Plastics Engineering 3E Episode 14 ppsx

Solutions to Questions 439 From 1 year isochronous at 0.5% strain, E = 370 MN/m2 so P, = = 227 N (2.12) h= - From 1 year isochronous curve at a stress of 6 MN/mz, E = 6/0.018 = 333.3 MN/m2 so s= =7mm (2.13) The maximum stress or strain is not specified so an iterative approach is needed. k2 x 370 x IC x 13S4 64 x l5P 3(3 + u)PR2 3 x 3.4 x 0.04 x 752 = 6.91 mm - 8u 8x6 3 x 0.6 x 5.4 x 0.04 x 754 16 x 333.3 x 6.912 From the 1 year isochronous for PP the initial modulus is 370 MN/m2 16 x 370 x 2S3 x 4 3(1 - u)(5 + v)P - 3 x 0.6 x 5.4 x 324 3(3 + v)PR2 - = 0.036 MN/m2 16Eh3S P= 3 x 3.4 x 0.0363 x 322 - - = 7.58 MN/m2 8h2 8 x 2S2 so U= but at so 7.58 0.0248 u = 7.58 MN/m2, E = - - - 305.6 MN/m2 16 x 2S3 x 4 3 x 0.6 x 5.4 x 324 P= (305.6) = 3 x MN/m2 3 x 3.4 x 322 8 x 2.52 U= (3 x lo-’) = 6.27 MN/m2 - 319.6 MN/m2 + P = 3.14 x low2 -+ u = 6.55 MN/m2 6.27 0.0196 but E = - - E = 321.2 MN/m2 + P = 3.15 x E = 321.2 MN/m2 + P = 3.15 x low2 MN/m2 MN/m2 + (I = 6.58 MN/m2 AD Pr Pr2 &e = - = -(2 - u), So AD = -(2 - v) D 2hE hE so (2.14) 3.15 x x 322 2.5 x 321.2 AD = (1.6) = 6.43 x mm bd3 12 x d3 12 12 I=-= - d3 From the 1 year isochronous curve, the initial modulus = 370 MN/m2 5WL3 384E6 5 x 150 x ZOO3 384 x 370 x 6 NOW d3= - - - So d = 19.16 mm, where W = wL My WL 150x200 = 5.1 MN/m2 c=-= I 16d2 - 16dz 440 Solutions to Questions At (I = 5.1 MN/mz + E = 342.3 MN/m2 + d = 19.67 mm (I = 4.85 MN/m2 -+ E = 346.2 MN/m2 + d = 19.6 mm (I = 4.88 MN/m2 -+ E = 346.4 MN/m2 + d = 19.6 mm (2.15) Once again an iterative type solution is required. = 0.0795E 1.5 ) W= {0.48(?) (;)“”> 2.4W 32 (I= - 0.267 From the 1 year isochronous curve, the initial modulus is 370 MN/m2 so W = 29.43 N + (I = 7.86 MN/m2 + E = 302.3 MN/m2 W = 24.03 N -+ 0 = 6.42 MN/m2 -+ E = 324.1 MN/m2 W = 25.76 N + (I = 6.88 MN/mz -+ E = 317.0 MN/m2 W = 25.2 N + (I = 6.73 MN/m2 -+ E = 317.4 MN/m2 W = 25.2 N -+ (I = 6.73 MN/m2 So W = 25.2 N (2.16) This is a stress relaxation problem but the isometric curves may be used. From 2% isometric, after 10 seconds, E = - 16*75 - - 837.5 MN/m2 0.02 =3N 837.5 x 34 x 10 128(1 + w)R3N - 128(1.4)53 x 10 - Ed4S W= Afterlweek, E= 8S5 - 427.5 MN/m2 0.02 so 427.5 x 34 x 10 128(1.4)53 x 10 W= = 1.55 N (2.17) From the 1 day isochronous curve, the maximum stress at which the material is linear is 4 MN/m2. This may be converted to an equivalent shear stress by the relation - 1.43 MN/m2 (I r= 2(1 + u) - 2(1.4) - Now If W = 4.5 N, 16 x 3 x 7.5 SO d3 = 16WR r=- ad3 ’ a x 1.43 d = 4.31 IYNII 16 x 4.5 x 7.5 ~r(4.31)~ T= = 2.15 MN/m2 Solutions to Questions 441 Equivalent tensile stress = u = 2.15 x 2 x 1.4 = 6 MN/m2. From the 1 day isochronous at this stress, E= ‘ -638.3 MN/m2 0.0094 128(1+ v)R3N K Ed4 E -_ - s= so KK where E2 = 638.3 MN/m2 and 1.4 0.002 = - = 700 MN/m2 So % change = +9.67% (2.18) At short times, E= ’*‘ - 1400MN/m2 0.004 Allowing for temperature, Ea = 1400 x 0.44 = 616 MN/m2 Overall strain, E = ET - E, = 0 ET = E, aAT = u/E = P/AE P = AEaAT = - n(10)2 x 616 x 1.35 x 4 x 40 = 261 N so (Note: u = (261 x 4)/~(10)~ = 3.3 MN/m2. Therefore since the strain is less than 0.5% no correction to the tensile data is needed for compressive loading). After 1 year Em = 370 MN/m2, Ea = 222 MN/m2 so 4 (Note: Once again the strain is less than 0.5% so no correction is needed for compressive loading). (2.19) = 0.417% AD 0.05 E@ = - =- D 12 From the 0.417% isometric curve, the stress after 1 year is 1.45 MN/m2 exceeded the stress in the pipe wall after 1 year. The pipe would leak if the hoop stress caused by atmospheric pressure (0.1 MN/m2) For PR 0.1 x 6 h 1.5 P = 0.1 MN/m2, a@ = - = - - - 0.4 MNlm2 Hence, leakage would not occur. 