ADVANCED ENGINEERING DESIGN - Tribology

1 ADVANCED ENGINEERING DESIGND esign for ReliabilityBook: ADVANCED ENGINEERING DesignEdition, ManualLast updated Feb 16, Anton van BeekDelft University of TechnologyMechanical Engineering2 DESIGN for lifetime performance and reliabilityRecent updates:Feb 27 Chapter 4, problem and updated, problem 6 Chapter 3, problem , and newChapter 3, problem 9 Chapter 3, Problem 10 Chapter 3, Problem 12 Chapter 3, Problem solution updatedChapter 3, Problem newChapter 3, Problem newMarch 20 Chapter 6, Problem and 5, Problem 1 Chapter 3, Problem solution updatedChapter 3, problem solution updated DESIGN for lifetime performance and reliability3 Problems Chapter 1 Problem : L10 service life Consider a quantity of 10 components that all fail within a year of service. Calculate the L10 service lifewith 90% reliability and 10% failure probability assuming a normal failure Problem : Tolerance field The diameter of a batch of shafts is normally distributedwith of the shafts within the tolerance field20 Then 95% of the shafts will have a diameterwithin a tolerance field of 20 mm A mm.

2 A) What is A? b) What is the coefficient of variation CV ? CV is defined as the maximum deviation of the meandivided by the : Driving torque interference fit An interference fit is realized with 20 H7/r6 hole/shaft tolerances. Thedimensions of the components are assumed to be normally standard deviation is calculated from the assumption that thetolerance interval is a 3 interval. Linear elastic deformation is to beconsidered which implies the torque that can be transmitted isproportional to the diametrical interference . The torque that can be transmitted, based on the mean value of the diametrical interference, is T50 [Nm].It is the torque with 50% failure probability. The torque that can be transmitted with 1% failureprobability is denoted as T1. The variation of performance, relative to the mean, is a measure of reliability. The coefficient of variationis defined as CV = deviation/mean. Calculate CV = (T1 T50)/T50. Problem : Driving torque tapered shaft hubs The torque T that can be transmitted by a tapered shaft hub connectionis proportional to the clamping force, the bolt preload Fi.

3 The preloadFi is proportional to MA / where MA is the tightening torque thecoefficient of friction in the screw assembly. The coefficient of friction is managed by using a proper thread lubricant and varies between Calculate the coefficient of variation CV =(T50 Tmin) /T50 whereTmin is the least torque that can be transmitted by the shaft for lifetime performance and reliabilityProblem : Interference fit with hollow shaft A gear is to be press fitted over a hollow shaft of 20 mm the interference = mm. The strain is that much thatthe hollow shaft will deform plastically. The tensile stress in the plastic regime varies much less with the strainthan in the elastic regime. The tensile stress in the plastic regime of the steel shaft is approximatedby linear interpolation between (g= ) = 350 MPa and (g= ) =450 MPa. Calculate the coefficient of variation CV =(T50 Tmin) /T50 whereTmin is the least torque that can be : Chain dimensioningThe illustration below shows a simpledrawing of a part made by the symmetrical toleranceinterval of A with 99% probability,assuming all tolerances are normallydistributed within the 3 interval andindependent.

4 Problem : Number of measurements needed to obtain a reliable estimation When the measurement of the coefficient of friction is repeated one will find a large variation. Considerthe measured values , , , and a) Calculate the 95% interval over which coefficient of friction may lie. b) What number of measurements are needed to estimate the mean with 95% reliability within for lifetime performance and reliability5 Problem : Estimation of service intervalFrom a series of experiments it isfound that a component life is =150 103 km and = 20 103 km. Acomponent reliability of 90% isspecified with L10, of 99% with the value L10 and L1 and theratio a1 = L1/L10. Problem : Conversion of MTBF to Reliabilitya)Estimate the MTBF for N=10 devices that aretested for Ttest=500 hours and during the test r=2failures )Estimate the probability that any one particulardevice will be operational at the time equal to theMTBF? c)Estimate the probability that the component willwork for 50% of the MTBF d) Estimate the percentage of the MTBF whereR(t)= : Fault Tree AnalysisConsider the fault tree with the component reliability given in the table below and calculate the failureprobability F(t) of the system for a service life CDE FGR(t) for lifetime performance and reliabilityProblem : Bearing reliability, deep groove ball bearingCalculate the operating reliability R(t=1000hr) of a deep groove ballbearing.

5 The calculated L10 life expectancy of the ball bearing is L10 =500A106 rev. The rotational speed is 4000 : The life expectancy of the ball bearings is related to the L10 basicrating life according (eq. , page 129). Problem : Reliability factor for Fatigue strength A power supply is cooled by 3 fans. The correct functioning of at leastone of the three fans is required to maintain sufficient cooling. Theoperating reliability of the system needs to be 99% for a service life hr, Rs ( hr) = The rotational speed is 4000 ) Estimate the required operating reliability Rj of the individual ) Calculate the required L10h of the individual ) Calculate C/PProblem : Reliability factor for Fatigue strength Data published of the endurance strength are always mean values. InNorton (2000) is reported that the standard deviation of the endurancestrength of steels seldom exceeds 8% of their mean. Estimate a correctionfactor for the endurance strength if a 99% probability is : Component reliability Calculate the component reliability of a drive shaft(motor shaft) loaded in the High Cycle Fatigue (HCF)regime with Ln=2A105 load the calculated fatigue life of L50 = 3A105load cycles and a standard deviation of =.

