homework7 - PROBLEM 5.44 A valuth reef 111.155 is leaded as...

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Unformatted text preview: PROBLEM 5.44 A valuth reef 111.155 is leaded as ShU‘M'I. Determine the force in members 3D,BE,and CE. .‘F SOLUTION TIME: _____'_ Ea: D: Ax: D 13;,» load Symmetry: A}; Ly: 34D N1 if {1J5 m} % 113+ [2 m)[241:a N] '. Melt in. mm fl 1 f‘ mfg: fl: 5 —{4 m}{$4cr N -12n N] = n FEB: 1320” N, Fm: 2.14 kN e «I tfilfidflflflh’fifl FBDSenti ABC: an : LEM]FEE—{2m]{fldflN—ilfl N}=1‘] _ n2er F». L5? N, FE: 2.4? EN T 4 % PM] — {2 m}{24fl N} = u FEE=§$EH FM=429NCd -mmmm:rhummmmmmm. Lam-2m Hum-1 PROPHIEI‘ARY.MA TEMAL. em 2130? The Mcfiq‘aw-Hiil Companies, Inc. All rights :esemd. Nu par! affirm Mum! may be disphym'. reproduma' or dime-wax In gal-Jam m- by any mus, "mm: the pn'ar miller: pmisshm afrhe pubfisfier, w med beyond .rFre Eirm'ma' dflrfibuflan m Imchm' and Edueamn-Pgmmm 5y ‘WfiGWHfoflr rflgir individuaf [flyrjuprfipfl‘rflflflfl {fyrm area sudden! wring this Mammfl yen are “Sing E: 'n'ilfiflflpe‘fim'ififi'fl. 351' mm“ 5.5.; I ._ A Fink muff [11153 is ioadcd as shown. Determine the force in mmnhcrs : ED, CD1 and CE. I rim a-IIII a : ":. .l'. J i L‘. . I I I ii§:\' I; I " :: '5 '-._ .43.“ .... ._ ~..~. -. ._ In .. Distance hem-rem loads = 1.5 m raw; Aft] By smmeny, Ay = KJ. =18kNI LEM!) = u: [1.5 £11}ch +[1.5m]I[_tikN}—{3m][_lé§kN—3kN} = u Fm = 22.5w T4 FCD=625kNT1 {35: u: lam-am—fim—%Ffifl+§{a25m}=13 Fw=293kN C4 PRGPMEI‘ARFMATEMAL It! 2m]? Tin: hIcGrnw-HHI Campanirs. Em. All flghls reserved. No parr afmif Manual my be imam-M, reproduced m‘ dis-tribmed in: anyfi-mr or by any means. “Winn! Hrs prior wn'zmn par-minim: of {FIE pHHr'sFlc‘r. or mad bqwnd h'JE J'J'miim' disrrbeI-fl'au 1-5 Peacha'rfi' MI! Mama” pmiflcd by :HcCrm-‘P-Hmfilr {heir fnn'r'n'dum’ marge prfii‘rfi‘hflffl-‘L fl'yvu area student using Hu': Manual. you are using it withammrmfl'mn. 3&5 PROBLEM 6.5? A vaulted mi truss 1'5 loadgd as shown. Damnmm the fares in members {4.3 fl}[3flfl1b}+{?.2 fi}{75fl 11:} + [11.52 fi}{?5fl lb} + {15.3 fi}{195u 1b} + [19.2 flHgm 11:] + {215.4 fi]{1fl5lfl 1b} . we” . ' +33.fifi 154:: — .=n “arm-- 11!: ‘!‘1?r;r&~r;r#§+r:—ua E“ I:' m A} ) A; = 2925 11:! I)’ (II: 2925 lb * 2(1501b‘J— 2[?5D 1h} — win It: — 9m} “3 —195fllb 44301}: + NJ. = n N}. = 30?:3113 FED Senfinu KIMN: g EM} = Cr: [12 Mij It: —1501bj-['2.4 fi]{3l]fl lb] —{1.3 MPH = {1 fix =|13mlb, 5K = 113:) kips T Q EM; = 0-. [11.52 mums 1b —1501b}-[E~.? Mann 11:} 44.32 fiJfiSD lb] + {3.96 fl][%fifl] = 3 FM = 4590.5.‘5 lb, Fm = 3.133. kips C «1 —~ 2;; = n: — —?§(~359n.9¢§ 1b] —]1—:FH - 113mm = n} \I .- fi, =—4461.41b, Fjg = 4.4451(1pcd FED Joint}: fir mm #é 2F; = a.- 113m 11: — $44514 lb] a F - = u 1 II 5,__ 5: H g £3390 H; F5: 2 “ELEM EH = 11351351" 'I I PROPRIETARY J'IMTEKIAL 'D 2'3"]? Tm! McGraw-Hill Campanies, [ILL Ail rights Tflitnl'cd. M: pan! qfl'his' Manual mqv EN? displayed ntprml'uc'm' gr djgrrmumd m any-jar.” 0,-- hr“ any meg-Ml wtham Infie- Fm'ar “Iran!” Wfiflflfifl'afl {If-me mhflflrfl, {If used hg'flfld Hrs I'r'mr'rw' IfJ'SfFfl-{mfl-Ul'l {I9 NRC-I'll?” Ema. Edi-lemurs pernrr'rmd by Mchw-Hmformer'r r'ndh-J'dmrmurse prepararr'mr. {T _1 m: are a swim: using {his Marmaf._vau am: using it H’flfiahfl'pcmrfsfian. F572 PROBLEM (3.6? The diagonal members in the eenter pnneIs of the toss shown are very slender and can act onlyr in tension; ouoh members are known as counters. Determine the force in member DE and in the counters which are acting under the given loading. 