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 - Cooling ... time Total compression cycle 0 .105 0.140 0 .100 2.230 2.575 Injection moulding Unload piece, opedclose machine Moulding cycle time Total injection cycle 0 .100 1 .goo 2.000 4.4 Thennoforming...

Plastics Engineering 3E Episode 13 docx

... chan- nels have a diameter of 10 mm and they are placed 40 mm from the cavity, then 2 x 40 x 10- ~ x 1.85 x io3 L= 11.5 x 4.2 x n x 10 x x (190 - 40) = 0.65 m ... the bar is given by u= - 1.4 MN/m2 lox 10 bd3 lox 103 I= - ~ = 833.3 IIUII~ 12 12 PcL2 140 x 2252 n21 n2 x 833.3 u 1.4 x 100 E=-= = 0.16% E 861.7 Now For buckling, ... this strain is reached at 104 seconds 104 3600 So time = - = 2.78 hours (Note: E < 0.5% so no correction to tensile creep data is needed). (2 .10) critical stress, P,...

Plastics Engineering 3E Episode 15 docx

... 2.85 10- 3 6.25 x 10- 3 1.28 x 3.68 x 8.5 x 10- 2 0.21 1.31 x 105 1.2 x 105 1.17 x 105 1.02 x 105 8.8 x 104 7.13 x 104 r = 1 x lo4 N/m2 and q = 8 x 104 Ns/m2 ... x 10- 5 0 0 -863.8 -863.8 -271 0 0 0 1.91 x 10- 4 b=[ 0 0 -3.49 x 10- 5 1.91 x 10- 4 -3.49 x 10- 5 0 0 -3.49 x 10- 5 1.91 x 10- 4 -3.49 x 10- 5 0 1 0 0 1.91 x 10& quot; ... seconds 10- 7 6 x 0.74 x 0.333 x 1.33 2.32 x 1.99 x 1.662 = 80.6 mm 100 x 106 3 x 104 For a Newtonian fluid, n = 1 = 101 .9 mm 100 x 106 x 0.74 1.2 x 102 112...

Plastics Engineering 3E Episode 1 pps

... R.J. (Roy J.) Plastics engineering. 3rd ed. 1. Plastics I. Title 668.4 ISBN 0 7506 3764 1 Library of Congress Cataloguing in Publication Data Crawford, R.J. Plastics engineering/ R.J. ... toughness have earned them a reputation as engineering plastics. Table 1.3 compares the relative merits of light metal alloys and nylon. PLASTICS ENGINEERING Contents Preface to the Third ... Edition Chapter 1 - General Properties of Plastics 1.1 Introduction 1.2 Polymeric Materials 1.3 Plastics Available to the Designer 1.3.1 Engineering Plastics 1.3.2 Thermosets 1.3.3 Composites...

Plastics Engineering 3E Episode 2 pptx

... Zinc 104 0 1420 1 410 1180 1280 1190 1200 106 0 1 I40 1380 1300 1400 1360 1630 1700 1400 1150 1200 1270 1370 1420 1340 905 1240 105 0 920 950 2100 1400 1300 108 0 1800 ... a ZY L 10- a 9 m E P 0 a Nylon 1 10 100 lo00 100 00 1OOOOO tl c C02 permeability (cc-25 pdm2-24 h-atm) Fig. 1.14 Permeability data for a range of plastics plastics are ... (polyester) EPOXY 104 0 1420 1 410 1180 1280 1190 1 200 106 0 1140 1380 1300 1400 1360 1630 1690 1400 1150 1270 1370 1420 905 1240 105 0 920 950 2100 1400 1300 108 0 1800 1800...

Plastics Engineering 3E Episode 3 potx

... = (ao - CY,)L. AT = (80 - 1 1 )10- 6(40)(80) = 0.22 mm x 100 = -0.55% 0.22 strain = 40 0.55 x - 100 stress = EE = -1.5 x 10 = -8.3 MN/m2 Thus there will be ... and the steel are 80 x 10- 6"C-' and 11 x 10- 60C-1 respectively. The modulus of the acetal at 100 °C is 1.5 GN/m*. Mechanical Behaviour of Plastics 71 The stiffness ... Example 2 .10 Mechanical Behaviour of Plastics 57 For some plastics, particularly nylon, the moisture content can have a signifi- cant effect on the creep behaviour. For such plastics, creep...

Plastics Engineering 3E Episode 4 doc

... from T1 to T2. 17.4(60 + 10) 51.6 + (60 + 10) - - 17.4(20 + 10) - 51.6 + (20 + 10) = -3.62 UT = 2.4 x 10- ~ Mechanical Behaviour of Plastics 123 F ttt 1-7 Fo7 ... (2.87) 102 Mechanical 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 - ... for 100 hours and removed for 100 hours then in equation (2.64), T = 100 hours and t, = 200 hours. Therefore after four cycles i.e. t = 800 hours. x=4 ~~(800) = ~~ (100 )...

Plastics Engineering 3E Episode 5 ppt

... bursting pressure, PB, as 1.0(8 x 10- 3)(32 x PB = = 5.6 MN/m2 3.0qi2 x 10- 3)2(Ir x 3.5 x 10- 3)1/2 2.19 General Fracture Behaviour of Plastics If the defect or crack in ... 1.93 (G) 'I2 - 3.07 ( $)312 + 14.53 ( $)*I2 (2 .105 ) -25.11 (G)'l2+25.8 ($)9/2} or (2 .104 ) (2 .106 ) (2 .107 ) Thus the basis of the LEFM design approach is that (a) ... in a sheet of finite width (b) Edge cracks in a plate of finite width (2 .101 ) (2 .102 ) Mechanical Behaviour of Plastics related to the stress intensity factor K by the following equation...

Plastics Engineering 3E Episode 6 pdf

... IO3 3 x lo3 5 x IO3 7 x lo3 10 x 104 15 x 104 Strain o 3.1 x 5.2 x 8.9 x 10- 3 9.75 x 10- ~ 9.94 x 9.99 x 10- ~ 9.99 x 10- ~ Use this information to predict the ... 2 s2 sc -2sc 2sc (2 -2) Tc = [ s2 c2 -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, ... of the cross-section. 2 .10 Show that a ratio of depth to thickness equal to 10 is the nod limit if buckling is to be avoided during short-term loading of plastics. What is likely to...

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Plastics Engineering 3E Episode 7 pot

... 7.28 x 104 3.44 x 104 4.17 x lo4 4.17 x 104 1.98 x lo4 3.42 x 104 7.28 x 104 3.44 x lo4 4.17 x 104 4.17 x 104 1.98 x 104 3.42 x 104 1 3.44~ 104 2.80~ 104 1.98 ... 2.80~ 104 1.98 x 104 3.44~ 104 2.80~ 104 1.98~ 104 1 1 7.28 x 104 3.44 x 104 -4.17 x 104 Q4= [ 3.44 x 104 2.80 x 104 -1.98 x 104 -4.17 x 104 -1.98 x lo4 3.42 ... Behaviour of Composites 1 2.30 x 10- ~ -2.24 x 10- ~ 9.55 10- ~ p = -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 -167.1...

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