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Suggested_Ch6_sol

Suggested_Ch6_sol - II Fruhlam 3.4 I 6.1 through 6.18 Using...

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Unformatted text preview: II Fruhlam 3.4 I 6.1 through 6.18 Using the method of Jomts determine the force in each member ' of the truss shown. State whether each member IS in tension or compression. WEE —"' 14 . _ .2 #8 4w +//25'46Aj§}- 533.2%” _ fj/PZ—ffi Fe' =+I1257131f _ [i3 fa - —140:/2~5"8” F' 'J'E-Hg” T ‘3— _ I +sz= 0; {MM/v)? — (/7543/V}(25~")-/' f2: _ I 5C: gang/y @=:}‘4€/V C 4‘ _ E '_ JGHr?‘ C: ficaflfifl _iffi=oz—/2o,é/ve/§c=o — I _ c ,5p=/Zo it! 535 *mm - " fie L=may ' 33:23:11? 64 +- +i27§=0: cW/fim-flé‘lz’wo (amber) 6.1 through 6.18 Using the method of Jomts determine the force in each member . _ of the truss shown State whether each member IS in tension or compression. .| I_gl_l_l_l | J._i_- _|_|_|_J_I I_.. r . - m E mam“: 2'.” =0: - A i'_- :2. 2a I! 14% |_|_- 41 44444444 JQFNH‘T— a: 152‘ 54: Magi/T 4 F0 ewe/ENC 4 15-71505 "3&20) +Fflgro F8 : +L/‘80.,£N -FR_~E‘4‘80’£NT <. _ | . 6.1 through 6.18 Using the method of joints, determine the force in each member _ of the truss shown. State whether each member is in tension or compression. F , op _ a _ i 1 #25149. Jab/+2] 1&4” / 7 Féo I . I ' - | 6.47 and 6.48 Classify each of the given structures as completely, partially, or — __ improperly constrained; if completely constrained, further classify as determinate | or indeterminate. All members can act both in tension and in compression. "*" " -4... i 3 SIMPLE rgusr (we LEWfl/y494’.fl5fi677w5 ”£17m: Fae/Muse Am ewgvrzya wgwflwwflwvwfimww (6) 74:42 ”7:56 r=6¢ ._ E E? @4- m+f‘:27) 77m 1: 1397 a J/MPILE 7721/55. WE PEOCEED m FomzflsruME @562; USA? M57300 or: Jom’I'S‘ To ExPean‘ FaRa? IN Efla/ Mafia-r: l/Y -- 75/2/14: OF G, fésotr 5mm EELow (@zol pa JD/m w smug/65) fl (Saws/5m SE7 0; 8,42 F046;: [5 cam/~50. 775703315: CoMFLEELY CONSTRAINED fir/0 firm/N475 ‘ EM = Pfl'fZT/fl LL)” (aware/4mm- 1' El E Ef T E ' ruhlim ESE — 6.58 The marine crane shown is used in offshore drilling operations. Detennine ’ — (a) the force in link CD, (b) the force in the brace AC, (c) the force exerted at A on _ 1 1 com. 1—I-l—HTPW. ”I"? II\III|\I!|HHI|,|,, .519 got-#1: AFDC ——.._.-“ W; (80000 Wm mm “ —75‘f-8 'éM 95MB :0 (ms. 8 fén/Mze) m FD(I§'m) :0 ‘ (a) £9 : MS MT «1; WE aka no}: fimt’ {RH/5’33 40 ,5 #881? 7L6”: if: /5 3/,— 45° 'M. We not: :‘1mfAcmd BC Mt W0 forte members; " Force Triangle: W/e hon/e (X: motfia 90’ = /5’0"-fi‘8,8/”— 6‘9" : 86Jq° _ 77 FM _,_ F6; /L/6> 15” JI‘nL/Bm— 5,1745“ ‘ “785.17" {19) gar/mag“ ‘. I411: {/038 :st.n 888/” W 1475, — + 78/.) éA/ +37%:O. (4J— (366057/5— W50 Ag :( ma fa M)cawa?.8°+ 784.8 WV é=+meémt 6.69 The cab and motor units of the front-end loader shown are connected by a _ vertical pin located 60 in. behind the cab wheels. The distance from C to D is 30 ' in. The center of gravity of the SO-kip motor unit is located at Gm, while the -- centers of gravity of the lS-kip cab and 16-kip load are located, respectively, at G8 and G1. Knowing that the machine is at rest with its brakes released, determine . (a) the reactions at each of the four wheels, (b) the forces exerted on the motor ' unitatCandD. . .I. ——T_ _ ._ 'I Ml —- 41‘ +14 — . "l— (a) Free Ber/LI: Ehh‘nfl Machine - . _l . ..