial cost in 5 - 7 times. Therefore the important direction both at designing, and at operation of automobiles is exact and authentic the estimation of the basic parameters of reliability of their details. In course work questions on forecasting parameters of average and residual resources of details of automobile engines are considered(examined). To the details limiting reliability of engines, details groups and - the mechanism which refusals, basically, are connected to deterioration first of all concern. Deterioration of details of the engine is influenced with set of factors, mainest of which are properties of rubed materials (physicomechanical, chemical), modes of operation (high-speed, loading, thermal), geometrical parameters (the form, the sizes, a roughness of a surface), greasing (quantity(amount), clearing, a supply). Definition of parameters of durability can be carried out on the basis of the data processing, received by results of natural supervision of group of automobiles which are maintained in the certain conditions. For the same purposes experimental materials by kinds of deterioration and characteristics of wear process of existing designs of engines can be used. In result for forecasting parameters of durability the correlation equations of durability of details of the automobile can be used. However both in the first and in the second cases it is impossible to avoid the mistakes caused by necessity of the account of all variety of factors, influencing on process of wear process of details of the automobile. The combined forecast therefore can be made, allowing to take into account advantages of the first and second variants of forecasting. At use of the diagnostic information while in service automobiles the most simple way of forecasting of a residual resource of details of the engine is analytical forecasting on sedate model.
2. The task for course work. While in service automobile engines details (rings, sleeves of cylinders, pistons) were replaced at excess of allowable deterioration of working surfaces. During supervision it was fixed N = 66 first replacements of details at the operating time given in table 2. We shall assume, that distribution of a resource of details before the first replacement submits to the normal law. It is required to find parameters of distribution (a population mean and an average square-law deviation(rejection)), to check up hypothesis about kind of the law of distribution, to calculate density of distribution, probability of non-failure operation and an average resource of a detail. By results of calculations to construct the histogram and curves of empirical and theoretical density of distribution of probabilities, and probabilities of non-failure operation of a detail. The initial data are placed in table 1.
TABLE 1. The initial data on the course project. The name 123 The automobile - 5410-61212 frequency of rotation cranked -12600 volume of the cylinder 9,0 twisting moment, *700 the piston, D1,20 the piston, S1,20 elasticity, 1051,0 the lock of a ring in a free condition, 0,188 thickness of a ring, t0,050 rings, b0,030* Hardness on : a ring, Sleeve, Piston NVk 700 / 100 90 microcutting 1,77 number of a box of transfers at dispersal For empty i?? i?? 3,1 2,4123, taking into account interest of movement as roads: in city In suburb Access roads 0,04 Uses of run? 0,68 resistance to movement: - City and suburban roads - Access roads 0,04 * * Speed of movement of the automobile, Va In city conditions, Va1 In suburb, Va2 On access roads, Va3 km / 25 ( 30 ) 35 ( 40 ) 5 ( 10 ) of the beginning of release of the engine, -1983 pressure, i 1052,35 pressure, 2 1051,01 the area in the lock of a ring, F2-02 1049,50 a deviation(rejection) of the initial area in the lock of a ring? F2-0 5175 the area of a backlash in the lock of a ring, F2-2 10442,6 degrees? 1,4 a deviation(rejection) of an error of diagnosing?? F2-1 19215 on -coupling 8100 weight 19100 6800 volume number on a lobby 3500 volume number on 3500 speed /80 - 100 number main 7,22 (6,53; 5,94) the Size 260R508 radius conducting 0,488 26,74 *2/40,6 - a ring ** In brackets the data are given for the empty automobile. 3. Calculation of parameters of distribution of resources of a detail of automobile engines. Item 3.1. Calculation of parameters of distribution of resources of a detail of automobile engines by results of their supervision in operation. Item 3.1.1. Parameters of distribution of resources of a detail pay off on the basis of processing the statistical information on the refusals observably(notice) in operation, and are used for development of strategy of maintenance of serviceability, an estimation of durability and non-failure operation of a design and need(requirement) for spare parts. Let's reveal the greatest lmax and the least lmin values of an operating time and we shall define(determine) width of intervals of grouping under the formula: ? l = (lmax - lmin) / 1 + 3,2*lg N, thousand kms, where N - the general(common) number of supervision, N = 66 TABLE 2. Values of resources l (are placed on increase), thousand kms. 66,3 132,5 156,4 164,1 180,3188,4 197,0 211,4 219,6 229,1 241,9 87,7 136,7 156,9 164?????????????????????????????????????????????????????????????????????????????????????????????, 2 (thousand kms) ? l =36,086??? 36?? Thousand kms. Item 3.1.2. We shall count up frequencies of hit of a random variable of a resource l in an interval of grouping. We shall choose initial l and final l values of size which undertake closer to integer lmax and lmin. l = 66; l1 =66 +36 =102; l2 =102 +36 =138; l3 =138 +36 =174; l4 = 174 +36=210; l5 =210 +36 =246; l6 = 246 +36 =282; l7 =282 +36 =318; l = 66 and l7 = l = 318 (thousand kms) l l1 l2 l3 l4 l5 l6 l 66 102 138 174 210 246 282 318 We draw a straight line and we break into intervals equal from 66 up to 318 thousand kms. Item 3.1.3. We shall define(determine) what quantity(amount) of resources gets in intervals and we shall define(determine) middle of these intervals. For convenience of using the data of calculations we shall bring in table 3.
