Machinery Components Maintenance And Repair Episode 2 Part 3 ppt
Metal Machining - Theory and Applications Episode 2 Part 11 ppt
... √* √* √* 33 00 93. 9
1
1.1 – – –
– 91 0 1 8 32 0 0 14 .2 0.8
K20 90.5 6 3. 5 32 6 0 13. 7 1.0 – – –
K05 20 √* √* √* √* 33 00 15.4 0.9 17 28 0 3. 0
K01 30 √* √* √* 33 00 15.7 0.8 – – –
– 80 6 4 10 32 0 0 15.7 ... meshing 20 3 4, 21 0
chip fracture criteria 20 9, 22 0, 25 2 3
comparison of approaches 21 0– 12
coupled thermal/mechanical analysis 20 5–6, 21 3
elastic example 1...
Metal Machining - Theory and Applications Episode 2 Part 8 pptx
... the results, a
34 6 Appendix 1
Childs Part 3 31 :3 :20 00 10: 42 am Page 34 6
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
D;
p
B
– 2kf
B
= p
C
+ 2kf
C
}
Geometry ... s
–
0
∂(de
–
)
– — ——— ——— d(de
kl
)(de
ij
)
0
3( de
–
0
)
2
∂(de
kl
)
(A1 .34 )
Constitutive equations for numerical modelling 34 3
Childs Part 3 31 :3 :20 00 10: 42 am...
Metal Machining - Theory and Applications Episode 2 Part 7 ppt
... s
m
3. OP is s
r
′. By vector 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
3
)
2
+(s
3
– s
1
)
2
(A1.6)
The yield criterion may be restated in terms of the principal stresses:
1
s
–
2
=—
[
(s
1
– s
2
)
2
+(s
2
– s...
Metal Machining - Theory and Applications Episode 2 Part 4 pptx
... 500f
0.46
d
0.810
+ 23 77(VS
1. 93
– 0.007ln V)
× (VB
0 .26
– 0.007ln V) (VN
–0 .33
– 0.007ln V)
F
f
= 629 f
0 .30
d
0. 720
+ 1199(VS
3. 58
– 0. 0 23 V
0 .27
)
}
(9.2b)
× (VB
–0.66
– 0 . 23 V
0 .27
) (VN
0. 03
– 0 . 23 V
0 .27
)
F
c
= ... analysis
Childs Part 2 28 :3 :20 00 3: 18 pm Page 25 3
strain rate for the plane-faced tool and type I and II tools with K = 1 .2. T...
Metal Machining - Theory and Applications Episode 2 Part 3 pot
... 1.0 2. 31 3. 89 1.0 2. 48 3 .22
Steel X 0.99 1 .25 3. 05 0.96 1. 43 2. 06
Steel Y 0.97 0.76 5.98 0.99 1.04 3. 04
Steel L 0.74 0 .38 8.78 0.88 0. 72 4 .21
Childs Part 2 28 :3 :20 00 3: 17 pm Page 24 4
With other ...
–0.000015(T 37 0)
2
a = 0.00 028 , m = 0.0016
Childs Part 2 28 :3 :20 00 3: 17 pm Page 24 2
24 6 Applications of finite element analysis
Fig. 8.16
co...
Gear Noise and Vibration Episode 2 Part 3 pot
... (V2r2-Vlr2)
+
[k !2/ r2
2
]
(s2r2-slr2-te)
= - F 23
[I3/r3
2
]
(A3.r3)
+
[D3/r3
2
]
(V3 r3) +
[D34/r3
2
]
(V3r3-V4r3)
+
[k34/r3
2
]
(s3r3-s4r3)
= - F 23
[I4/r4
2
]
(A4.r4)
+
[D4/r4
2
]
... equations
and
take
rl=r2
[Il/r2
2
]
(Al.r2)
+
[Dl/r2
2
]
(VI
r2) +
[D 12/ r2
2
]
(Vlr2-V2r2)
+
[k !2/ r2
2
]
(slr2-s2r2+te)
-
Q/r2
[I2/r2
2
]
(A2.r2...
Gear Noise and Vibration Episode 2 Part 3 doc
... (V2r2-Vlr2)
+
[k !2/ r2
2
]
(s2r2-slr2-te)
= - F 23
[I3/r3
2
]
(A3.r3)
+
[D3/r3
2
]
(V3 r3) +
[D34/r3
2
]
(V3r3-V4r3)
+
[k34/r3
2
]
(s3r3-s4r3)
= - F 23
[I4/r4
2
]
(A4.r4)
+
[D4/r4
2
]
... equations
and
take
rl=r2
[Il/r2
2
]
(Al.r2)
+
[Dl/r2
2
]
(VI
r2) +
[D 12/ r2
2
]
(Vlr2-V2r2)
+
[k !2/ r2
2
]
(slr2-s2r2+te)
-
Q/r2
[I2/r2
2
]
(A2.r2...
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 ... frequency
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
30 0...
GMAT exam success Episode 2 Part 3 ppt
... constraint and the fact that one of your read-
ers is a computer, it’s best to stick to the old standard.
23 2
The writing process can be divided into four steps:
1. Planning
2. Drafting
3. Revising
4. ... making
effective changes.
Be Flexible
24 3
embrace their social responsibilities and empower their leaders and employees to serve local and
world communities will reap reward...
Gear Noise and Vibration Episode 1 Part 3 pptx
... gears.'
Proc.
Inst.
Mech. Eng.,
Vol
178, 19 63- 64, Part
I, pp
20 7 -22 6.
2.
Maag Gear Handbook (English version) Maag, CH8 0 23 , Zurich,
Switzerland.
3.
Harris, S.L.,
'Dynamic
loads
on ... by
taking moments about
one end and is
approximately
(r-l)
2
/2r
+
\!2r
which
is
[(r-l)
2
+l]/2r
from one
end. This
has a
minimum when
r
is
V2
and the
c...