-L- ' I - -\:- Dimensions _ :' —]:—!- -. in inches 2 ‘iL-i' .‘Jj = :q. T: Hemmt .-....n‘_;_! _4‘ _- ;_ ‘ m! _eac‘n {mammal e — _: ._ "l ‘..__L_ .___ . ‘r' i —; a; . 35 -|— 5 = fiéac'h'om at — -— ‘ each rear wheel, _ . .|._ I I | . :_ I :-—.——| . . :- I I I I . -+“; imwz (warms )—r(-/8klrs)(35) + gnaw“) — ,~ (50t"r5)(/?0 )= 0 ' £6 : 52.48 K1375) 3:26-29 KI‘PSL saw/(est 4 +1 El; :0: T—l6kif’5 +Zfl '— f8 MP5 +52% kilos—505p =0 1214:3152 bps fl = /5,76 kip T < .. C (307 + 2.8 (85) -(50 krfs)(//o ) = 0 __ 3'06 :5500 4505) . (0; firm/ling me 8:25.2im‘ps _ 3:: c: 559» 85(521/8) =/o%.2 _ ZECIquéqklfS)I Ig:3H96klP§‘—‘ 1’5: gm): 1) ~3eé :0' 131 $4.6 mend“ _ x +?EG=D' Z(,?é.2‘i)—5.O--D€i :0 :72 tiflg'krfd; ‘ 6.97 Using the method of joints, determine the force in each member of the truss 3"" H’I’ Fri-e. 3.991;: 2"“!!! En‘l'l'f‘fi. Truss.- w-W. Ila?" flEMfa: Eflfm}_{fig&mg*fifl H' +33%; _='£¢~£H'\} fitmflT‘ "*IF’D' 3r 1mm énwm... :_:_I+TZFID: EP+%(MAF)—M j Egbfhfiy 15%— flag}; 5 fl |fl—Hfi- WTHJ. &.=fi_ {we F;3H-MT :5 5.3 56-17%” ‘ '&= Isz c. +T Z13“): £59049”)- {as—e” = 0 (due!) J__ 6.108 A 400—kg block may be supported by a email] frame in'eac'h of the four ways — shown. The diameter of the pulley is 250 mm. For each case, determine (a) the force components and the couple representing the reaction at A, (b) the force exerted at D on the vertical member. Jr 1 l I, a, I_I_ _I_.-_5_|_I —I-—|—I—I- -| toad: W:Uf001g)(4.31m/5&): 3421f N’ "—l' Problem 6.108 I_ _ . Hf, ~ (7848NX2M) = 0 M,,:/56% Mm . _ MR- =J57I7kamj ‘ ”Fitz 8047 : Hem bcr 356 We note, fiat DE is 4 filo—Face y‘ncmbtr.' 7348/1 *5 EMB‘O: (gEmVPX/m)—(7895N)(2m):a — 55:22 F17 N- ' (5) F4"?! o’h Vori‘t'cal Member: — '9': 2mm my '4 ..|— —.— — — llgnnnllgglllll! “HUBE- (01) Pm _E-l_uuu, magma LEFg-o: Ax: o - .- H’EF‘Vfufit flg—371’fN201- __(3‘?2‘+ N)(2.42‘5m3:0 M41-5337 M'm I QIBIBLf/(‘N’m5 ‘ Fm 5130111: PHH'QT in Eli. =0: CZ—(B‘rzv macaw": 0 012377477" +¢ZF>=OI Cg —3qZV N - (3‘729‘N)qusfl=6 -~q @667N ’- W aurinIIaIaag-mggl —- ... I BET-i I- ' - - I» -- In . -I _Q’ I .IO'.‘ . If?“ I“ '— —-I ll ask—+4 g; rmM Meme; BEé — II- I _m E‘s: If E +951M8=o; ' I III I Lg/fm. 23,:sem (r COMM“) E.- l D —(Eééqu)(2m)=0 I: III I. .FDErlB‘iHaN I l (5)133th on VCF'ITm/ membci" T —-I8 95 ICNZJIS'O ‘ - - - _ _ ' i —TT| |. I #I I I— I 2125 (a) Free Eve .: Frame _ - $2550; 367m: 0 ' 4:3fisz—> ‘ _ Jr‘IZfzo'; 1E3 L—SWVN =0 ., %:3.72er1‘. 4 _‘_ EMA :0; . _ MAMMZHN )(2.7zérm) (3721/NX2125-m):0 — N .- : _ N-m H M 255% m_/_14.:.§735I<N'm) 4 — 3725‘ N —(3929N)(2m):‘o F135 :/_/0‘77N (b)[email protected] Oh VCV‘ITI‘MI/ Wlf’mj’fil': 3 : (MD M V50 4 ...
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