Plastics Engineering 3E Episode 2 pptx

... 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....

Ngày tải lên: 13/08/2014, 09:20

... 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....

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... 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...

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... 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....

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... 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...

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... -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...

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... = -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...

Ngày tải lên: 13/08/2014, 09:20

... 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:...

Ngày tải lên: 13/08/2014, 09:20

... 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...

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... 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 -...

Ngày tải lên: 13/08/2014, 09:20

... 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...

Ngày tải lên: 13/08/2014, 09:20

... 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). .....

Ngày tải lên: 13/08/2014, 09:20

... 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...

Ngày tải lên: 13/08/2014, 09:20

... 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...

Ngày tải lên: 13/08/2014, 09:20

... (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...

Ngày tải lên: 13/08/2014, 09:20