TABLE 3. Definition of frequency of hit of resources in the given intervals. No intervals (thousand kms) Middle of intervals (thousand kms) Frequency of hit in an interval, ni166 - 1028432102 - 13812063138 - 174156154174 - 210192175210 - 246228216246 - 28226437282 - 3183001 Item 3.1.4. Definition of parameters and characteristics of the normal law. The density of probability f looks like (l) normal law: _ ____ _ _ _ f (l) = 1/(?????????????? exp [-(li - a) 2 / 2?? 2], where _ _ a and? - parameters of the normal law of distribution; exp (z) - the form of representation of number in a degree z: exp (z) = ez ) We shall calculate a population mean a under the formula: _ r __ a = 1 / N*?? li * ni, where i=1 r - quantity(amount) of intervals; N - the general(common) number of supervision; li - middle of intervals; ni - frequency of hit in intervals. = 1 / 66* (84*3 + 120*6 + 156*15 + 192*17 + 228*21 + 264*3 + 300*1) = = 1 / 66 *12456 = 188,72727?? 188,73 (thousand kms) ) We shall calculate a root-mean-square deviation(rejection)?? Under the formula: _ ________________________ ?? =?? 1 / (N - 1) *??? li - a) 2 * ni, (thousand kms) _ _____________________ ?? =?? 1 / (66 - 1) *??? li - a) 2 * ni, = 46,2898?? 46,29 (thousand kms) ) We shall calculate values of empirical density of distribution of probabilities f (li) on intervals of an operating time: f (li) = ni / (N *? l), ) We shall calculate normalized and deviations(rejections) of middle of intervals: _ _ _ _ yi = (li-a)/?, ) We shall define(determine) values of theoretical density? Distributions of probabilities f (li) under the formula: _ _ f (li) = (1/????? f (yi), where ___ f (yi) = (1/? 2?) * exp (-yi2 / 2) The received values of calculations in items(points) in, , we shall reduce in table 4. TABLE 4. The table of calculations of empirical and theoretical density of distribution of probabilities and normalized and deviations(rejections) of middle of intervals. n i \ yif (li) f (li) f? (li) n1-2,2620,00130,03330,0007n2-1,4850,002 50,13330,0029n3-0,7070,00630,32780,0071 n40,0710,00720,40,0086n50,8480,00880,28 570,0062n61,6260,00130,10890,0023n72,40 40,00040,02220,0005 ?) By results of calculations we build in figure 1 histogram: an empirical curve, distribution of density of probabilities f? (li), a theoretical curve of distribution f? (li) and a leveling curve. Rice 1. The histogram of middle of intervals, a curve of distribution of density of probabilities f (li), a theoretical curve of distribution f (li) and a leveling (bending around) curve. Item 3.1.5. Check of the consent between the empirical and theoretical (normal) law of distribution by criterion?? : .) We shall define(determine) a measure of a divergence?? Between empirical and theoretical distributions: r ??????? ni - ni `) 2 / ni `, where? i=1 ni and ni ` - conformity of empirical and theoretical frequency of hit of a random variable in i- an interval. For convenience of calculations criterion?? We shall define(determine) under the formula: r _ _ _ ? 2 = N *? l *???? f (li) - f (li)] 2 / f (li), i=1 ? 2 =5,12 .) We shall calculate number of degrees of freedom m (thus intervals in which frequencies ni less 5- we shall unit with the next intervals): m = r1 - k - 1, where r1 - number of intervals received at association; k - quantity(amount) of parameters of the law of distribution. The normal law is two-parametrical and is defined(determined) by a population mean and an average square-law deviation(rejection), i.e. k=2. m = 4-2-1 = 1 .) On values?? And m we shall define(determine) probability of consent P (??) theoretical and empirical measurement P (??) = P (5,12) = 0,0821; (??) > 0,05, empirical distribution means will be coordinated to the normal law of distribution. Item 3.1.6. Definition of estimations of parameters of reliability of a detail: ) We shall calculate value of average resource R at the normal law of distribution which is numerically equal to a population mean and, therefore R = and = 188,73 (thousand kms) ) We shall calculate probability of non-failure operation of a detail on intervals of an operating time under the formula: _ _ r P (li) = (N-?? ni / N), i=1 P (l1) = (66-3/66 = 0,95; . P (l7) = (66-66/66 = 0 ) We shall construct a curve of probability of non-failure operation of detail P (li) depending on its(her) operating time l in figure 2. Rice 2 Diagram P (li) a curve of probability of non-failure operation of a detail depending on an operating time l. Item 3.2. Calculation of parameters of distribution of resources of a detail on the correlation equations of durability. For data gathering on operational reliability of units of the automobile it is required 5-6 years, therefore the estimation of durability of new models of engines is made on the basis of the analogy, the accelerated tests and models. One of directions of forecasting is development the models representing correlation dependence of a line of regress between sizes, describing a level , and a parameter of a resource of a considered(an examined) detail. For details of the engine the given approach is realized as the correlation equations of durability: = + (R - *n)-1, where To criterion ; And, In, With - factors; R - an average resource of a detail; n = -0=1980-1970=10 - the predicted period (-year of the beginning of release of the engine, 0-1970 a reference point of the predicted period). The criterion pays off under the formula: = k*k*S (pR + 0.1D2*pi*b-1*r-1), The average resource pays off the equation: = - 25,2 + 81840 / (R - 2,75n), where k - specific criterion of physicomechanical properties of a ring; k - specific criterion warmly intensity; pR - specific pressure upon a wall of the cylinder from forces of elasticity of ring ;