6 Problem : Stress concentration factor "A chain is only as strong as its weakest link, regardless of the strength of the stronger links". Do youagree and what do you think about the reliability if the failure mode is fatigue? DESIGN for lifetime performance and reliability7 Problem : System reliabilityCalculate the failure probability F(100 hr) of twocritical components of a system connected in product catalogues it is found that the reliabilityof one component is specified with = 150 hr and = hr, the other component is specified with = 120hr and = hr. Hint: first step is to calculate R(t) of both : System reliabilityA heavy duty motorized frame features a quad drivesystem using two high power DC motors and four drivebelts. All four belts are required to maintain optimalcontrol. From field testing it is found that the service lifeof the belts under heavy duty operating conditions isnormally distributed with a mean = 200 hr and astandard deviation of =.

7 Calculate the operatingreliability of the set of 4 belts for a service life of 150 : System reliability There is a rule of thumb that says that the bearing load P relatedto the dynamic load rating of the bearing C is:Normal loaded bearings P= loaded bearings P= a motor drive equipped with two ball bearings. One ofthe bearings is loaded with P= , where C is the dynamic loadrating of bearing type 16004, C= kN. The motor rotates withn=1400 rpm during 8 hours a day, 5 days a week it is 1920hr/year. Calculate the life expectancy [years] of this bearing with1% failure probability. Hint: First calculated L10 and L10h. The life expectancy Lna of ball bearings is related to the L10 basic ratinglife according Lna =a1 L10 (Table page 129).8 DESIGN for lifetime performance and reliabilityProblem : System reliability Consider a motor drive equipped with two ball bearings. The lifeexpectancy of the bearings is calculated as L10h=12,000 hr andL10h=16,000 hour respectively.

8 Calculate the system (motor)reliability for a service life of 10,000 hour. a) The life expectancy Lna of the ball bearings is related to the L10basic rating life according Lna =a1 L10 (Table page 129) wherea1 is the reliability factor derived from a statistical Weibulldistribution .b) It is assumed that the failure distribution f(t) of the bearings isbest fitted by a Weibull Failure Distribution function with shape factor = (Eq. page 12).Problem : System reliabilityThe operating reliability of a production line needs to beestimated. The production line consists of 6 identical pickand place units with similar operating conditions. The most critical components of the individual pick and placeunits are identified using an FMEA procedure. The operatingreliability of these components are established and finallypresented in a fault value of the operatingreliability of the production line R(t).Which of the components would youselect to improve its operatingreliability by 5 percent in order toimprove the system reliability?

9 DESIGN for lifetime performance and reliability9 Problem : System reliabilityA monitoring unit is applied to register as a function of timeunexpected machine standstill which is caused by a specificcomponent in the pick and place units. The service life t (hr)of this component is derived from the monitored data andlisted in the table life t (hr) of the critical component590 420 520 480 490 510 450 480 Calculate the required maintenance interval in order toreplace this component in all pick and place units in reliability of the individual pick and place units should beat least R(t)= What is t?Problem : Precision and accuracy Repeatability (precision) is the error between a number ofsuccessive attempts to move the machine to the sameposition. Repeatability can be represented by the intervalwhich contains N% of the measured positions. Accuracy is the difference between the intended positionand the mean of the measured positions.

10 Theroot mean square deviation (RMSD) or root mean squareerror (RMSE) is a frequently used measure for accuracy. Theaccuracy can be improved by adjusting the offset. Theprecision remains the same. Resolution is the smallest possible movement of a system when actuated. Also known as the step resolution of an instrument is the smallest increment that the gage an intended position of xtarget = 40 mm and a range of measured x data: , , mm. Calculate the achieved accuracy and the repeatability within the95% confidence interval. What is the resolution of the caliper shown above?10 DESIGN for lifetime performance and reliabilityAnswers ) )The mean value of the diameter of the shafts =20mm. The interval is a 3 interval which resultsin = The 95% symmetrical interval, with on either side, corresponds to z= and A= = The CV value is / = )20H7: Dmax = 20 mm + 21 m, Dmin = 20 mm + 0 m, Dmean = 20 mm + m, D= 21/6 m20r6 : dmax = 20 mm + 41 m, dmin = 20 mm + 28 m, dmean = 20 mm + m, d = (41 28)/6 mInterference mean 50 = m m = 24 m, Interference SD The minimum value of with 1% failure probability Coefficient of variation, probabilistic: thus, Tmin= TmeanThe worst case scenario is obtained with the maximum bore diameter and the minimum shaft minimum value of (Deterministic): and )The mean value of the coefficient of friction 50 = The obtained preload F, with given tighteningtorque M, is minimal when the friction in fastener will have the highest ratio (worst case).