442 Solutions to Questions (2.20) This is a stress relaxation problem, but the question states that creep data may be used Then from a 1.6% isometric taken from the creep curves it may be determined that the stress after 10 seconds is 15.1 MN/m2 and after 1 year it is 5.4 MN/m2. (a) Initial pressure at interface = p = (ha)/R = 3.02 MN/m2 Thus normal force at interface, F = p x 2xRL = 3.02 x 2x x 5 x 15 = 1.423 kN So axial force, W = pF = 0.3 x 1.423 kN = 0.427 kN (b) Similarly, after 1 year for u = 5.4 MN/m2 the axial force W = 0.153 kN. (2.21) (a) As illustrated in Example 2.7, the extraction force, F, is given by F = xppDW so 1.2 103 = 4 m/m2 = ~(0.24)(40)(10) Treating the bush as a thick wall cylinder subjected to this value of external pressure, then Benham et al. show that at the outer surface of the bush, the stresses are: radial stress, a, = -p Solutions to Questions Hence, the hoop strain, &e, is given by 443 AR ~a, &e = - = - - - REE The negative sign simply means that this is a reduction. 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 the necessary contraction to have easy assembly. This suggests a cooled tempem- of -123°C. (c) At the bore of the bush, Benham et al. shows the stresses to be 2k2p a@ = k2 - 1’ a, = 0 Ar =- a0 vu, 2k2 P _- - r E E E(k2 - 1) hence -2( 1.14)*17.5(4) 2 x 103(1.142 - 1) A, = = -0.3 ITM Therefore the internal diameter of the bush will be 35 - 2(0.3) = 34.4 mm. (d) If the long-term modulus of the nylon is 1 GN/m2 then for the same interference conditions, the interface pressure would be reduced to half its initial value ie 2 MN/m2. This means that the separation force would be half the design value ie 600 N. (2.22) If the acetal ring is considered as a thick wall cylinder, then at the inner surface there will be hoop stresses and radial stresses if it is constrained in a uniform manner: hoop stress, ae = p { z} radial stress, a, = -p where p is the effective internal pressure k is the ratio of the outer to inner radii Hence, the hoop strain, E@, at the inner surface is given by Ar ae a, &e = - = - - v- rEE Solutions to Questions 444 so, =${(=)+"} k2 + 1 When the bobbin is cooled by 60°C the effective constrained contraction of the inner surface of the acetal will be Ar = (a,, - as)rAT Ar = (80 - 11)10-6(10)(60) = 0.0414 mm so, 0.0414 = - P(W { (-) 1.52+ 1 +o.33 } 3 x IO3 1.5'- 1 p = 4.24 MN/m' hence, uo=p{m}=4.24(~) k2 + 1 =11MN/m2 (2.23) Plot log i vs. loga and straight line confirms the Power Law with A = 3 x lo-", II = 0.774 EI - EO = it = Ad"' so = 0.95 x lo-* + (3 x 10-'1(5)0.774)(9 x lo6 - 1 x IO6) = 1.033% (2.24) A plot of log E against log t is a straight line for u = 5.6 MN/m2 So E(t) = At" where A = 0.238 and N = 0.114. After 3 days (= 2.59 x Id seconds) strain = 0.988% ET 0.988 E(T) - ~lp(t) - 0.988 - ER(~) F, = - (2 x 