Plastics Engineering 3E Episode 3 potx
Plastics Engineering 3E Episode 1 pps
...
45
48
61
66
71
74
81
84
95
95
10 3
11 0
11 6
11 9
12 1
12 1
12 7
13 1
13 4
13 6
13 7
13 8
14 0
14 2
14 2
14 3
14 5
14 5
14 7
14 8
15 0
15 2
15 2
15 4
16 8
16 8
16 8
xiv
Preface to the ...
G
0.97-
1. 34 1. 1 1
-
1. 21
-50 -13 0 -40 -13 0
40D
to
72D 30A
to
45D
43
to62
60
to
70
21 to
45 25
to
45
E
E
E
F
G
P
E
PIE
1....
Plastics Engineering 3E Episode 2 pptx
...
0 .25
0 .2
0 .2
0 .2
0.15
0.14
0 .23
0 .22
0 .24
0. 52
-
-
0 .2
0.5
0 .25
Ob2
0 .2
0 .22
1.18
-
-
-
0 .20
0.15
0 .24
0 .25
0 .25
0.16
0.14
0.17
0 .2
0 .2
0.0 32
0.0 32
90
12
111 ...
30
25
70
45
70
115
62
24 0
75
180
55
185
65
105
84
72
33
70
40
10
32
25
50
14
72
40
70
2. 2
2. 8
2. 6
2. 9
1.7
1.3
3.0
2....
Plastics Engineering 3E Episode 3 potx
... by
bd3
where
I
=
-
VL3
ti=-
3EI
12
ti=-{
FL3
1-ptana
}
3EI
p
+
tana
Insertion force
F= (Lfd)l
{
:-:E:}
(2.21)
For
polypropylene in the situation given, Fig.
2. 23
shows ...
18
mm.
60
Mechanical Behaviour of Plastics
Thickness
=
1.2
mm
Fig.
2.14
Acetal
bottle
cap
given by
3pR2(1
+
u)
-
3
x
0 .37 5
x
( 13) 2(1 .33 )
-
8h2
8(
1
.2)2
o...
Plastics Engineering 3E Episode 4 doc
...
above,
a4
=
10
MN/m2 at
r4
=
300
s
SO,
~(350)
=
0.003
=
0.3%
(iv) and in the same way
~ (45 0)
=
0.0 04
=
0 .4%
The predicted strain variation is shown in Fig. 2 .43 (b). The ... Behaviour
of
Plastics
0.1
0.08
Sp
65
-
B
0.06
0. 04
0.02
0
0
50 100
150
200
(SI
Fig.
2 .46
Variation
of
strain
with
time
Stress
(MN/m*)
0
100
-
Time
(s)
Fig....
Plastics Engineering 3E Episode 5 ppt
... Steel
5
1.2-2
0. 35- 1.6
0.1
-0.3
5- 7
6 .5
3 .5- 6 .5
0. 25- 4
0.4 -5
0.3-0.8
1.3-1.4
0.01
-0.02
100
8
2-4
4
0.9-1.6
0.3-0 .5
5- 7
1
0 .5- 5
3
1-2.6
3-4 .5
0.7-1.1
1-4
0. 75
140 ...
0.014-0.023
0.0 05- 0.008
0.12
0.1 25
0.0 25- 0. 25
0.06
0.02-0 .5
0. 15- 0.2
0.02
0.03-0.13
0.01
0 .5
17
6
0.2-0 .5
0.02-0.06
14
16
5- 100
3.6
0.4-2...
Plastics Engineering 3E Episode 6 pdf
...
-sc
]
Then
1.12.10-~ -6. 3.10-~ -1.87.10-~
-i.87.10-~ 5.89.
io -6
4.35.10-~
s=
[
-6. 3. 2 .65
.
5.89
Directly by matrix manipulation
E,
=
1.1 26.
cy
=
-6. 309
*
yxy
=
-1.878
- ... nylon
66
Weight fraction,
Wf
property
0
0.10 0.20
0.30
0.40
0.50
0 .60
Density
1140 1210 1280 1370 1 460
1570
1700
Tensile strength (GN/m2)
0.07
0.09
0.13 0.18 0.21 0.23...
Plastics Engineering 3E Episode 7 pot
...
=
-2.09
x
10-~
1.49
x
10-~
2.01
10-~
3 .77
x
10-4
5.34
x
10-5
-2.03
x
10-4
1
1.05
x
lo3 235.2 -1 67. 1
239.2 -79 .2
-1 67. 1 -79 .2 233.0
[
D
=
[
235.2
2.15
x
10-3
-1.60 ...
104
14. 37: io4
1.41
io4
A
=
1.41
x
lo4
1. 27
x
104
0
6.65
10-~
-5 .73
10-~
-6.81
-6.81
x
-2.05
x
1.02
x
1
-1.46
x
io3 604.5 -1. 67
x
io3
-16 .7
x
lo3...
