Page 380 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
358
Steps Computation Results Units
19. Compute the According to and B/d, from Table 12.4, by f = 0.00582
coefficient of friction interpolating, we have
C = 4.349
f
f = C =4.349 × 0.001339 = 0.00582
f
20. Select the According to the ratio of width to diameter B/d, and C = 0.0541
Q
coefficient of oil flow ratio of eccentricity = 0.676, from Table 12.3, we
have the coefficient of oil flow C = 0.0541.
Q
7.85 × 2
3
3 −1
21. Calculate oil flow Q = C d = 0.0541 × 0.001339 × Q = 2.56 × m s
Q
rate 3 −5 3 −1 0.15 10 −5
× 0.15 = 2.56 × 10 m s
−3
22. Compute the Assuming the density of lubricant = 900 kg m , Δt = 17.54 ∘ C
temperature rise of specific heat capacity of lubricant is
∘
−1
the lubricant from c = 1800 J (kg C) , heat transfer coefficient of
−1
inlet to outlet lubricant is = 80 W (m 2 ∘ C) . From Eq. (12.26),
s
f p

Δt =
d
2c C + s
B Q v
0.00582 × 1.511 × 10 6
0.001339
=
2 × 1800 × 900 × 0.15 × 0.0541 + ×80
∘ 0.075 0.001339×7.85
= 17.54 C
Δt 17.54 ∘
23. Compute the oil t = t − = 50 − = 41.23 C
i
m
2
2
inlet temperature Δt 17.54 ∘ ∘
t = t + = 50 + = 58.77 C < 80 C
o m 2 2
∘
Since normally t = 35–45 C, the inlet temperature
i
is acceptable
24. Select tolerance According to the diametral clearance Δ= 0.20 mm,
and fit the tolerance and fit is selected as F6/d7 according to
GB/T 1801-1979. The diameter of the bearing bore
is 150 +0.068 the diameter of the journal is 150 −0.145
−0.185
+0.043
25. Compute the Δ max =0.068 − (−0.185) = 0.253mm Δ min = 0.188 mm
minimum diametrical Δ =0.043 − (−0.145) = 0.188mm Δ = 0.253 mm
clearance and the min max
maximum diametrical Since Δ= 0.20 mm is between Δ min and Δ max ,
clearance therefore the tolerance is acceptable
26. Check the At the maximum clearance h min = 0.02943 mm
Δ
minimum film = max = 0.253 = 0.0016867
thickness at the d 150
maximum clearance From
F 2 17000 ×(0.0016867) 2
C = = = 1.5212
p
2 vB 2 × 0.027 × 7.85 × 0.075
According to C and B/d,from
p
Table 12.2, we have = 0.7674
From
d 150
h min = 2 (1 − )= 2 × 0.0016867 ×(1 − 0.7674)
= 0.02943 mm
Since h > [h], the design at maximum clearance is
min
satisfactory.
(continued)