2.59 - 2.59) x 16 t, = =1 2.59 x l@ So from (5) Fr = 1 + ti - (tR + 1)" 0.988((2)0."4 - 1) = Er(t) = 0.0813% Alternatively: Strain after 2 x 2.59 x 16 seconds = 1.069%. Now recovery may be regarded as reversal of creep, So residual strain = 1.069 - 0.988 = 0.0812% 15 x 153 10.5 x 10.53 10.5 450 + - - (10.5Q = 3454.1 mm4 12 12 0 909 (2.25) I = ___ - 12 04 12 I = - = 3454.1, D = 14.27 IIUII For solid beam, Solutions to Questions 445 Weight of solid beam = 14.27’ x loe6 x 909 = 185.1 g Weight of foamed beam = (4 x 2.25 x 12.75 x x 909) + (450 x 10.5’ x = 153.9 g 96 Saving = 16.9% My WL2 x7.5 I 24 x 3126.3 7 x 3126.3 x 24 2502 x 7.5 From the 1 week isochronous, E = - = 466.7 MN/m’ (2.26) u=-= W= = 1.12 kN/mm 7 0.015 WL4 1.12 x 2504 6= = 7.8 mm 384EI - 384 x 466.7 x 3126.3 (2.27) Weight of solid beam = 909 x 12 x 8 x 300 x Weight of composite = (909 x 2 x 2 x 12 x 300 x 10-6)+(500 x h12 x 300 x h = 7.27 nun, so composite beam depth = 11.27 mm. The ratio of stifhesses will be equal to the ratio of second moment of area x lo3 = 26.18 g bd3 12 x 83 12 12 I. = - 512 mm4 solid - 12 x 11.273 12 x 7.273 - 12 12 Icomposite = (g)’ 12 x 7.273 12 = 1163.45 + 1163.45 Ic/Is = ___ - - 2.27 512 (2.28) (a) Solid Consider a flexural loading situation as above, so W 48EI s L3 cx EI _- 446 So Solid (b) Foamed Solutions to Questions Weight = p, x H x 1 x 1 = p,H = 12p, ES Ratio = -(12) PS EI = Ef (g) = (E) E, ( T) 1 x 123 12 1.5 Weight = pf x H x 1 x 1 = 12pf = -p, Ratio = E, - ( T) 12 x 1.5 = (8)- ES Ps PS (c) Composite b’ = (2) (I) Weight = ps(4 + (8 x 0.44)) x 1 x 1 = 7.56~~ 120.3ES E, Ratio = - - - (15.9)- 7.56~~ PS So ratio foam: solid: composite = 8:12:15.9 (2.29) (a) Sandwich Beam of Minimum Weight for a given stiffness KWL3 deflection, S = - KI where K is a factor depending on loading and supports. For Es~,, >> E,, EI Ebdh2/2 2KWL3 S= E&,,bdh2 Solutions to Questions 447 Total mass m = pCoEbhL + p*,,bdL From deflection equation (1) d=- 2KWL3 Esbnbh26 so To minimise the weight with respect to h, then amlah = 0 4panK WL4 =O am - = p-bL - ah Esbn6h3 or Pcmh = 4skh weight of core = 2 x weight of skin (b) Sandwich Beam of Minimum Weight for a Given Strength Assuming the core does not fail in shear, failure occurs in bending when the stress in the skin reaches its yield strength cy, ie M a, = - bdh where M = bending moment. Mass of beam m = p-bhL + psbnbdL Using M bhu, d=- To minimise the weight with respect to h, blah = 0 448 Solutions to Questions or (2.30) (a) Maxwell p-h = bd weight of core = weight of skin. 12 cU) = - + 12x50) = 1.26% 2000 90x109 EllJo = - l2 + - l2 106(100) = 1.933% 2000 90x109 00 but - = 0.6% e 6150 = 1.933 - 0.6 = 1.333% (b) Voigt 4 (2.31) Maxwell Strain (50 s) = Kelvin Strain (50 s) (2.32) From graph UO/~ = 0.01 so At E = 0.0089, t = 3000 s 2 0.01 ~=-=~ooMN/~~ 0.0089 = 0.01 { 1 - e-Moo(t/q)} [...]... 44 .145 N (i) loss of energy due to friction, etc = (44 .145 )(0.3 - 0.29) = 0.44 J (ii) energy absorbed due to specimen fracture = (44 .145 )(0.29 - 0.2) = 3.973 J impact strength = 3.973 = 165.5 kJ/m2 (12 x 2)104 (iii) Initial pendulum energy = 0.25 x 44 .145 = 11.036 J 463 Solutions to Questions Loss of energy due to friction + specimen fracture = 0.44 + 3.947 = 4. 414 J Remaining energy = 11.036 - 4. 