Plastics Engineering 3E Episode 15 docx

474 Then using Solutions to Questions (b) This laminate is unsymmetrical. In the above analysis, D2 and D3 are reversed and this has the effect of making [BIZ 0. The matrix values are. 0 1.18 x 104 [ 3.51 ; 10'' 1.19 x 10'' A = 1.19 x 10'' 8.88 x lo3 a = -7.15 x 2.07 x 0 B=( 0 0 -271 ) 9.74 x 10-5 [ 5.75 ; 10-5 -7.15 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" 467.9 159.3 D= (159.3 118.5 : ) d = -5.36 x 0.016 0 0 0 1.58.2 7.30 x 10-3 [ 4.31 ; 10-3 -5.36 x 10-3 Using the above equation relating strains and curvatures to N and M we get Ex = 2.3 x = -2.86 x yxy = o K~ = 0, K~ = 0 and K~~ = 7.67 x mm-' The difference in this case therefore is that we start to see some twisting of the laminate due to the non-symmetric nature of the lay-up. Solutions to Questions 415 (3.17) This is a non-symmetric laminate. The Q matrices are given in the text and the A, B and D matrices are determined from 3 A = CQf. (hf - hf-1) f=1 -Ex- - a11 a12 ai6 bii b12 bib- - N, - CY a21 a22 a26 b21 bn b26 Ny Nxy YXY - ab1 a62 a66 bbi b6z bM Kx ~II b12 bla dii d12 d16 M, where Nx = 30 N/mm - Ky hl a2 hb d21 d22 d2b My -Kxy- -bbl b62 BM d61 db2 da- -Mxy- 476 Solutions to Questions 1 -2.68 10-3 236.0 47.7 -29.3 D= ( 47.7 50.3 -13.9 -29.3 -13.9 47.2 8.37 x lob3 -8.00 x 4.65 x 4.65 x lob3 -2.68 x low3 0.036 -8.00 x 0.031 (3.18) (a) Maximum Stress Criterion The local stresses are given by 62T 312 - = 20.1, 01 a2 r12 - = 0.52, - = 1.36 61 T As the second term is less than 1, a tensile failure in the 2-direction would be expected. (b) Maximum Strain Criterion From the data given 5 = 2.49 x w3, 22T = 3.777 x io3, = 5.84 x 10- gZc = 5.66 p12 = 7.14 10-~ The 1-2 strains are given by "' - 1.36 IT g2T - = 24.9, - = 0.5, - - -91 -92 nz Hence once again a failure is predicted in the 2-direction. (c) 'hi-Hill Criterion The Tsai-Hill equation gives Therefore failure is predicted. (3.19) (i) For an applied stress of a, = 1 MN/m2, the stresses in the 1-2 direction are given by GtT &T 212 01 02 r12 - = 1867, - = 200, - = 92.4 Hence an applied stress of a, = 92 MN/m2 would initiate a shear failure before tensile or compressive failures occurred. Solutions to Questions 477 (ii) Maximum Strain Criterion From the data given $IT 32T ilT = - = 0.035, i2T = - = 5.56 x El E2 $IC 32c El E2 ?12 y12 = - = 0.01 GI2 dlc = - = 5.5 iZc = - = 0.011 Also, [I] = s [ so [ 3 = [ EJ] 10-5 ?12 - = 92.4 ;IT i2T - = 2082, E1 E2 YlZ - = 253, Hence once again a stress of a, = 92 MN/m2 is the limiting condition. (iii) Tsai-Hill Criterion Letting X be the multiplier for a, ( xz)2 - (x+)? + ( z)2 + ( F)2 = 1 Solving for X gives X = 83.8. Hence the limiting condition is a, = 83 MN/m2. (3.20) Let N, = 1.2 N/mm (ie a, = 1 N/mm2). By calculating the matrix [a] we get (for 30" ply). 