1 Elk n Fan—-l-—an-—-i——a nut-He n 50 LU TIDN FED Truss: Q EMA = o; [32 em), - {24 n}{12 hips} — [in one tops] ..[email protected]'AG 52F} .. _ J - _ a 1,- - _ : #{e n}[n Rips]=fl, H} :15 kins I TEF}.=EI: A],—6kipS—9kips—12k_ips+151tips=fl nw =12 tops} —.—EFT=[}: AI=G Since only BE can provide the donnward force necessary for equilibrium. it must be in tension, so CD is sleek. Fm= t] 1211. = El: 12 kips — IS kins —%FHE =13 PM =1ono kips T «I Since onlyr EF can provide the downward force necessary for equitibtinm~ it must be in teneion, so DC? is slack, Fm = [I {wigs}, fibre: . . _ 3 FBI) Section FGH: | 5F}- = D: 15 “P5 ' ‘3 “PS ‘ g Fee = n F FEF = kills T ‘ Knowing that Fm = FM. = CI, inspection of joint D gives i”; Ff“ * See: note before Problem 6.54. PRBPRIEIARY MA TERML it: EMT The MeGraw-Hill Companies. Inn. All rights reserved. No part of this Manna! may he dismayed. reprm'ueee' or nit-meme in ertyfonn or by any means. without the prior written permi'sston ofrhe punts-her. ow rn'ed beam-Jed the Matted non-om” to mother-.5- and WWW-9 PWIT-‘E-I'I' 1'3? fitcfiml-HIHIHF their I'M-divided! courseprepnretton. {firm are HRH-ITEM un'ng no Mantra}. you are [Ht-Hg i! Manner permits-ton. £84 FRDBLEM 5.?3 Far the frama: and wading shown, dttartninc the components ofalI fumes lam stating on member DECF. «— — —_— —- ".a f" SD LUTIDN FED F rnmt: Q EMA = u: {0.25 m]D_‘ - {0.95 m][43fl M} = 0 01:1324Ni—4 Nata that EE- is a tam-farce mamberfi E5 = E” —*E.FI=LJ: +132q3~£+£l=m Ez=1324N—~i Ey=1324NT1 - Q EMS = n: [0.54) m](1824 N} — {ms m}C + {1195 m}{480 N} = u C=3324N11 {211;0; -Dj,+IE{Z4N—IHE4N+4E{}N=D Dy=4SUN14 PRUPRIETARY MA TERMI- E‘ 2007 Tim MDGI'HW-Hi” Cnmpanius. [LIL All righxs resumed. NI: par: qffirr's Manual ran-1.1.: be .I'l'fSpJ'flrg'fl'. reJIJrrJduced or disrfibured h: ajwjbfln m‘ by any mmm'. H'J'Ikrmr tire prior “finer: pmnr'xsirm raft-’11: pHNr'shyr. m' turd halyard die Hmflen' dia'rrr'hwfm: rn reachm and Hummus“ permit-fed by Mchw-Hiflfar fire]? J'rdipidufi! course preparation. Him-pr arr r: Nun'er using :hifflrfmrmfl, ya!“ are using I! uifiourpcrmiifian. 2496 PROBLEM E.Bfl A circular ring of mifius 8 in. is pimmd at A and is suppoflad by red RE. which is fitted with a collar at C that can be moved along the ring. For the pflsitinn when H = 355, dclcmfluc {a} the force in rod EC, Eb) The reaction at A. SGLUTIDN FEDRing: {a} 1; EMA =0; {3 in.}[FHCcm35G}—{E En.}{61b}=fl EEC = 1.3245 1b. Fm. = 132 lb 0 «I {:5} —~ 3+; = u; A} —{T.32461b]c0535° = u I = a; A_1I.