Page 381 - Analysis and Design of Machine Elements

Steps Computation Results Sliding Bearings 359
Units
27. Check the At the minimum clearance: h min = 0.03225 mm
minimum film (1) check the minimum film thickness:
thickness and the
temperature rise = Δ min = 0.188 = 0.001253
at the minimum d 150
clearance From
F 2 17000 ×(0.001253) 2
C = 2 vB = 2 × 0.027 × 7.85 × 0.075 = 0.8399
p
According to C and B/d,from
p
Table 12.2, we have = 0.6568
From
d 150
h = (1 − )= × 0.001253 ×(1 − 0.6568)
min 2 2
= 0.03225 mm
Since h > [h], the design at minimum clearance is
min
satisfactory.
(2) Check the temperature rise
According to and B/d, from Table 12.4, by
interpolating, we have
C = 4.6277
f
f = C =4.6277 × 0.001253 = 0.005799
f
According to and B/d, from Table 12.3, by
interpolating, we have
C = 0.0536.
Q
The temperature rise is
f
p
∘
Δt = d = Δt = 18.75 C
2c C + s
B Q v
0.005799 × 1.511 × 10 6
0.001253 = 18.75
0.15 ×80
2 × 1800 × 900 × × 0.0536 +
0.075 0.001253×7.85
The input temperature is
Δt 18.75 ∘
t = t − 2 = 50 − 2 = 40.62 C
m
i
Δt 18.75 ∘ ∘
t = t + = 50 + = 59.38 C < 80 C
o m 2 2
Both t and t are acceptable.
o
i
28. Final design Lubricating oil L-AN46 B = 50 mm
result t = 50 ∘ C
m
journal:
150 −0.145 mm
−0.185
R = 0.0032
z1
Bore:
150 +0.068
+0.043
mm
R = 0.0063
z2