Machinery Components Maintenance And Repair Episode 2 Part 4 ppt
Metal Machining - Theory and Applications Episode 2 Part 4 pptx
... V
0 .27
)
}
(9.2b)
× (VB
–0.66
– 0 .23 V
0 .27
) (VN
0.03
– 0 .23 V
0 .27
)
F
c
= 1862f
0. 94
d
1.11
+ 26 77(VS
0 . 24
– 0.05ln V)
×(VB
0 .23 –
0.05ln V) (VN
0.16
– 0.05ln V)
where F
d
, F
f
, and F
c
are ... conventional,
(b) and (c) type I and II tools,
K
= 1 .2
Childs Part 2 28:3 :20 00 3:19 pm Page 26 2
where F*
t
, F*
r
and F*
z
are the specific cutting forces in t...
Dimensioning and Tolerancing Handbook Episode 2 Part 4 ppt
... ) ( )
2
1
22 222 222 2
22 222 222 222 2
22 222 222 2
0300.)1(8.0060.)1(8.0070.)1(8.
0075.)1 (2. 0050.)1(8.0070.)1 (2. 0050.)1(8.
0075.)1 (2. 0030.)1(8.0 020 .)1 (2. 0155).1 (2.
0300).1 (2. 0060).1 (2. 0070).1 (2. 0075).1(8.0050).1 (2. 0070).1(8.
0050).1 (2. 0075).1(8.0030).1 (2. 0 020 ).1(8.0155).1(8.
+−+
++++
++++−
++−++++
+++++−=
ems
t
t
ems ... = . 04 72
1
)...
Metal Machining - Theory and Applications Episode 2 Part 11 ppt
... [W/mK]
P01–10 50 16 20 6 8 < 2 6900 16 .2 1 .2 20
P05 25 49 16 15 8 12 < 2 7000 14 .2 1.8 20
P01–15 48 16 20 5 11 –* 7000 15.7 –* 20
P05 Total carbide: 94 Total metal: 6 –* 6100 17 .2 1.8 –*
P10 Total ... a
1
)
2
a
1
+ a
2
————— a
1
≤ x < ————
(a
2
— a
1
)
2
2
SF(x, a
1
, a
2
) =
{
(A7.4b)
2( x – a
2
)
2
a
1
+ a
2
1– ————— ———— ≤ x < a
2
(a...
Metal Machining - Theory and Applications Episode 2 Part 8 pptx
... 3(s
xx
′
2
+ s
yy
′
2
+ s
zz
′
2
) + 6(s
2
xy
+ s
2
yz
+ s
2
zy
) ≡
(s
xx
– s
yy
)
2
+ (s
yy
– s
zz
)
2
+ (s
zz
– s
xx
)
2
+ 6(s
2
xy
+ s
2
yz
+ s
2
zy
) = 6k
2
or 2Y
2
(A1 .26 b)
which is the same ... the results, a
346 Appendix 1
Childs Part 3 31:3 :20 00 10: 42 am Page 346
p
A
+ 2kf
A
= p
B
+ 2kf
B;
p
C
+ 2kf
C
= p
D
+ 2kf
D
(A1.19)
p
A
+ 2kf
A
= p
D
– 2kf...
Metal Machining - Theory and Applications Episode 2 Part 7 ppt
... addition
s
r
′
2
= s
r
2
–3s
m
2
=(s
1
2
+ s
2
2
+ s
3
2
) – 3s
m
2
(A1.5)
After substituting for s
m
from equation (A1.1),
3s
r
′
2
=(s
1
– s
2
)
2
+(s
2
– s
3
)
2
+(s
3
– s
1
)
2
(A1.6)
The yield ... stresses:
1
s
–
2
=—
[
(s
1
– s
2
)
2
+(s
2
– s
3
)
2
+(s
3
– s
1
)
2
]
= Y
2
or 3k
2
(A1.7)
2
330 Appendix 1
Fig. A1 .2 Geometrical representation...
Gear Noise and Vibration Episode 2 Part 4 pps
... 5*cos(1 .44 88
-(i-l)*0.1);
yl(i)
-
40 .7 84
-
5*sin(1 .44 88
-(i-l)*0.1);
end
fori=16:N;
%
involute
ra
=
(i-16)*0. 02;
xl(i) =40 .7 84* (sin(raHi-16)*0. 02* cos(ra));
yl(i) =40 .7 84* (cos(ra)+(i-16)*0. 02* sin(ra));
end
for
... i=(N+l) :2* N
; %
Image
in
x=0
other
flank
x2(i)
=
-
xl (2* N+l-i);
y2(i)
-
yl (2* N+l-i);
rot
1=0 .45 413;
xl(i)
=x2(i)*cos(rotl)
+y2(i)...
Gear Noise and Vibration Episode 2 Part 4 pot
... 5*cos(1 .44 88
-(i-l)*0.1);
yl(i)
-
40 .7 84
-
5*sin(1 .44 88
-(i-l)*0.1);
end
fori=16:N;
%
involute
ra
=
(i-16)*0. 02;
xl(i) =40 .7 84* (sin(raHi-16)*0. 02* cos(ra));
yl(i) =40 .7 84* (cos(ra)+(i-16)*0. 02* sin(ra));
end
for
... i=(N+l) :2* N
; %
Image
in
x=0
other
flank
x2(i)
=
-
xl (2* N+l-i);
y2(i)
-
yl (2* N+l-i);
rot
1=0 .45 413;
xl(i)
=x2(i)*cos(rotl)
+y2(i)...
Gear Noise and Vibration Episode 2 Part 6 ppt
... the
pinion
at
position
2. So
8,
=
Fpi,
+
Fp
12
and
5
2
=
Fp
12
+
Fp
22
.
As the
transmission error
is the sum of 81 and
S
2
T.E.
=
F(p
n
+2p,
2
+
p
22
)
giving
F.
Once
F ... analysis
of the
input
and of the
output
and the
ratio
(complete with phase)
is the frequency
transfer
function
required.
14
12
50
100
150 20 0 25 0
300
350
40 0
F...
Process Selection - From Design to Manufacture Episode 2 Part 4 pptx
... strategies and PRIMAs 24 3
//SYS21///INTEGRAS/B&H/PRS/FINALS_07-05-03/07506 543 76-CH0 02- 1.3D – 22 4 – [35 24 8 /21 4] 9.5 .20 03 2: 05PM
.
Infrared Brazing (IRB): uses quartz-iodine incandescent lamps ... jigging and fixtures.
7.13F Soldering process.
22 6 Selecting candidate processes
//SYS21///INTEGRAS/B&H/PRS/FINALS_07-05-03/07506 543 76-CH0 02- 1.3D – 24 0 – [35 2...
342 TOEIC Vocabulary Tests Episode 2 part 4 pptx
... inattention; forgetfulness; abandonment
definition (c) neglect
48 8
Questions Index
PHOTOCOPIABLE © www.english-test.net
Answers 1 24
TOEIC Vocabulary / Word by Meaning / Test # 1 24 (Answer Keys)
A1 adj. ... portions; to cleave
definition (b) slice
47 9
Questions Index
PHOTOCOPIABLE © www.english-test.net
Answers 122
TOEIC Vocabulary / Word by Meaning / Test # 122 (Answer Keys)
A1 v. to...