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HW9%20Solutions - CEE 4?70 | Introduction to Composite...

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Unformatted text preview: CEE 4?70 | Introduction to Composite Materials | HW #9 l Partl :T=0;P= ??? ** See Matlab code for additional calculations Calculation for: Laminate [i-«Iflslflgls Carbon Fiber Polymer Composite t ¢ 0 - t i 0 o a 5 2 t D 26 t l 0 -- t 0 - t «45 E t +45 210 T ill—L ‘ p ARE—l m = cos 9 n = sin 6 E1 = GP“. E2 = GPI‘I 1.712 : 612 = 5 GPG. = E1 _ 31251 _ 52 __ G v _ I91252 11 1 _ FREE 12 —1 H vuvfl 22 —‘1 _ vuvfl as 12 21 — El 011 012 0 m2 n2 2mn m2 212 mn [Q] = Q12 022 0 [T1] = n2 m2 —2mn [T2] = n2 1712 —mn '3 0 055 —-mn mu m2 m n2 —2mn Zmn m2 — n2 [Q] = [T1]_1[Q][T2] [fl] = 3 3] [15’] P P r = 0.030m NI = — 2m : 2n:(0.030m) : 530516}? CEE 4770 | Introduction to Composite Materials | HW #9 | Helen Liu Na: 5.3051613 N3, = NI}, t=0.25x10‘3m 202—5: 5512—4: 222—31: z3=—2t 242—16 25:0 [A] = Zléfla — 2H) R 1 [80] : [AlfillN] [5]): = Gr1 3168.51D for 0” layer 52 : [T1][a]x : —93.5P T12 0P 51 _ 630.8955)" for + 45° layer 02 = [lanes]I = 61.8525P T12 —206.1118P {TI 630895513 fer — 45“ layer 5’2 = [T1][cr]x = 61.8525lJ T12 201(23111813l Solve for P values using Tsai—Wu Criterion X6 = ~1250 MPa XT = 1500 MPa Y6 = —200 MPa YT = 50 MPa 5 = 100 MPa F_1+1 F__1+1 F_ 1 F 1 1 1—Xr Xe 2—Yr Ye 11*— XrXc 22— YrYe F1 : —1.33€ -‘ F2 3* 1.59 — 8 F11 : — F22 = '— F66 : "" for F151 + F252 + F1153 + FREQ-22 + Ffifirfz "' 1lF11F22(0-1J2) = 1 for 0” layer: (—1.33e — 10)(3168.5P) + (1.5e ~— 8)(—93.5P) + (5.5325e — 1‘:’9)(31Ei8.5P)2 + (1.09 — 16)(—93.5P)2 + 0 — 1/(55325‘9 -- 19)(1.0e — 16)(3168.5P)(—»935P) 2 1 P = 470.65 kN ml— 253.13 kN for + 45“ layer: ("-1.33e — 10)(630.8955P) + (1.5e — 8)(61.8525P) + (5.53259 -~ 19)(630.8955P)2 + (1.03 — 1s)(1531352513)2 + (1.0e — 16)(—206.1118P)2 — (5.5325e — 19)(1.0e -—~ 16)(63U.8955P)(61.8525P) = 1 .' P 2 384.90 k er — 569.99 M! CEE 4770 | introduction to Composite Materials | HW #9 | Helen Liu for -~ 45” layer: (—1.33e — 10)(630.8955P) + (1.58 — 8)(61.8525P) + (5.53255- — 19)(630.8955P)2 + (1.0a — 16)(61.8525P)2 + (1.09 — 16)(206.1118P)2 — 1/(553259 — 19)(1.Ue — 16)(630.8955P)(61.8525P) : 1 P = 384.90 kN or —- 569.99 kN Partll:P=O;T=??? [ii] = i2 5 [if] : _ N : =——— r 0030m I? anz 2H(0.030m)2 2 176.8388? U1 0 for 0“ iayer[32 = [T1][o]x = 0 ] 1'12 220151" 01 271040?” for + 45” Zayer['-’Tz] I [T1J[o]x 2 —1659UT] T12 O '51 —27104UT for — 45°layer 52] = [T1][o]x = [ 1659GT ] T12 0 Solve for P values using Tsai-Wu Criterion X6 = —1250 MPa XT = 1500 MPa YE : *200 MPa YT = 50 MPo s = 100 MPa F_11F*_1+1F_m1 F_1 F__1 1“XT X6 Z‘YT YE “— XTXC 22— YTYC 66—52 F1 = '— F2 : 1.58 *" 8 F11 : _‘ F22 : — F66 _ _‘ for 2D: F101 + F252 + F11o12+ Fzzoz2 + Ffifirfg —- 1/F11F22(fl-10-2) = CEE 4770 | Introduction