+f_?.324-51b]sm35° — 611:: = 0 AF: 1.?93T51b A = (3.26 lb ilfifi?” *1 PRGFRIETARY MATERHL Ea 20m The MEGuw-Hfl] Companics: Inc. Ml rights reserved. No pm? rar'IFrrLs Mmmu.’ may: be displayed. repmrfuczu’ rIr Jami-Burned in aayfom or by any meme: wr'rhuu: :J'ra prim- “mm: Fermi-don (If Hm midis-her. ar um! Eneywrad Hie J‘J'mfleu' difrrr'hmfan m teacher: and Murmurs pal-mier by Mchw—Hflffarrher‘rmuffin-Mara! come Mpmalffis. fryer-r are a student usingrhir Manual. Ivan are using 1: “Mum permission $95 SDLU TIDN FBI) Frame: PROBLEM $.53 Determine all the forces exerierl en member A! “men a eleelcwise eeuple ef magnimde lEfl lie-ft is applied te the frame {a} at peth D. {in} at peint E. Here: In analysis of entire frame, leeatien el" M is immaterial. New else that .43, BC, and F G are iwe force members. {EMH={}: {a filly—ISUIb-fl =e. i},=3e.e 1e 1 «I TEfi.=U‘. lfllb—%Ffl=fl. Ew=TEIle 11721.50 «I .. '12 . _ Q = U: '— Plug] = [11 FW = H i 12 m Ex=fli Fm—ETSEh—T21b=fl, FHC=144fllb —-4 As Hbfive.‘ = fl yields FAB = 373.0 lb F 22.6" 4‘ Then: . . '3‘ L EMC 2 e: [2.5 fiJfi—grae lb - a}..G]—1ae1b-rr = e 12 —- an = a: Fat — Erma lb} = 0, F3: m ran lb —-— 4| PRGPRIETARYMATERIAL ’3- ml]? The MeGi-ew-Hill Companies, lire. All rights reserved. M: pm': effiris Manual may EIL- ri'i'xpleyed. repairer-nerd er a'i'flri'hmed r'n ear-farm or by any means. wr'rheu: :J're prior Heme-n Imp-milrsfrm rgr' rhr' plrh eaimiorswrméi‘red Ep- Mefirew-Hrlffer weir individual rmeprapumrien. {firm are e mien: wring We Malawi. fisher. er used hen-and .'."u.- limiteddi'flrilruifmr re rancher-i- and _-.-rir.r are airing i'r H'fIhrJIJEJLh‘rflrl'Sfl'flfl. 9M PROBLEM 6.96 The cab and motor units of the front-end loader shown are connected by a a- vertical pin located fill in. behind the cab wheels. The distance From C to D is 3D in. The center of gravity of the SCI-kip motor unit is located at Gm, while the eentem of gravity of the lS-kip cap and let-kip load are located, mepeotively, at G: and GI. Knowing that the machine is at I‘BSI. odth its _ brakes released, determine [a]: the reactions at eaeh of the four wheels, (it) the foreee exerted on the motor unit at C and D. 50 LUTIDN {a} FED Entire maebine: EM: = fl: {95 in'i “'5 H135} — {35 mills kills] {145 in] -(12o in.) {it} kips] = o, e = 25.241 kips e = 215.2 lope l «I —15 kips + 2A - lE leips + 2(26.241kip5]— 5E] Rips = l}. Nil-r1 A our“; E 1?: 5ier airbag-1:41 ‘3‘ = I i {b} FED Motor unit: (EMS: I11: [30 injti‘.r + {85 in_}[52.482 kips} — {1 1t] in.]{5fl lo'pe} = {J Ex = 34.534 lope, 1‘:l = 34.6 hips -— 4 —* Eli: U: Dx— 34.534 kipe = l}. DI: 34.6 1:11:15 —-r '1 the}: o: 52.4e2 kipe — 5o kipe — o”: e. or: 2.42 hips 1 4 PflflPflEJ—iifli’ ENTER-[AL 'E' 2913? Th: MfirflW-Hiii Companies, inc. All rights. reserved. No par! oftiiis' i'ld'unuai may be dispiaye‘ci. reproduced or disrr-ibwd in entire-m or by any means. WHth file pfim‘ wrmen penm'eeior: of fine puoiisher. or rue-1f mam: me ir'mr'reci n'r'.-.-rrfbuzi.:-n m inc-Frem- iran educators pea-rifled ilj' Mefim Hl-Hr'iiflr' riii‘ir individuai mum! preparation. ifluu are r: madam using {iris Mdmmi,_t'au are Hing” wr'rimur Ernrr'ssion. Elle} ...
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This note was uploaded on 04/13/2008 for the course MAE 206 taught by Professor Annahoward during the Fall '08 term at N.C. State.

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homework7 - PROBLEM 5.44 A valuth reef 111.155 is leaded as...

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