Plastics Engineering 3E Episode 8 ppsx
...
TI2
El
=
1.144
E2
=
1.1 28
io?
y12
=
-1. 688
x
10-~
~ 583 ,
?I2
-
188
i2T
-
=
1659,
-
El
E2
Y12
;IT
Thus once again, an applied stress of
188
MN/m2 would cause shear failure ... Westport, CT
(1 980 ).
Folkes, M.J.
Short
Fibre Reinforced Thermoplastics,
Research Studies
Press,
Somerset
(1 982 ).
Mathews,
F.L.
and Rawlings, R.D.
Composite Materials:...
Plastics Engineering 3E Episode 9 pdf
...
lo00
lo00
loo0
loo0
lo00
loo0
825/25/150
8251501125
3001401660
1050
1 390
91 0
1360
92 0
96 0
1050
1070
93 0
30
5
60
1
140
60
5t
1
It
2
0.75
0.25
2
0.5
0.3
1.6
0.4
0.25 ... been considerable
266
Processing
of
Plastics
Controlled
temperature
(1)
Fig. 4. 19( a) Sheet extrusion (1)
thick
sheet
(2)
thin sheet
Fig. 4. 19( b) Pipe extrusion
(1)
r...
Plastics Engineering 3E Episode 10 docx
... volume of the sheet is given
by
Aihj
=
150
x
100
x
2
=
3
x
lo4
mm3
The surface area of the final product
is
Aj
=
(150
x
100 )
+
2 (100
x
60)
+
2(150
x
60)
=
4.5
io4
mm2 ... would leave
a
volume of
(3
x
104 -2.52
x
lo4)
to
form the walls. The
area
of the walls is
A,=(2 ~100 ~60)+(2~150~60)=3~
lO4~*
320
Processing
of
Plastics
CharQino area
-...
Plastics Engineering 3E Episode 11 pdf
...
E.G,
Extrusion
of
Plastics,
Newnes-Butterworth,
1976.
Schenkel,
G,
Plastics Extrusion Technology and Practice,
Iliffe,
1966.
Fisher,
E.G,
Blow
Moulding of Plastics,
Iliffe,
1971. ...
(
dr'
dr
-
dr
))
350
Analysis of polymer melt flow
V=
211
dz
Now at
y
=
0,
V
=
VO
so
1dP H
211
dz
vo
=
(z)
Substituting
aP/az
in the expression for
V
g...
Plastics Engineering 3E Episode 12 ppsx
... check the
value
of
Bi.
In
this case
20
x
1.5
x
10-3
Bi
=
=
0 .12
0.25
Analysis of polymer melt flow
5 .12
Analysis
of Heat Transfer
during
Polymer
Processing
Most polymer ... data on the thermal diffusivity of their
plastics but in the absence
of
any information a value of
1
x
lov7
m2/s
may
be used for most thermoplastics (see Table
1.8
and Table 5.1). .....
Plastics Engineering 3E Episode 13 docx
...
Orientation can
be
introduced into plastics such
as
polyethylene and polypropy-
lene (both semi-crystalline) by cold drawing at room temperature. Other brittle
plastics such as polymethyl methacrylate ... use
of
these rules:
Structure of Plastics
423
Fig.
A. 12
Qpical
illustration
of
spherulites
Twisted
tddcd
chain
growth
from
nucleus
Fig.
A .13
Structure
of...
Plastics Engineering 3E Episode 14 ppsx
...
so
4x
20
1 .142 +1
AR=m{
(1 .142 -1)-0’4}
AR
=
0.29
ITM
(b>
AR
=
a.R.AT
=
143 °C
0.29
loo
x 10-6
x
20
AT
=
Hence the nylon bush would need to
be
cooled by
143 °C
to achieve ...
specimen
fracture
=
0.44
+
3.947
=
4. 414
J
Remaining energy
=
11.036
-
4. 414
=
6.6218
J
=
0.15
m
6.6218
So
height of swing
=
-
44 .145
(2.59)
K
=
p(m~)&a...
Plastics Engineering 3E Episode 15 docx
...
(4.16)
so,
mass 150
x
m3
1
200
volume
=
-
=
P
150
x
10-3
x
109
1 200
XR~H
=
=
1.77
mm
150
x
106
H=
a
x
1200
x
(150 )2
=
19
s
3
x
104
x
(150
x
-
3qv2
Also ...
Calendering, 313, 315
Capillary viscometer 369
Carbon fibres, 28, 39, 168, 181, 189,
Carreau, 352
Celluloid,
2
Cellulose, 1
Centrifugal casting, 337
Charpy impact, 152 , 155 , 1...
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