The problem needs to be solved using probalistic design using normal distrubtion table Z A deep groove ball bearing with the following particulars:Bearing number = 6207; d = 35 mm; D = 72; B = 17 mm; Basic dynamic capacity (C) = 20 kN.The bearing runs at speed 2000 rpm. It is required that the bearing should survive for a life period of(10 000 + 500) hours. Estimate reliability of the bearing, assuming that the coefficient of variation ofthe dynamic capacity is; (Vc = 0.27)N.B:P Li sC; Where P is the equivalent bearing load and L is the life in million revolutions

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Answered: A deep groove ball bearing with the…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A ball bearing subjected to a radial load of 4000 N is expected to have a satisfactory life of 12 000 hours at 720 r.p.m. with a reliability of 95%. Calculate the dynamic load carrying capacity of the bearing, so that it can be selected from manufacturer's catalogue based on 90% reliability. If there are four such bearings each with a reliability of 95% in a system, what is the reliability of the complete system?
....The rating life at 90% reliability for a ball bearing is 15 x 106 revolutions. If the bearing is subjected to an equivalent radial load of 2000 N, which of the following is its dynamic load rating? Group of answer choices 4506 N 4932 N 1890 N 1665 N
A 02-series single-row deep-groove ball bearing is to be selected from Table 11–2 for the application conditions specified in the table. Assume Table 11–1 is applicable if needed. The value of radial load Fr = 8 kN, axial load Fa = 2 kN, design life LD = 2.40(108), ring rotation factor V = 1, and reliability RD = 0.98. Assuming that the values taken from Table 11–1 are X = 0.56 and Y = 1.63, determine the equivalent radial load rating for this application. What is the equivalent radial load rating for this application in kN?
Estimate the life in hours of a 02-30 mm angular-contact ball bearing subjected to 20,000 revolutions per cycle. Each cycle consists of a radial load of 15 kN for 6 min, and then a change in load to 25.05 kN for 4 min. Determine the expected life of the bearing, in millions of revolutions, for each load, assuming the bearing has a C10 value of 20.3 kN. The expected life of the bearing for 15 kN load is ___ × 106 revolutions. The expected life of the bearing for 25.05 kN load is ____× 106 revolutions.
A certain application requires a ball bearing with the inner ring rotating, with a design life of 26 kh at a speed of 455 rev/min. The radial load is 2.5 kN and an application factor of 1.2 is appropriate. The reliability goal is 0.92. Find the multiple of rating life required, xD , and the catalog rating C10 with which to enter a bearing table. Choose a 02-series deep-groove ball bearing from Table 11–2 and estimate the reliability in use.
A certain application requires a ball bearing with the inner ring rotating, with a design life of LD = 15 kh at a speed of nD = 480 rev/min. The radial load is Fr = 1.4 kN, the axial load Fa = 0.4 kN, and an application factor of af = 1.2 is appropriate. The reliability goal, R = 0.99. a) Determine the multiple of rating life required, xD, and the catalog rating C10FR, where FR is the calculated bearing load in [kN] with which we enter a bearing table. Specify a 02-series angular contact ball bearing from Table 11–2, b) Estimate the reliability (R) and multiple of rating-life (XD) in use.
i) Explain the static rating of rolling contact bearings. ii) Write any three technical reasons why ball and roller bearings need to be lubricated. b) Design a self-aligning ball bearing to be used in a turbine to carry a combined radial load of 4466 N and thrust load of 1420 N. The basic dynamic load rating of the bearing is 31 kN at 547 rpm. Take k=3 for all types of ball bearings. Take the value of the shock load factor as 1.3. The axial and radial load factors are 1.8 and 1.2 respectively and the rotational factor is 1. Calculate: i) Basic equivalent dynamic load in N ii) Design equivalent dynamic load in N iii) Expected life of bearings in hours
A ball bearing subjected to an equivalent dynamic load of 4,000 N is expected to have a life of 8.3 x 108 revolutions at 99% reliability. Determine the basic dynamic load rating of the bearing based on a reliability of 90% in kN.
b) Design a single row deep groove ball bearing with basic dynamic load rating of 33 KN to be used in a turbine to carry a radial load of 1584 N. The expected life of the bearing is 5591 hours at 529 rpm. Take k=3 for all types of ball bearings. Take the value of the shock load factor as 1.8. The radial and axial load factors are 1.3 and 1.6 respectively and the rotational factor is 1. Calculate: i) Expected life of bearings in millions of revolutions ii) Design equivalent dynamic load in N iii) Basic equivalent dynamic load in N iv) Axial load acting on the bearing in N
Thory of machine: Select one of the following bearing for the load conditions. Radial load, F=3000 N, Axial load, F = 1600 N, RPM = 1000. (1) Deep Groove ball bearing dynamic load rating=47000 N radial bad factor, X=0.56 Axial load factor, Y= 1.83 (i) Cylindrical roller bearing Dynamic load rating 28000 N radial load factor, X=1.0 axial load factor, Y=0.0
It has a full-slide type bearing rotating at a speed of 1975 rpm.A load of 5693 N is applied. SAE 40 oil in bearing at 60C inlet temperatureplanned to be used.Shaft and bushing diameters in bearing with tolerance data 41−0.010 respectively+0.000mmand 41.125−0.000+0.100 mm.Average radial clearance mounting for this bearing with bush length 20.50 mmCalculate the following;a) Bearing characteristic number,b) Minimum lubricant film thickness,c) Eccentricity value,d) coefficient of friction,e) Average oil film temperature andf) Find the Tmax temperature value.
Calculate the basic rating life at 90% reliability for the following cylindrical roller bearing and conditions.Fr 5.05 kNFa 3.12 kNBearing: NU 1010 ECP (on pp 522 in the SKF catalogue)Clearance: Normal Give the answer in millions of revolutions.

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