Solutions-Set 04 - *4—4 Delermine[he dispLacement nf E...

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Unformatted text preview: *4—4. Delermine [he dispLacement nf E with respect In C of the mpmlte shalt in PrnhA—l Pf. {— 23mm) MIL. = E = W = —[J.'.I‘111 mm The negative Sign indiea [es end 3 moves [awards end C. Aluminum Cum-er Steel H]. = m III-Fa If“: [25. III-Fa I; = 2130 CH :1“ = 55-! mm1 Am: = I"? mm1 A“, = 39 mm1 [IS I: N E k N '9 he! i— T t N A .1— 1'] 1mm '5 2; IN 4325' mm—lrmfl mm -—:HIEI mm— q 1» 1‘ W F“ u 2.: Ha 15 m 4—11. The 1111M is supf111rte11 I131 the [em 3114 stainless steel wire: that are 1:11nnee1e11111 the rigid members A E 111111 DC Determine the vertical [fi11f1l11eement [1f the 2.541311 1119111 if the members were [1riginally h11li1:[1nL11l when the l11311 was applied. Each 1Mire ha11 a enm-fieetianal area [1l' 1 F1 mmz. Internal Forces 1‘11- Ifle wires. Fame}, 1; +2111 =11: ngt.2f|—2.ifti.91=i] Fag-=IH'1'511N +1 EFF = [1: PM + 1.1115 — 2.3 :11 PM = [1.52:1 1:11 1:1;[11111 q +2119 = [1. Fepftifl] 411122111131 =11 F1:1.-=11.2[11-131:11 +TEF_u = [1: F.1;.;;+ [1.211113— [1.1125 =11 F1;;:-=[J.41fi'1‘ 1111 Displacement 50 = % = [1.12145 111111 .F' 1'. . . ~ = fl = —” Emjz‘ffl’flfl] "1" = [1111111112 mm 5,1; 11.116116; “- [w Sh= [LEHJii-Eit mm 311' = 1111-1114] — 111151162 = [DUNE mm Ewe“ [1.63111111111541111131 6.4...”- = — = — = [111111111 AME tfift [13111 [F1 mm 11.1 = .11, — 11.1,” = [1.11112h [1.1111113 =11.21[132 mm 11551.35 1.3151111110511119] ~ = = =[1.111[1111 “*5 11ch 11111113111111] mm 51 [111111111 J_ _ M— u . 31—11:.335 51 = 1121113; + 11.5335 = [LB-F1 mm 1 mm |_1 .11 121211111 mm | 1...:- ....... ———- 1-~-~ 1: : '~-..__ : IlJm Eu CUE-£63m 0.5111 0.9m 01 [M H." m HEN Octaber—OI- 15 10:02 PM £1.ng fig; $.57 ____ S 400w .-.Ej| 4 a 93 GA!»- i’l 400 M Nat-é; I!" = 400 'MW * 5 I l 1:; z 800 mm, in; | it “:— 7?(25:)2 WW} AW 3 7: (55.29?) W: : 5AM Fem EM : .734 “03 Mfi. 5,1? =- 200 mag m COL) Us) _.E4j ‘ "fir-CAD) MM; fl; MM a WWW J3. j (3%? G affirm) : 5:9 (5+; [SJ 3 g? : : 4OOXJUEK4DQ ' [5:7 Afitfic kw K g " if 5‘“ ((6-4 +Lz) FQAJC L»: F“ r M + E Rm; Em Am 1% L--:————V_____,,/ '1 __Y__ _} aim if: F, OLUJ— A71 * fit Cf; an]; 9» “’7 5”" ,4 94119,»; 0115 a AB @Jfl/QKW WM vmfipggk AM Eflfi A“: g“, 'I‘I : SEC/$1: avg/a a—Ffir’f’fl AS? Last? E? (4‘4; Eu.— F‘“ I”? Am fie a, g : 8.20m; m F r H as w E; : 46’ W W (2 J 3*? sac .J’ E: ’5: {£1 W M ///4m 1:13 4—65. Initiall].I the A-Efi belt shank fits snugly against the rigid caps E and Fen the fitifi‘J-Tfi aluminum sleeve. If the thread r11'1J1e hell shank has a lead eft rnrn.and the nut is tightened é of a turn. delerrnine the average ntrrmal stress develeped in the huh shan}: and the slees'e_The diameter at halt shank isd' = fittrnrn. Equaflen er Equlllhrittm: Refening tn the free-bed} diagram (if the cut part [It the assemth shuwn in Fig. .5. —T EFF = n: F, — Ft =1] [1] Cempmlhllltg Equation: 1v.'.-'hen the nut is tightened 354 (if a turn. the unennstrained hell 1nl'ill be shortened h} fie, = %f]j = [iJ'S mm. Referring la the initial and final pesitien 01' the assemhl}.I shtrwn in Fig. b. 5a — 51:, = 5F: F inn F 4:74] m5 _ _L = + iftitifiz‘rflfltiif‘itfl frees? — [Jflzitfififljfttfi 1.49935. + F; = 1 an arm [2] Striving Eqs. 11‘; and {2}. F5 = F‘, = SEEHEIJfi-N Norma] Stress: F iiifi an .]fi if, = f = _— = 45.9 MP3 Ans. ' “inter? — 11.22% F Siifi H3115 rr¢.=—b-—=tfitJMFa A115. .115 _ F 't. — [F.[Jfi‘ 4 f i Ans: tr; = 45.9 MPa.erE, = 13“ MPa *I—EI. The mnler red CD of [he assemhly is healed frnm [LT mm T. = 31°C in T2 = IEHJ'C using eLeclrical maistanoe heating. Al the IU'IMEL' temperature 'i"1 lire gap between i." and lire rigid bar i511? mm. Determine Uri: lurce in mils AB and EF caused by the infleas: in Iemperalurc. Rods AB and EF are made uf Meel,and each has a cum-arclinnal area of IE mm3. CD is manic nf aluminum and has a crusa- scuinna] area of 3?:1 mml. Est = 2411] GPa. Ell = Ti] GPa,anrJ [M = Hiram: an = {Md — $5., — [1.11m F {rub pm we Uri-l ’“— = B 111- 1511 [:24 — 4 — rum? {marm'inzmmmfi ( 61”: H J {3131(111'51fiumng} L. 12F, = 123nm — 511423.? m wry-3:11; F—ZF‘,‘=II] {Ta Striving Exp; (1] and {2] yield-r. F“ = FE; = F“ = 423 kN A115. FED = F = [HSer ...
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