Question

5) [10] A journal bearing has a journal diameter of 75 mm with a unilateral tolerance of - 0.075 mm. The bushing bore has a diameter of 75.15 mm and a unilateral tolerance of 0.1 mm. The busing is 75 mm long and supports a 2585 N load. The journal speed is 750 rev/min. Find the minimum oil film thickness and the maximum film pressure for SAE 20W lubricant at an operating temperature of 65 °C for the minimum clearance assembly.

Answer 1

75 mm TS)5 mm W = 258 S N Solution: Given dater - journal diameter, dmax bushing bore diameter, bmin length of the bushing I=75 mm journal speed N= 750 rew/min lubricant operciting temperature, T = 65°C Now the minimum radical clearance can be given boty ! Cmin bmin dmax SAE 20W = 75.15-75 2. 2 Cmin = 0.075 mm Now l 1 75 75 dmax Now radius of the journal can be given by r= dmax 플 75 = 37.5 mm

Now Clearance ratio can be given by:- 7 = Soo Cmin 37.5 0.075 Now we will Now the pressure can be given by: P= W 2585 = 0.460 N or 0.460 mpa edmax (75)*(75) mm2 obtain the viscosity of SAE 2016 temperature of from. the standard dentar u= 15 mPas Now the bearning characteristic number of the can be given by : ail.coppershanging to operating Sommerfeld number

Sa 8 (min) u N P where N= 750 rev min 750 rev = 12.5rer 60 sec. so s = soo 00) 15x10-3 X 12.5 0.460x 106 s=o. 1019 corres 3 Now from the standard dentar s=o.log and dmax I 1 we Japonding to hoz 0.34 or ho (10.34) (c) с. CAS AS C= (min 0.075mm or ho = (0.34 )x(0.075) So, minimum film thickness, ho = 0.025s.mm Am

to on the s = d.lolg and from the Standard deter, corresponding find I 1 7 dmax --- P 0.4 Pmax = 4 P Pmax Pmax = so 1.15 mpg 0.460 0.4 Pman So, Ismpa the press ssure, maximun film Au