414. .. this with t = A-Bu t = toeu0IRT B = y/RT and Solutions t Questions o 460 so to= 1.225 x 1015 = 2.22 x 10-l2 s 150 x Id (8. 314 x 293) - A (2)- Also y = BRT = 0.467 x 8. 314 x 293 x = 1137.6 x m3/mol so at u = 44 MN/m2, T = 40°C t = 2.22 x 10-12exp 150 x I - 1137.6 x d x 44 x 106 8. 314 x 313 = 1.06 x 1 s 6 (2.49) 3 years = 0.95 x 108 seconds From Fig 3.10 creep rupture strength = 8 MN/m2 Using a safety... redrawn from the creep curves q= 0 E 2 4.3 : x 10" Ns/mz 00 = - = 140 0 x lo6 N/mz $ EO So from text using the fact that for a strain of 0.48, a 2 5.5 0 1 MN/mz (from creep data) = aoe(-o/'lt = 5.5e( -140 0~106x900)/4.3x1012 = 4.1 m,, /,2 1.0 0.75 C 'e s 0-5 0-25 lime (2.35) From the graph below, and the theory of the 4-element model, 4.2/0.003 = 140 0 MN/m2 $I = UO/EI = 4.2 00 00 h = Retarded Creep = - - -... The stiffness terms in the global directions are (for h = 2.4 mm) 1 E, = all h’ 1 E, = a z z h’ E, = 5 .14 x 104 MN/m2 E , = 5 .14. 104 MN/m2 1 G,, = - G,, = 1.96.104 MN/m2 a66 h’ a2 1 v, = -, a1 I a12 v, = a22 ’ and the strains may be obtained as E, V, = 0.308 V, = 0.308 cy = -5.98 = 1.94 x (3 .14) Using the S and Q matrices to calculate 8 A= f=1 yxy and D, (hf - hf-,), = -1.12 10-7 for each layer,... Solutions to Questions a = A-' -3.22 x -2 .14 x 3.71 x -1.58 x mm/N 1.58 x 10-5 5.27 x 10-5 d := D-' 4.7 10-5 -9.68 x 10-6 -6.42 x 10-5 d = -9.68 x 1.11 x -4.73 x (Nmm)-' -6.42 10-5 -4.73 x 10-5 1.58 x 10-4 1 1 1 a12 -a12 v, = -up , =E , = - E , = - G -all h ' a22 a22 h' x y - a66 h all , E, = 31.9 GN/mz, E, = 13.5 GN/m2, G , = 9.5 GN/m2 1 1 1.56 x a = -3.22 x -2 .14 10-5 [ [ v, = 0.206, , vp = 0.087 It... = 11.036 J 463 Solutions to Questions Loss of energy due to friction + specimen fracture = 0.44 + 3.947 = 4. 414 J Remaining energy = 11.036 - 4. 414 = 6.6218 J 6.6218 So height of swing = -= 0.15 m 44 .145 (2.59) K = so [ p(m~)'/~ (t)+ 10.6 (t)'- 21.7 1.12 - 0.23 Wh 1.75 x lo6 = where so [f (R x 100 x 5 x 10-6 (3' + io x 1 0 - 3 p (31= P = 4.89 kN K , = D& (2.60) + a, = (E)1 K = 8.15 mm from equation... 0 0 ) 104) SO x=l t'x " [(y) - ($ - l)"] x=lO [(4x)"- (4x - l)"] = 0.108% ~ ~ ( 9x 6 = 1.1 lo4) x= 1 (2.44) From question (2.20) at 5.6 MN/m2 the grade of PP may be repnsented by E ( t ) = 0.238(t)0. 114 So, after lo00 seconds, ~ ( t= 0.523% ) After 10 cycles in the given sequence (t' = 1500 seconds) x=lO [ ( 1 5 ~ ) ~-"(~ 5 ~ l)0."4] = 0.691% 1 - ~ ( 1 5 104) = (0.523) +OS23 x x= 1 Then if a straight... Equivalent modulus after 900 cycles = 5/0.01 = 500 MN/m2 Solutions to Questions 458 1 C 'E s 3i _ -100 101 i o ~ 103 105 104 i o ~ 107 Total creep time (2.45) For the PP creep curves in Fig 2.4, n = 0. 114 at Q = 7 MN/m2, ~ ~ ( 2 1 6 x 104) = 0.99% after 11 cycles of creep (10 cycles of load removal), t = 66 hrs = 2.37 x 1 6 s (using equation in text or from computer p g r a m ) ~ ~ ( 2 3 71 6 ) = x . 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. MN/m2 So % change = +9.67% (2.18) At short times, E= ’*‘ - 140 0MN/m2 0.004 Allowing for temperature, Ea = 140 0 x 0.44 = 616 MN/m2 Overall strain, E = ET - E, = 0 ET. - 1’ a, = 0 Ar =- a0 vu, 2k2 P _- - r E E E(k2 - 1) hence -2( 1 .14) *17.5(4) 2 x 103(1 .142 - 1) A, = = -0.3 ITM Therefore the internal diameter of the bush will be