The journals in a high speed oil engine are 80 mm in diameter. and 40 mm long. The radial clearance is 0.060mm. Each supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is about 65oC. Using Raimondi - Boyd charts analyze the bearing under steady state operation.

Answer: S = 0.284

Explanation:

dа ta: d = 80 mm;

l = 40 mm;

c = 0.06 mm;

F = 9kN;

n = 3600rpm = 60 rps

SAE 40 oil

T = 65°C

Therefore:

p = F / ld

= 9 x1000 / 40 x 80

= 2.813 MPa

μ = 30 cp at 65°C for SAE 40 oil

S = r^2 x μ x n / c^2 x p

S = (40) ^2 x 30*10^-3 x 60 / (0.06) ^2 x 2.813*10^6

S = 2880 / 10,126.6

S = 0.284

l/d = ½,

h o / c = 0.38

ε = e / c = 0.62

h o = 0.38 x C

= 0.382 x 0.06

=0.023mm

= 23µm

e = 0.62 x C

= 0.62 x 0.06

= 0.037 mm

Viscosity temperature curves of SAE graded oils

(r / c) f = 7.5,

S = 0.284

l / d = ½

f = 7.5 x (c / r)

= 7.5x (0.06/40)

= 0.0113