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...

Ngày tải lên: 13/08/2014, 09:20

... projected area is (~/4)(90)~ = 63 60 mm2 = 6. 36 x m2. Hence, clamp force per cavity = 120 x 6. 36 x The projected area of the runners is 4 x 40 x 6 = 960 mm2. Assuming that the ... loo0 825/25/150 8251501125 300140 166 0 1050 1390 910 1 360 920 960 1050 1070 930 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 ‘EVOH...

Ngày tải lên: 13/08/2014, 09:20

... E.G, Extrusion of Plastics, Newnes-Butterworth, 19 76. Schenkel, G, Plastics Extrusion Technology and Practice, Iliffe, 1 966 . Fisher, E.G, Blow Moulding of Plastics, Iliffe, 1971. ... often presented graphically as shown in Fig. 5.11. 2 .6 2.4 2.2 0 c E2 UJ C I 1.8 a, 3 v, 1 .6 1.4 1.2 1 0 1 2 3 4 5 6 7 8 Recoverable shear strain Fig....

Ngày tải lên: 13/08/2014, 09:20

... here. The reader is referred to standard textbooks on linear optimal control, such as [1], [11], [ 16] , and [25]. 50 2 Optimal Control Theorem D If the control u o :[t a ,t b ] → Ω is optimal (in ... provided λ o (0) > 0. Using the optimal open-loop control law u o (t)=−bλ o (t)=−be −at λ o (0) , 46 2 Optimal Control 2) If the optimal control u o (t)attimet lies in the interior of the c...

Ngày tải lên: 12/08/2014, 16:21

... R.J. (Roy J.) Plastics engineering. 3rd ed. 1. Plastics I. Title 66 8.4 ISBN 0 75 06 3 764 1 Library of Congress Cataloguing in Publication Data Crawford, R.J. Plastics engineering/ R.J. ... 202 2 06 208 218 223 2 26 232 232 234 2 36 238 240 245 245 2 46 2 46 25 1 252 257 259 262 264 278 278 279 285 297 298 299 30 1 302 18 General...

Ngày tải lên: 13/08/2014, 09:20

... 1 I .5 0.01 1 3.51 7.95 4.0 6. 78 13.15 17 .6 21 .6 8.34 5 .68 6. 73 10.1 3.12 1.53 3.0 I .47 1.73 0.73 2.34 1.07 6. 67 4 .62 3. 46 1.82 9. 16 3.82 0.77 0.81 1.05 1.02 1.23 1.09 ... 2 .6 2.5 1.5 2 .6 3.0 0.2 1.2 0.5 3.0 0.007 3 .6 9.0 11.0 8 40 65 2 30 60 3 70 60 4 4 1 .6 70 3 0.8 12 100 60 60 8 150 80 400 150 200 80...

Ngày tải lên: 13/08/2014, 09:20

... month (2 .6 x lo6 seconds) at this stress, the strain is obtained from Fig. 2.15 as 2.2%. Hence E = 22 x 1 06/ 0.022 = 1 GN/m2. pp 0.375(13)412(1 - 0.332) 6= 64 0 - 64 x ... dominates. 14 12 3 lo t E= E 6 4 2 0 1 2 3 4 5 6 7 a 9 10 Dh Fig. 2.32 Optimisation of Corrugation depth 66 Mechanical Behaviour of Plastics As a practi...

Ngày tải lên: 13/08/2014, 09:20

... deflection, 86. This is shown as line (ii) in Fig. 2 .63 (b). The elastic stored energy would then be U2 = i(F + aF) (6 + 36) (2. 86) From equations (2.85) and (2. 86) the change ... T2(= 60 °C) and the shift factor at TI(= 20°C) and then subtract these to get the shift factor from T1 to T2. 17.4 (60 + 10) 51 .6 + (60 + 10) - - 17.4(20 + 10) -...

Ngày tải lên: 13/08/2014, 09:20

... cyclic loading 1 56 Mechanical Behaviour of Plastics Table 2.3 Charpy calibration factor (0) alD 0. 06 0.10 0.20 0.30 0.40 0.50 0 .60 0 Values S/D=4 S/D =6 S/D=8 1.183 1.715 ... polyester LDPE MDPEMDPE Nylon 66 Polycarbonate Polypropylene copolymer Polystyrene UPVC Glass Mild 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...

Ngày tải lên: 13/08/2014, 09:20

... all a12 a 16 bll b12 b 16 a61 a62 a 66 b61 b62 b 66 811 812 8 16 dll d12 d 16 a21 a22 a 26 b21 b22 b 26 821 822 86 d21 d22 d 26 861 862 866 d61 d62 d 66 where N = ... 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 60 4.5 -1 .67 x io3 - 16. 7 x lo3 -792 60 4.4 B=...

Ngày tải lên: 13/08/2014, 09:20