Machinery Components Maintenance And Repair Episode 2 Part 1 pps
Gear Noise and Vibration Episode 2 Part 1 ppsx
... Fig.
9. 12 )
with 19 :29
at
input
and
23 : 31
at
output.
For a
complete
meshing cycle
the
layshaft
would have
to do 19 x 31
revs
and 589
revs would take rather
a
long time
and
require ... of
waveform.
Analysis
Techniques
14 9
1
0.8
0.6
0.4
0 .2
0
-0 .2
-0.4
-0.6
-0.8
-1
6 8
window
length
L
10
12
Fig 9.7
Finite length records showing
end
effects....
Metal Machining - Theory and Applications Episode 2 Part 5 ppsx
... regenerative
28 2 Process selection, improvement and control
Fig. 9.9 Unconditional chatter limit
Childs Part 3 31: 3 :20 00 10 :38 am Page 28 2
f
mach
(1 – n
1
)
n
1
1
C
c
t
ct
+ C
t
n
1
C
opt
= C
c
t
load
+ ... C
p
(x)
29 2 Process selection, improvement and control
Fig. 9. 12 Fuzzy optimization of cutting conditions; only three constraints 1, 4 and 10 are considered...
Metal Machining - Theory and Applications Episode 2 Part 1 potx
... cutting
process (Part 1) . Trans. ASME J. Eng. Ind. 10 0, 22 2 22 8.
Usui, E. and Hirota, A. (19 78) Analytical prediction of three dimensional cutting process (Part 2) .
Trans. ASME J. Eng. Ind. 10 0, 22 9 23 5.
Usui, ... 19 5
Fig. 6 .18 The dependence of
η
c
on (a)
λ
s
and (b)
r
n
, for machining a carbon steel (after Usui and Hirota, 19 78)
(a)
Childs Part 2 28:3 :20...
Bearing Design in Machinery Episode 2 Part 1 pps
... 0:877
DT
m
¼
8:3P½R=Cðf Þ
10
6
Q
nRCL
1 Àð0:5ÞQ
s
=Q
¼
8:3 25 Â0:877
3 :29 Â0: 519 5
¼ 10 6
C
b. Maximum and Average Oil Temperatures:
Maximum temperature:
T
max
¼ T
in
þ DT ¼ 20 10 6 ¼ 12 3
C
Average temperature:
T
av
¼ ... highly dependent on the film
thickness, particularly under high loads.
For turbulent fluid films, the equation is
T
max
À T
1
¼
f p
2
N
2
D
3
2gc
p...
Gear Noise and Vibration Episode 2 Part 1 docx
... Fig.
9. 12 )
with 19 :29
at
input
and
23 : 31
at
output.
For a
complete
meshing cycle
the
layshaft
would have
to do 19 x 31
revs
and 589
revs would take rather
a
long time
and
require ... resulting Fourier analysis
and add
bands
in
groups.
If the
original record
was for 10 s the
bandwith would
be
0 .1
Hz and
adding
10
bands would widen
the
band...
Gear Noise and Vibration Episode 2 Part 4 pps
... there
is a
20 8
Chapter
12
end
for
th
=
1: 8;
%
rotate
for
other
8
teeth
xl((th *2* N
+l):(th+l) *2* N)
=
xl(l :2* N)*cos(0.69 813 *th)+yl(l :2* N)*sin(0.69 813 *th);
yl((th *2* N
+l):(th+l) *2* N)
=-
xl(l :2* N)*sin(0.69 813 *th)+yl(l :2* N)*cos(0.69 813 *th);
end
saveteeth9
... tip
relief
and
deflections
for
contact ratio
of 2.
21 2
Chapter
12
50
10 0
15...
Gear Noise and Vibration Episode 2 Part 5 pps
... noise.
22 8
Chapter
14
1. 08:
1. 06
1. 04
1. 02
tangent
from
-0 .1
position
base
circle
0.98
L
•-
<
1
-—
— -
—
-0. 12
-0 .1 -0.08 -0.06 -0.04 -0. 02
0
Fig
14 .5 Expanded view
of
involute near ... Monitoring
23 9
at
c
c
o
2
*
4>
0
U
0)
£
.*:
(0
4)
0,
n
u
-50
-10 0
-15 0
-20 0
-25 0
-300
-350
Tooth
Tooth
Tooth
Tooth
Tooth
Tooth
Tooth
Tooth
Tooth
Tooth...
Dictionary of Engineering Episode 2 Part 1 ppsx
... handle.
2.
An
nance and repair of machinery or equipment.
arm with a striking head for sounding a bell or
{ handho
¯
l}
gong.
[
MECH ENG
]
A power tool with a metal
hand lance
[
ENG
]
A hand-held ... to 1/ 4 U.S. liquid pint, or to
getter sputtering
[
ELECTR
]
The deposition of
1. 1 829 411 825 ϫ 10
Ϫ4
cubic meter.
2.
A unit of
high-purity thin films at ordinary vacuum levels...
Process Selection - From Design to Manufacture Episode 2 Part 1 ppsx
... transferring the partly
6.1F Manual assembly process.
18 0 Selecting candidate processes
//SYS 21/ //INTEGRAS/B&H/PRS/FINALS_07-05-03/0750654376-CH0 02- 1. 3D – 16 5 – [35 24 8 / 21 4] 9.5 .20 03 2: 05PM
5 .2 Electrochemical ... depending on part complexity,
typically Æ0.5 mm.
1 82 Selecting candidate processes
//SYS 21/ //INTEGRAS/B&H/PRS/FINALS_07-05-03/0750654376-CH0 02-...
Engineering Mechanics - Statics Episode 2 Part 1 ppsx
... d+()
T
BC
db−()
2
i
2
+ c
2
+
+ A
x
+ 0=
i−
T
DE
ak−()
2
i
2
+ j
2
+
i
T
BC
db−()
2
i
2
+ c
2
+
− A
y
+ 0=
j
T
DE
ak−()
2
i
2
+ j
2
+
c
T
BC
db−()
2
i
2
+ c
2
+
+ W− 0=
M
Ax
T
DE
j
i
ak−()
2
i
2
+ j
2
+
+ ... are
T
BC
10 0 lb= T
BD
10 0 lb= A
x
10 0 lb= A
y
10 0 lb= A
z
10 0 lb=
Given
F
1
cos
θ
()
F
2
+
()
cT
BC
a
c
a
2
b
2
+ c
2
+
⎛
⎜
⎝
⎞
⎟...
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