Engineering Tribology Episode 2 Part 9 ppsx
Engineering Tribology Episode 2 Part 9 ppsx
... Coatings,
Wear, Vol. 1 52, 19 92 , pp. 127 -150.
TEAM LRN
446 ENGINEERING TRIBOLOGY
88 J.C. Angus, Diamond and Diamond-Like Films, Thin Solid Films, Vol. 21 6, 19 92 , pp. 126 -133.
89 B.C. Oberlander and ... Graphite Slider, Wear, Vol. 198 , 199 6, pp. 20 8 -21 5.
93 R. Ahmed and M. Hadfield, Rolling Contact Performance of Plasma Sprayed Coatings, Wear, Vol. 22 0, 199 8,
pp. 80 -...
Engineering Mechanics - Statics Episode 2 Part 9 ppsx
...
y
w
2F
H
⎛
⎜
⎝
⎞
⎟
⎠
x
2
=
dy
dx
w
F
H
x= h
w
2F
H
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
= F
H
w
2h
a
2
⎛
⎜
⎝
⎞
⎟
⎠
2
=
tan
θ
max
()
wa
2F
H
= cos
θ
max
()
2F
H
4
F
H
2
wa()
2
+
=
T
max
F
H
cos
θ
max
()
= F
H
2
wa()
2
4
+=
wa
2
a
2
16h
2
1+=
Guess ... y−()=
Σ
M
= 0;
Mwy
y
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wb
b
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wa
a
2
⎛
⎜
⎝
⎞
⎟
⎠
+ wby− 0=
My() wby
1
2
wy
2
−
1
2
wb
2
−
1
2...
Handbook of Micro and Nano Tribology Episode 2 Part 9 pot
...
Handbook of Micro/Nanotribology.
Ed. Bharat Bhushan
Boca Raton: CRC Press LLC, 199 9
© 199 9 by CRC Press LLC
© 199 9 by CRC Press LLC
Preface
Second Edition, 199 9
The first edition of ... (alk. paper)
1. Tribology Handbooks, manuals, etc. I. Bhushan, Bharat, 194 9- .
TJ1075.H245 199 9
621 .8
′
9 dc21
98 -24 466
CIP
This book contains information obtained from authent...
Engineering Tribology Episode 1 Part 9 pps
... -0. 690 -0.348 - - -
7 W [N] 1.7575×10
4
-0. 793 2. 033 2. 596 1.0 42 0.884 -1.510 - - 1.108 -
8 H [W] 2. 791 5× 10
3
-0.5 79 1.530 0.873 2. 500 1.6 42 -0 .22 5 - -
9 ε 1.0516× 10
2
0. 399 -1.040 1.3 72 -0.5 39 ... 2. 5 12 0.563 0. 528 -1. 090 -0.383 - - - 1.385 -
2 H [W] 3 .93 07× 10
3
-0.706 1.577 0.477 2. 240 1 .28 7 0 .24 9 -0 .20 4 - 1
. 324
- -
3 ε 1 .26 66× 10
-2...
Engineering Tribology Episode 2 Part 1 doc
... w
y
*sin(x*)
(5 .99 )
The squeeze term ‘w*’ can be included in the parameter ‘G’ of the Vogelpohl equation, i.e.:
=
h*
1.5
∂
2
M
v
∂x*
2
+
(
R
L
)
2
∂
2
M
v
∂y*
2
= FM
v
+ G
FM
v
+
∂h*
∂x*
+ 2w*
(5.100)
Damping ... by:
h* = h*
static
+ ∆x*cos(x*) + ∆y*sin(x*)
(5 .90 )
∂
2
h*
∂x*
2
()
=
∂
2
h*
∂x*
2
static
+
∂
2
∂x*
2
[
∆x*cos(x*) + ∆y*sin(x*)
]
(5 .91 )
The...
Engineering Tribology Episode 2 Part 2 pdf
... LRN
27 4 ENGINEERING TRIBOLOGY
Equating flow through the orifice to the total lubricant flow through the bearing (6. 19) :
πd
2
2
p
s
− p
r
2
(
(
1 /2
C
d
=
p
r
h
3
η
B
Rearranging:
πd
2
η
2h
3
B
p
s
... these bearings are shown in Figure 6 .2.
TEAM LRN
26 8 ENGINEERING TRIBOLOGY
1
2
5
10
20
B
0 0 .2 0.4 0.6 0.8 1.0
C/B
0.4
0.5
0.6
0.7
0.8
0 .9
1...
Engineering Tribology Episode 2 Part 3 pps
...
p
max
=
3W
2 a
2
=
3 × 5
2 (5. 799 × 10
−5
)
2
= 7 09. 9 [MPa]
p
average
=
W
πa
2
=
5
π(5. 799 × 10
−5
)
2
= 473.3 [MPa]
· Maximum Deflection
δ= 1.0 397
W
2
E'
2
R'
()
1/3
= 1.0 397
5
2
(2. 308 ...
p
max
=
3W
2 a
2
=
3 × 5
2 (3.65 × 10
−5
)
2
= 1 791 .9 [MPa]
p
average
=
W
πa
2
=
5
π(3.65 × 10
−5
)
2
= 1 194 .6 [MPa]
· Maximum Deflection
δ= 1...
Engineering Tribology Episode 2 Part 4 docx
... 1.03 39
R
y
R
x
()
0.03
0. 02
()
0.636
= 1.03 39
0.636
· Simplified Elliptical Integrals
= 1. 39 82 = 1.0003 +
0. 596 8R
x
R
y
= 1.0003 +
0. 596 8
× 0. 02
0.03
= 1.77 19 = 1. 527 7 + 0.6 023 ln
R
y
R
x
()
= 1. 527 7 + 0.6 023 ln
0.03
0. 02
()
· ...
p
max
=
3W
2 ab
=
3 × 50
2 (2. 32 × 10
−4
) × (1.75 × 10
−4
)
= 588.0 [MPa]
p
average
=
W
πab
=
50
π (2. 32 × 10
−4
) × (1.75...