Engineering Tribology 2E Episode 4 pptx
Engineering Tribology 2E Episode 4 pptx
... ratios.
0 0 1
0 .48 2 0.2
0 .46 6 0 .4
0 .45 3 0.6
0 .44 2 0.8
0 .43 1 1.0
0 .42 2 1.2
0 .41 4 1 .4
0 .40 6 1.6
0.399 1.8
0.393 2.0
0.387 2.2
0.381 2 .4
0.376 2.6
0.371 ...
h
1
= h
0
(K + 1)
(4. 44)
Substituting into (4. 43) the constants ‘
h
’ and ‘C’ in terms of ‘K’ are:
= 2h
0
(K + 1)
h
¯
(K + 2)
C =
1
h
0
(K + 2)
(4. 45)...
Engineering Tribology 2E Episode 5 pptx
... discussed in
Chapter 2.
y
z
41 °
42 °
43 °
45 °
46 °
48 °
47 °
49 °50°
45 °
44 °
42 °
40 °
37°
35°
30° 27°
Shaft Oil film Bush
36.5°
37°
41 °
43 °
45 °
48 °
49 °
50°
39°
41 °
43 °
37.5°
Oil supply groove
Shaft
Oil
film
Bush
Attitude ... Vol. 102, 1980, pp. 43 9 -44 4.
43 B.A. Gecim, Non-Newtonian Effects of Multigrade Oils on Journal Bearing Performance, Tribology
T...
Engineering Tribology 2E Episode 10 pptx
... Tribology, Vol. 113, 1991, pp. 327-3 34.
116 H.A. Spikes and A. Cameron, A Comparison of Adsorption and Boundary Lubricant Failure, Proc. Roy. Soc.,
London, Series A, Vol. 336, 19 74, pp. 40 7 -41 9.
117 ... 1970, pp. 1 74- 178.
89 A.W. Batchelor, A. Cameron and H. Okabe, An Apparatus to Investigate Sulfur Reactions on Nascent Steel
Surfaces, ASLE Transactions, Vol. 28, 1985, pp. 46 7 -47...
Engineering Tribology 2E Episode 1 pot
... ENGINEERING TRIBOLOGY
9 .4 Summary 44 2
References 44 2
10 FUNDAMENTALS OF CONTACT BETWEEN SOLIDS 44 7
10.1 Introduction 44 7
10.2 Surfaces of solids 44 7
Surfaces at a nano scale 44 8
Surface topography 44 9
Characterization ... condition 143
Load capacity 144
4. 5 Journal bearings 146
Evaluation of the main parameters 146
· Bearing geometry 146
· Pressure distribution 148...
Engineering Tribology 2E Episode 2 ppsx
... 000
35
40
45
50
60
70
80
90
2
2.5
3
3.5
4
4.5
5
6
7
8
9
10
15
20
25
30
35
40
45
50
60
70
80
90
1.9
1.8
1.7
1.6
1.5
1 .4
1.3
1.2
25
30
35
40
45
50
60
70
80
90
100
150
200
250
300
350
40 0
45 0
30
35
40
45
50
60
70
80
90
100
150
200
250
300
350
40 0
100
120
FIGURE ... 3.52
ISO VG 5 4. 14 4.6 5.06
ISO VG 7 6.12 6.8 7 .48
ISO VG 10 9.00 10 11.0
ISO VG 15 13.5 15 16.5
ISO VG 2...
Engineering Tribology 2E Episode 3 potx
... w
h
dxdy
(4. 13)
LUBRICANTS AND THEIR COMPOSITION 79
T
ABLE 3.3 NLGI grease classification [28].
000 44 5 - 47 5
00 40 0 - 43 0
0 355 - 385
1 310 - 340
2 265 - 295
3 220 - 250
4 ... 17, 19 74, pp. 283-289.
49 F.G. Rounds, Some Effects of Amines on Zincdialkyldithiophosphate Antiwear Performance as Measured in
4- Ball Wear Tests, ASLE Transactions, Vol. 24, 198...
Engineering Tribology 2E Episode 6 pdf
... constant, i.e.:
222 ENGINEERING TRIBOLOGY
Substituting (5 .40 ) and (5 .46 ) into (5 .47 ) yields:
∂
∂x
⌠
⌡
h
0
M∂z + U
∂
∂y
⌠
⌡
h
0
N∂z
()
∂p
∂x
+
⌠
⌡
h
0
M∂z
(
∂p
∂y
)
= 0
(5 .48 )
This equation can ... bearing is c = 0.00 04 [m], then from equation (5.87) the value of
flow ‘Q’ is:
Q = 0.5
× 6.8 × 0.2 × 10 × 0.00 04 = 2.72 × 10
-3
[m
3
/s] = 2.72 [litres/s]
2 34 ENGINEERING TR...
Engineering Tribology 2E Episode 7 doc
... applied to distinguish between these surfaces: if the centre of curvature
2 54 ENGINEERING TRIBOLOGY
1
3
4
0 0.1 0.2 0.3 0 .4 0.5 0.6
Misalignment parameter
t
Dimensionless critical frequency
2
FIGURE ... was
published by Grubin in 1 949 [1]. It was found that the combination of three effects:
2 84 ENGINEERING TRIBOLOGY
p p
kk
σ
1
σ
1
σ
3
W
Body A
Body B
Interface
45 °
FIG...
Engineering Tribology 2E Episode 8 ppt
... × 10
−3
)
= 55 .4 [MPa]
p
average
=
W
4bl
=
5
4 × (5.75 × 10
−6
) × (5 × 10
−3
)
= 43 .5 [MPa]
· Maximum Deflection
= 2 .40 × 10
−8
[m]
δ= 0.319
W
E'l
[]
2
3
[)]
+ ln
(
4R
A
R
B
b
2
= 0.319
[]
2
3
[)]
+ ... steady-state component [45 ].
The flash temperature theory was originally formulated by Blok in 1937 [46 ] and developed
further by Jaeger in 1 944 [47 ] and Archard in 1958 [...
Engineering Tribology 2E Episode 9 potx
... Tallian and J.I. McCool, An Engineering Model of Spalling Fatigue Failure in Rolling Contact, II. The
Surface Model, Wear, Vol. 17, 1971, pp. 44 7 -46 1.
336 ENGINEERING TRIBOLOGY
If the contacting ...
b =
πlE'
()
4WR'
0.5
=
4 × (5 × 10
3
) × (6 × 10
−3
)
π × (5 × 10
−3
) × (2.308 × 10
11
)
()
= 1.82 × 10
4
[m]
0.5
· Thermal Diffusivity
χ=
ρσ
K
=
46 .7
7800 × 46 0
=...
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