1.235 [ F] =a [ = [ -102483 x 10-5 Then [ 3 = T, [ 3 = [ 3&] x Ek ?12 - ~2379, - - - 3624, - =606 E1 E2 Y12 ZIT Hence the limiting condition is a, = 606 MN/m2. Also, [:]=Q,["] and [::]=To[:] TXY YXY r12 TXY 478 which gives Solutions to Questions 0.787 [ = [ -O.CW] MN/m2 TI2 -0.148 Hence the Maximum Stress Criterion also predicts 6, = 606 MN/m2 as the limiting condition. For the Tsai-Hill Criterion, letting X = multiplier on stress a,, then solving gives X = 583. This predicts a stress of ax = 583 MN/m2. If this is repeated for all the plies then we get Table 1 Maximum permissible values of a,, to cause failure in each PlY MaKstressorMax. 'l$ai-HIu Strain e = 0" a, = 1201 MN/mz a, = 1067 9 = -30" a, = 606 h4N/mz a, = 583 e = 30" a, = 606 m/m2 ax = 583 The limiting condition is thus ax = 583 MN/m2 as predicted by Tsai-Hill. (3.21) K, = a(~a,)'/~ (E tan (s)) 'I2 '/2 = %rac)'/2 Wb (E tan (g)) P 112 43 = - (Z x 5 x (1.05) 30 x 5 P=49kN 13.5 1 (3.22) CSM: K, = a(n~,)'/~ -+ a, = (=) - = 9.06 x 10-3 m a 2((9.06 x 10-3)-5.35 - (lo00 x 10-6)-5.35) N= 3.3 x 10-'8(80)12.7fi.3s(-10.7) = 3 x 16 cycles Solutions to Questions 479 Woven Roving: a, = 34.9 x N = 1.14 x lo6 cycles so in a similar way (4.1) QD = 1/2n2D2NH sin4cos4 = 1/2n2(50 x 10-3)2(100/60)(2.4 x 10-3)sin 17.7~0s 17.7 = 1.43 x m3/s nDH3 sin2 4 (f) 1% QP = n (50 x (2.4 x 10-3)3 sin2 17.7 - 12(200) = 1.67 x m3/s Total flow = QD - Qp = 1.26 x m3/s The shear rate in the metering zone will be given by . y=-= - vd ZDN COS4 - TT (50 X w3) (1OO/60)COS 17.7 H H 2.4 x 10-3 - KP (4.2) For the die, Q = - - ZrPR4 8Lv for the extruder. 6n x 200 x x lo00 x 50 x 100 = 171 h4N/m2 60 x (2.4)2 x 0.3191 Pmax = Qmax = 1/2n2D2NH sin4cos4 = 1.436 x m3/s From graph, operating point is P = 43 MN/m2, Q = 1.075 m3/s New operating point is P = 85 MN/m2, Q = 1.075 m3/s (4.3) Extruder: Qmax = 1/2n2D2NH sin 4cos 4 = 1/2n2(25)' (2) (2)sin 17'42'cos 17"42' = 2.98 x m3/s 480 Solutions to Questions Die: 6~(25)(500) (g) (400 x lod6) 22 x tan 17’42’ - - = 123 MN/mz Q=2{$}P Hence the characteristics shown may be drawn. The operating point is at the intersection of the two characteristics. Output, Q = 2.13 x m3/s (4.4) but Q = 1/2lsdNH sin qi cos qi - nDH3 sin’ qi - (13 Solutions to Questions 48 1 so 0 nDH3 sin’# Qq Q = 1/2dNH sin4cos4 - 12q KIL AH sin 4 cos 4 1 +BH3sin2q5 nD A = 1/2x’D2N and B = - 12LKl where so =O dQ _- (1 +BH3sin’4)AH(2cos24- 1) -AHsin~cosQ,(2H3Bsin~cos~) d4 - (1 + BH3 sin2 4) (1 + BH3 sin’ 4)(2 cos’ 4 - 1) = 2H3B sin’ COS' 4 (1) dQ dH (1 + BH’ sin’ 4)A sin 4 cos 4 - AH sin 4 cos 4(3BHz sin’ 4) (1 + BH3 sin2 4) Also - - - =O so Using (2) in (1) (1 + BH3 sin’ 4) = 3BH3 sin2 4 (2) (3BH3 sinZt$)(2cos2(b - 1) = W3Bsin24cos’4 (3BH3 sin2 COS 24) = W3B sin’ COS' Q, 2cos24 = 2cosz4 -+ 3(1 - 2sin24) = 2(1 - sin24) thus 482 which can be solved to give 4 = 30" Also (4.5) (4.6) (4.7) (9 [l + BH3 sin24] = 3BH3 sin2 4 Solutions to Questions F = nR2P0 (L) = 1r(0.25)~(50 x lo6) m+l = 2.26 MN AB =loom CBA 150 3 3 How ratio = - = - = 50 A Clamp force =lrR2Po 100 (ii) (iii) = 1.06 MN 1/41rd -0.250 c.s.a. s.a. IrD DC = Dtan lo", BC = D/cos 10" C.S.U. = d + D2 tan 10" = D2(1 + tan 10") s.a. = 20 + 2D/ cos 10 + 20 tan 10" c.s.a. oz(1 + tan 10") - s.a. 2D(1 + (cos + tan lo") = 0.268 D FG = (D/4)tan3Oo c.s.a. = d(l.5)/4tan30° s.a. = 6(D/2) = 30 - 0.216 D c.s.a. _ s.a. 8tan30 Solutions to Questions 483 (4.8) but c.s.a. = d/2 s,a. = 20 + D = 3 D c.s.a. OZ s.a. 60 = 0.167 D _- 2n x 0.15 x 5 60 = 0.078 m/s Vd = 275m = Q = 2000 kghour = 0.3% x m3/s = 5.04 mm 0.396 x Q = HWVd SO H = 1 x 0.078 x = d(5.04 - 4.5)150 = 9 mm Now w = 1.786 (ie x/H) so P, = (3.57 - 0.64(1.786 + 5.77(0.33))) 3 x 1.5 x 104 x 0.078 4.5 x 10-3 = 0.93 MN/m2 (4.9) For Ho = 0.8, x = Jm = 10.9 mm, w = 5.45 P,, = 24.3 MN/m2 (same equation as above) For For For H~ = 1.2 mm, P, = 4.9 m/m2 Ho = 1.6 mm, P, = 0.86 MN/mZ Ho = 1.9 mm, P, = 0.073 MN/mz (4.10) Initial volume of sheet = ($)4=750Z Since the dome formed is a free surface it may be assumed to have a constant thick- ness, t Final volume = (T) t t = 7roz -b t = 2 mm nOZ 2 - (4.1 1) CY = tan-' ($) = 80.54' L = 35 um, to = 3 mm H = 3Otanb = 180 = 0.58 mm a- 1 rb [...]... (0.513 10-9 (s) (s) (z) = 14.4 MN/m2 = exp (4 .15) = exp where a = k/pC, = thermal diffusivity and H = wall thickness = 1/B -28.4 x 8.6 x lo-% 300 - 250 = exp 300 - 23 3 x 10-3 x 230 t = 484.4 s = 8.07 minutes mass 150 x volume = -= 1200 P (4.16) X R ~ H= so, H= Also 3qv2 8nFH4 t =- m3 150 x 10-3 x 109 1200 150 x 106 = 1.77 mm a x 1200 x (150 )2 3 x 104 x (150 x = 19 s 8a x 106 x 100 x 103 x 1.774 x (1200)2... 424 Oxidation, 26, 27 Parison, 268 Parkesine, 2 PBT, 8, 11 15 PEEK, 7, 8, 15, 27, 28, 174, 177, 181, 189 328 Permeability, 35 Perspex, 2 Index 504 PET, 8, 11, 15, 269, 283 308 Phase lag, 110 Phenol formaldehyde, 17,326 Phenolics, 8, 17 Pigments, 3 Plane strain, 126, 132, 154 ,428 Plane stress, 130, 132, 154 , 427 Plastic zone, 133 Plasticisers, 3 Plastics, 1, 3 Plate constitutive equations, 197, 210 Plug... Breaker plate, 250 Brittle behaviour, 135 Buckling of ribs, 79 Bulk modulus, 57 Calcium stearate, 3 Calendering, 313, 315 Capillary viscometer 369 Carbon fibres, 28, 39, 168, 181, 189, 192, 228, 233, 328 Carreau, 352 Celluloid, 2 Cellulose, 1 Centrifugal casting, 337 Charpy impact, 152 , 155 , 157 Chemical resistance, 5 Chopped fibres, 329 Chopped strand mat, 329 Clamping forces, 293, 326, 401 Clamping systems,... 