Question

fast please

CHAPTER 8 8.1. A 120 central partial bearing, 67mm in diameter and 100mm long, has a minimum film thickness of 0.017mm. SAE 20 oil (see Figure 8-2) is used; 1-884rpm and ric-1000. If the oil film has a temperature of 72.5°C, calculate the load the bearing is carrying.

Answer 1

SOL Given data:- mm 120° central partial bearing Diameter CD) = 67 mm, Y= 33.5 mm length (L) = 100 mm minimum film thickness Chal=0.019 mm 884 ypm r/c oil film temperature (T) = 12.5°C Therefore, we know that CU) - Speed cn) - 1000 Tangential velocity cu) - Tan 60 TI X67x884 60 = 3101.17 mml sec U = -3101.17 mm/sec . Given Ž= = 1000 33-5 с 1000 .:( = 0.0335 mm Cclearance of joumal bearing) Given, minimum film thickness chol .: eccentricityce): 1- ha = 0.017 mm 0.017 0.0335

Eccentricity CE) = 0.4925 of from the given valve temperature (71) = 72.5", SAE 20 Cgraph average values of viscosity for She- lubricant oils] Absolute viscosity cu) = 11.84 centipoise = 11.84 x103 45 = 0.0000 000 1187 N-6/mm Therefore (€) ratio 100 67 = 1.4925 1.4925 4 = from the graph of load friction characteristicą value of won [g]based on E = 0.4925 and £ 51.4925 Mo 1.90 = 1.90 HU 0.0355 j 1.90 33.5 (0.0000000 1187) (3101.17) WI = 62.282 N/mm Total Load acting per axil Length W = W,XL = 62.282 X 100 wiW- 6228.2 N Therefore, the total load cw) for the fournal bearing W = 6228.2 N