Plastics Product Design Part 7 pps
Handbook of Materials for Product Design Part 7 ppt
... modulus, 106 psi 0°
262.2
177 .6
7. 00
271 .5
181.3
7. 11
251.9
173 .8
6.85
219.6
181.8
6 .78
224.2
1 87. 9
6.82
216.1
170 .6
6 .75
193.2
165.0
6.95
196 .7
173 .8
7. 18
198.2
154 .7
6.80
Bearing
ultimate stress, ... at 4% elong. ksi 0°
77 .3
20.5
80.9
21.5
74 .8
19.1
68.8
17. 7
71 .4
19 .7
65.3
17. 2
53.6
17. 0
54.6
18.5
52.1
15.8
Interlaminar shear
ultimate stress, ksi 0°...
Handbook of Materials for Product Design Part 8 ppsx
... 17.1 0. 98 3. 98 0.25 1.25 3 .89 14.1
Nitrile rubber
(39% ACN)
5.1 14.0 0.72 3.45 0. 18 1.07 5.60 22.1
Polychloroprene
(CR-G)
2.96 9.97 0 .88 3.50 19.2 55 .8
SBR (23%
styrene)
17.3 41.5 12 .8 34.0 ... processing
parameters for Hytrel, and for a number of other materials in this
chapter, can be found on the producers’ Internet home pages.
* Hytrel is a registered trademark of...
Handbook of Materials for Product Design Part 11 pps
... mathe-
matical design of the tapering cross-sectional profile of the trough
which, in combination with the incline of the top of the trough, assures
that the volume flow of glass over the pipe is uniform ... furnace off to cool overnight.
On an industrial scale, a continuous belt lehr is often employed for
annealing. One of the several suggested heat treatment schedules for
th...
Handbook of Materials for Product Design Part 16 pps
... Method for Determination of Coefficient of Linear Ther-
mal Expansion of Plastics
D3386 Test Method for Determination of Coefficient of Linear Ther-
mal Expansion of Electrical Insulating Materials ... Testing of Materials 13.31
D0790 Test Method for Flexural Properties of Plastics
D0882 Test Method for Tensile Properties of Thin Plastics
D0952 Test Method for Bo...
Handbook of Materials for Product Design Part 18 ppsx
...
common reinforcement configuration 5.49
compacting 5.58
component design phase 5.40
curing 5.57
damage tolerance 5.85
design allowables 5.38
design needs 5.40
drilling 5.74
effects of absorbed ...
Micromechanics 5.33
Midplane symmetry 5.80
MIL -Handbook- 5 5.36
MIL -Handbook- 17B 5.38
MIL-STD-810 5.96
Modulus of elasticity 2.41 2.84
steel 2.86
Modulus of rigidity 2....
Modern Plastics Handbook 2011 Part 7 ppsx
... Books,
Boston, 1 972 , p. 222.
114. Strong, A. B., Plastics, p. 295.
115. Ibid., pp. 2962 97.
116. Petrotheneđ Polyolefins, p. 84.
1 17. Berins, M. L., Plastics Engineering Handbook, p. 1 17.
118. Wire ... Monitor Cavity Filling Pressure,” ANTEC’ 97,
19 97, p. 3659.
172 . Mueller, N., private correspondence, 1999.
173 . C-Mold Design Guide, p. 144.
174 . Belofsky, H., Plastics: Pro...
Plastics Engineered Product Design Part 2 ppsx
... have
22
Plastics Engineered Product Design
on. The chemical structure and nature of plastics have a significant
relationship both
to
properties and
the
ways they can bc processed,
designed, ... to produce openings without side action
movements
f
PARTING
LINE
SIDE
MEW
OF
PART
PARTING
LINE.
32
Plastics Engineered Product Design
Pressure is ofken involv...
Plastics Engineered Product Design Part 9 pps
...
coefficients on the composite optimum design?
as design variables for
the
multilayered-filamentary composite?
composite optimum design?
31
8
Plastics Engineered Product Design
conductive coatings ... appropriate
to
3
14
Plastics Engineered Product Design
Missile in flight includes the use
of
plastics
330
Plastics Engineered Product Design
_u_I...
Plastics Engineered Product Design Part 12 ppsx
... one of the broad product headings gets one into the listings
of
the individual companies themselves, where, if available, one can
448
Plastics Engineered Product Design
-
12.
Eliminate ... processes
Materials
Primary equipment
Mold design
Process controls
Auxiliary equipment
Troubleshooting
Testing
420
Plastics Engineered Product Design
metalworking perm...
Plastics Product Design Part 2 pdf
...
Polytetrafluoroethylene
116
130
175
21
5
26 0
26 0
400
330
24 0
26 6
347
41
9
500
500
755
626
The
T,
is dependent on the processing pressure and the time under
heat, particularly during a slow ... going from the product design
through production. Examples of potential cost
reductions include.
(1)
redesign product with thinner walls
to
reduce
production cos...
Plastics Product Design Part 3 doc
... 2,500 2.160 2.400 2.510 116 104
100
Nylon
6 Glass beads
30
1,250 0.140 0 .32 0 0 .36 8 8 93 391 34 0
5,000
0.290 0.650 0.750 862 38 5 33 3
Reinforcement Strain
(%)
hours
(I@
psi) hours
carbon ... 6/10 Ferrite 83 2,500 0.4 63
0.507
0.568 540 4 93 440
5,000 0. 638 0. 732 0.952
784 6 83 525
Polypropylene Carbon powder 2,500 1.100 1.140 1.970 114 87 63
5,000 6. 230 6.9...
Plastics Product Design Part 4 ppt
...
-
v120,)
E22
1
e,
=
-
6,
G
(2 -42 )
(2 -43 )
(2 -44 )
(2 -45 )
(2 -46 )
Substituting
Eq.
2-35, 2-36,
and
2-37
into
2 -44 , 2 -45 ,
and
2 -46
results in,
e,
=
-
(cos2
6-
v12 sin2 ...
=
E22
1
e,
=
-
(sin
@cos
6)ol
G
(2 -47 )
(2 -48 )
(2 -49 )
Finally, substituting
Eqs.
2 -47 , 2 -48 ,
and
2 -49
and
2 -41 , 2 -42 ,
and
2 -43
into
Eqs.
2-29,2...
Plastics Product Design Part 5 potx
...
b-b 1
0.2
x
5
x
106 0.2 85 0.2 85
x
IO6
1 150
psi
131 5
psi
0.2
x
5
x
106
+
0.1
x
3
x
106
0.2
x
5
x
106
+
0.1
x
3
x
106
0.2 85
0.1 35
0.2 15
0.3 65
0.326
x
IO6 ...
3,100
3,000
2,400
19,300
1,200
1,400
3,100
2 ,50 0
2,800
55
69
61
72
82
62
30
35
25
70
2 .5
12
12
5
20
12
6
a
a
75
75
60
5. 4
60
90
600...
Plastics Product Design Part 7 pps
...
1, 070
1,325
1, 575
1,800
2,500
3,200
1.50
2.50
4.44
6.25
8.06
9. 87
12.56
16.00
20.06
24.50
28 .75
32 .75
44.81
56.50
1.60
2.63
4.62
6.56
8.48
10.35
13.18
16.80
21. 07
25 .73 ...
6800
441
0
640
2910
460
510
1 070
a60
1800
9640
4420
1650
1 570
1060
21 200
59000
11.5
7. 5
1.1
4.9
0.8
0.9
1.5
3.0
16.3
7. 5
2.8
2 .7
1.8
35.9
100.0...
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