6 N/m2 PH T=-+P= 2L 3.2 x 1 6 x 2 x 30 = 19.2 MNlm' 1 m3/s 489 Solutions to Questions (5.8) Flow rate = 3 d m i n Area = 2 ( (15 Q= + 15) + n(8.3)) 3 x 112 x 10-6 60 p = - -6Q T H ~ (30 - m2 = 112 x = 5.6 10-6m3/s 6 x 56 x + 26) x 22 x 10-9 = 150 S-' from flow curves, at 9 = 150 s-l, r = 1.5 x 2Lr AP=-= H 105 N/m2 2 x 10 x 1.5 x 105 = 1.5 MN/m2 2 Power = P Q Power = 1.5 x lo6 x 5.6 x = 8.4 Nmls (=... Polystyrene, 2, 4, 15, 34, 59, 130, 149, 283, 295.418 Polyurethane, 16, 17, 26 Polyvinyl chloride, 3,4,7, 15, 33,44,59, 135, 149, 269, 283, 295, 308,417 Positive forming, 307 Power law, 293, 351, 396, 399,403 Pressure bag moulding, 33 1 Pressure flow, 254, 234 Pressure forming, 308 Profile production, 264 Pseudo-elastic design, 53 PTFE, 1, 15, 28, 32, 33, 417 Pultrusion, 330, 337 PVC, 3, 4, 7, 15 33, 44, 59... 173, 391, 393 Thermal diffisivity, 3 1, 391 Thermal strains, 61 Thermal stresses, 61 Thermofonning, 306, 309 Thermoplasticrubber, 10, 16 Thermoplastics, 3 Thermosetting plastics, 5 , 7, 17, 170, 304, 328 505 Tie-temperature superposition, 116 Toughness, 26, 154 , 155 , 283 Tow, 328 TPR, 10 Tracking, 32 Transfer moulding, 326 Transverse modulus, 179 Triaxial stresses, 149,427 Tsai-Hill criterion, 233 Twin... 6 x 15 x = 120 s-L 0.75 x 10-3 y=-= H From the flow curves, t = 1.3 x Id N/mz G = 3 x 104 N/m2 So yR = r/G = 4.4 and SO BH = 1.85,BT = 1.36 New H = New V = 15 x 0.7s x 10-3 = 0.41 x 1.85 x 1.85 x 1.36 = 37.7 x 6 x 37.7 x 0.41 x 10-3 so, y= T h i s gives T = 2.2 x so, yR = 5.5 m m/s = 552 s-' 105 N/mZ, = 4 x 104 N/m2 G and BH = 2.08, = 1 4 5 BT 5 0.75 x 1-3 = 0.36 x lod3m 2.08 = 45.39 x New V = 15. .. Polyarylether, 7 Polycarbonate, 4, 9, 10, 16, 26, 34, 283, 295, 308 Polyester, 5, 15, 17 Polyether sulphone, 16, 28, 59 Polyetheretherketone, 7, 15 Polyetherimide, 7 Polyethernitrile, 7 Polyethylene, 3, 4, 7, 12, 32, 269, 283, 295, 392,416 Polyimide, 7 Polymerisation, 2, 3 Polymers, 1, 3 Polymethylene methacrylate, 4, 15, 59, 417 Polyphenylene, 5 Polyphenylene oxide (PPO), 8, 9, 16 Polyphenylene sulphide,... 0.95,1. 415, 2.03and 1 2 , 8 Also for q/qR A ' = 1, AAT -BAP - = - - 28'32 - 2 9 MN/m2"C 7 AT 9.54 Similarly for the others, ( A P ) / ( A T )= 1.17,1.87,3.07 1 9 and 3 For acrylic Index ABS, 3, 11, 16, 20, 26, 33, 131, 135, 149, 283, 295, 308 Acetal, 4, 14, 59, 61, 76, 149, 418 Acrylic, 4, 15, 33, 34, 59, 90,124, 136, 149, 283, 308 Activation energy, 136 Alloys, 11 Aminos, 17 Amorphous, 4, 15, 137,... 2) = 112 x Q= 3 x 112 x 10-6 60 y = - -6Q TH2 - m2 = 5.6 = lob6 m3/s 6 x 56 x (30+26) x 22 x = 150 S-' from flow curves, at 9 = 150 s-', r = 1.5 x 10s N/m2 2Lr 2 x 10 x 1.5 x 1 6 AP= = 1.5 MN/mZ H 2 Power = P Q = 1.5 x 106 x 5.6 x = 8.4 N d s = 8.4 W 490 Solutions to Questions From flow curves, at 9 = 150 G = 3.05 x 104 N/m2 YR = T/G = 4.92 SO, From graphs for swell ratio BST = 1.41, BSH = 1.99 Thus . (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. (5.8) Flow rate = 3 dmin 489 Area = 2 ( (15 + 15) + n(8.3)) = 112 x m2 3 x 112 x 10-6 60 = 5.6 10-6m3/s Q= p= 6Q = 150 S-' 6 x 5.6 x TH~ - (30 + 26). _- 2n x 0 .15 x 5 60 = 0.078 m/s Vd = 275m = Q = 2000 kghour = 0.3% x m3/s = 5.04 mm 0.396 x Q = HWVd SO H = 1 x 0.078 x = d(5.04 - 4.5 )150 = 9 mm Now