to Composite Materials | HW #9 | Helen Liu far 0“ layer: (1.09 — 16)(22015T)2 = 1 T = —4542.4 Nrn er 4542.4 Nm for + 4501ayer: (—1.33e —— 10)(271040T) + (1.5e — 8)(——16590T) + (5.53259 -- 19)(271040T)2 + (1.09 — 16)(—16590T)2 — 1/ (5.53256 — 19)(1.0e H 16)(271040T)(—16590T) = 1 T = 4909.0 Nm 0 — 2046.0 Nm for -— 45“ layer: (—1.33e — 10)(—271040T) + (1.5e — 8)(16590T) + (5.53259 — 1-‘~3)(—271040r)2 + (1.03 — 16)(16590T)2 — ./(5.5325e — 19)(1.0e — 16)(—271040T)(16590T) = 1 .- T = 2046.0 Nrn. —- 4909.0 Nra Part III : PH 2 ? frem Part I fer +45 Layer; T = ??? P 384.90 kN E = Wei—— = 192.45 RN 2 192450 N m 3168.5(192450) 60977898 for 0” layer 02] = —93.5(192450)] = [—1.79941e7] T12 0 + 22015 T 22015?" fer + 45” layer 52 61.8525(192450) — 165903" 1.1903587 — 165901" —206.1118(192450) —3.96662e7 01] 630.8955(192450)+ 2710407] [12141698 + 2710407 T12 fer — 450 layer 61.8525(192450)+ 16590?" = 2081118092450) 1.1903597 + 16590?" 3.96662e7 J1 630.8955(192450) — 2710401" 121416.98 — 2710401" 0'2] : ] T12 Solve for P values using Tsai-Wu Criterion X5 = —1250 MPa XT = 1500 MPa YE = —200 MPa YT = 50 MPa 5 = 100 MPa F_11F_1+1F_1 F_1 F 1 1_XT XC 2—K!" YE “— XTXC 22— YTYC '56—'52 F1 : — F2 : "““ 8 F11 = *"' F22 : 1.09 — F56 = — for F151 ‘1' F202 '1' F115? ‘1“ F2202? + Ffififfz “" 11' F11F22(5152) I 1 for 0” layer: (—1.33e — 10)(6.09778e8) + (1.5e ~— 8)(—1.79941e7) + (55325.? w 19)(6.09778e8)2 + (1.09 — 16)(-—1.79941e7)2 + (1.09 — 16)(22015T)2 — 1/(553259 —-- 19)(1.0e — 16)(6.09778e8)(—1.79941e7) = 1 CEE 4770 I Introduction to Composite Materials | HW #9 | Helen Liu T = —4612.91 Nm or 4612.91 Nm for + 45” layer: (—1.33e — 10)(1.21416e8 + 2710407") + (1.5e — 8)(1.1903597 - 165907‘) + (5.5325e — 19)(1.2141698 + 271040702 + (1.06 - 1e)(1.1903597 — 16590T)2 + (1.0:? — l15)(—3.‘~9(€~.u662e7)2 — 1/(553259 -— 19)(1.0e -— 16)(1.2141698 + 271040T)(1.1903597 — 165907“) 2 1 LT = —1490.95 Nm or 4413.8 Nm f0?" -— 45“ layer: (-—1.33e — 10)(1.21416€8 — 2710407“) + (1.59 —~ 8)(1.1903597 + 165907“) + (5.53253 — 19)(1.2141698 — 271(3407‘)2 + (1.09 —- 16)(1.19035e7 + 16590732 + (1.09 — 15)(3.9666227)2 - 1/ (5.53253 — 19)(1.0e — 16)(1.2141688 -- 27104UT)(1.1903597 + 165907") 2 1 T = —1761.23 Nm 0?" 3736.44 Nm Part III ** See attached contour plot of T vs. P non-failure region r. 'L If. h i; I j I. I | .u . . . Fania. ‘4 nflnfl’j *“I-I-u- fl ‘- ‘1' II. I Ir *1;- r %"‘ F ' -' - -£ :- i - .fl. ..- ..- .. t 311* I|q-:-—.--—._______,__ "VI-'n-I-I-Iw'i— E .nlr I 'n: '1; 1 . F ,_ ' 1. H. R U '* .—.- Il'x I? 5; q I“ J" 1 t " - i" __ r H H- I ‘ IL f” .r. ._L H“. r '. r - ' .- ' .r E 1' . I "'II- :'E 'u :_r "I "I “all -I i') I _ Pllt._-.i_i% _ 1_. _ II] I ...
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HW9%20Solutions - CEE 4?70 | Introduction to Composite...

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