Truss_8solvedProblems

Truss_8solvedProblems - (0'10 Déq'Cfi/VVWLC (33am, {m...

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Unformatted text preview: (0'10 Déq'Cfi/VVWLC (33am, {m eugm WW )0 éFx=OJ@:O 1?; I? 2M6:o 9 51 :1 F7 (0'40 (143:0 Zfilfp Zero fam WEW: J03??? H -—-—> z O dcwd' D [D 2‘: A97 :0 9L (lot FDSMZLOZ” —-——=7 FD FD Zfix :0 ~AD + FD (05 22.6220 ---=> AD dom+ L; 75 A6 mum? wave-51,34” {fa—<1? 3:9 I :Fx : O /74 003%,)!" /% V\\/\) A9324} I I d\\y . “O alFi” 40%}S5 )0? _ {9/}? t O t ZIgW “F6 =é2H ‘HD‘ZQ 305 (Ix/Wm : AB): Ecificpr Z4 yifsTa/mm - + :o—vrwm ’EF +AEw§SL§4°:Q ——-=7.EF 3 24,0 kiFS' CUM,ny Fig. P6.10 I\ N.“ N00 b’MQ mgvrm W/&%3% 2 Join)? N “4 NM=0 W JaerM~=> ML‘O : W émfiawficb rO‘g I ‘ 6kN T WV zf The members CJ and CF of the loaded truss cross Am B C =- 3.00 kN C but are not connected to members BI and DG. Compute the forces in members BC C1 C1 and C] = “"22 kN C HI ., ' ’ ' CI=5.00kNT ' HI =-§:Q:56=kN=T b\\ 4 kN 10 kN 8 kN , . 2*} =0 "7 3X: (003’: (205 Problem 4/41 if“? t t ’r ALBA ‘51 4 ’0 3 r ZMF=O (sew :00») + 46%) +3, 02) + 49mm tZ)-éaa>s€90 (4) = o J 3 [4,2 m A .4 1...,— .. B we”; AL. 6:55?!“ 2!? :0 3 a E «(pass/1020 H41 'JC 3M6 =0 _ 4L :Smaafl (Ml’ésmfiw) ’42— Lg; : Marzz kK,Qi 2M130 (5,2 ~142Xs) — 36%) + 344) + 162 smCaCB) =0 Bc=3k&c zMgo IH (4) 4;,_.:’1’/(3j — I4. 240:3{9 +5.2. 4 =0 26:0 _ ’51 + £41 ’4‘ ((01st +I(_smoL : 0 31C: £61442. +4+l025m3§7j 4/10; Problem 4/16 JEianD ELK E .. 32 D 2FY=O=DE 300 UZM‘l’ C. 500 ,’>< —-’ _ Zl—X, ~O 213:0: lSO- DESMQ 4/16 The signboard truss is designed to suppott a hori- zontal wind load of 800 lb. A separate analysis shows that ~3— of this force is transmitted to the center connection at C and the rest is equally divided between D and B. Calculate the forces in members BE and BC. l$ ‘ :%+6=L(9,bn 735 (800% 150 g, (500): 500 500 ISO—*9 DE = 535 (cw-P0 956 'DC D; = 300 Gyms) —4E. :O—fir CE:S§O (cows?!) . gamma = 5000356 - SE 517129 (1b) , l I .1 4 J ’2 Determine the force in each member of the loaded truss. g '\ zmfizo ‘39 ’T T —4(s)—a(cz)+poz)=o A ‘1 (a D D: 5.5 I. 4kN 6kN g ZFY=O=A+SIS, f't/fl __,,A= ifs— .vProblem4/2 AB film‘l'Ai , 5F 26:0:45~ FAB %)=O—> FB:5‘,(,Z§ (MFA 2 a=o= a: - mew» a; = 3.275 (was S 1% {Ton/d” F: Bali—7E: 2F“! :0 "’ TFEE 9" 4 -’ (MS-7"“) 1?, =0 —» a5 = 3.375 (mum) 4 0:3er 3 «. F5; 25, =0 = :4 + 5.425%) f FEE (372‘ =0 » 525 = —o.<>ot ~> F =O,951(+ms2 55 F Fee a; = o = same) ’ (—900 + F350 —> as: “W “is “Wm BF 02% c: 41's 2F; =0 =4.I2s—ED(—§> —-' ED. =Ia-L7Sccmw) F FLD £1111 =0 3 ‘FEC‘F (0375 1754': 5-5 (+996) 5; 6.275” szt D: EDA; 2a =0 : é.%7s[§) — 1:5,) ——*» FED= 4, xzsémsv 4/6 Calculate the forces in members CG and CF for the truss shown. 67/; _ 1 m, a C 2 T") Dx 3 , me = z; *6: 250,6!" fl 3 I ZMD=O 3 0. 7 Problem 4/6 \ FéF 2(4) +4][a>—— FEFégsme) :o . .o I FE: H93 (Comer) {row 9 g, 704;? é E,=o =.é7~ gFQ—z) —>rgp=o.94 (ms) F 2 1:; =0 = Io.&7-Fw—gp¢~z. ZMJO = z(4)+4(a)~1>xcs) PF 1’ ED: laoo Gems) X: 30.07 ZMA=O= Dy (6) — 2(2) -> D1=0.(a7 FL? /I 9/ ' ‘ (IBM'F F: F%4g. 25,: O rig; we +94 sm(e+45" cw v F =1.oo m )GLN) Hf}? c? (56 V X / F 10.00 Jon/Fla 6?: 97} i =0 = poo—Fécsmé lit:- Z-Z‘l ($045) lék“) F GC. Loo l m 6-25 Determine the force in members EF and EG of the truss shown when P: 35kN. V 17 o ‘ if) 2 MP :0 35(nz)+35(%)+35(4)— gammfiasfio new gas 250 kN COMFY -SS (4)'3‘S(%) + 1:5,:[603303(éggflmgdgygfl 3 ...
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Truss_8solvedProblems - (0'10 Déq'Cfi/VVWLC (33am, {m...

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