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PROBLEM 6.2 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in fension or compression. SOLUTION Ereg body: Entire truss edi SF =0: 8,20 ee JEM, = 0: 005.75 ft) (945 IbX12 )=0 , el c=720] = 12 8 ——»? Cc + art HZ8,=0: 8, +4+7201b-945b=0 B,=225t| Free body: Joint 2: K 5 . N pal 0º E AÊ fes Ep =35Ib CA “E, Bus de Ta Esc =3001b 74 Free body: Joint C: ae 1 815 E =70lb Cd ob ap oo ETA Fa =300lb T(Checks) Fac PROPRILTARY MATERIAL. & 2010 The MoGraw-Hill Companies, Inc. AI vights reserved, Vo pars af this Mamial may be displaved. ed ow disavibunedl im aus form or bye amp means, without the prior wititen permission cf the publisher, or used Devand the limited ution 10 teachers und educolnrs permilted by McGraw-Hill for their individual course preparation. Iyor are a student using sis Memul, mam are using it without peratisston, 738 PROBLEM 6,3 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compressian. SOLUTION Free body: Entire truyss + ER,=0: €,=0 €,=0 Dem, =0: (92ENJGm)+ CAS m)=0 C,=-128EN C,=128kN] +HÍLF, 0: B-192kN-I28kN=0 B=320kNÍ Free body: Joint A: E He q TN RNA -400k . A Res Fi =ADOkN € 4 EA Fr =240EN 4 Fe B=3204M ree body: Joint C: cH240kN—O0 e = 42,72 kN FecLBEN TA «le, = en ENJ=1.28kN =0 (Checks) PROPRIETARY MATERIAL. & 2010 The McGraw-Hill Companies, Inc. AN rights reserved, veprmiinecl or distributed in amp form vm by amp meaurs, without the prior wriitem permission of the publisher, or user! Beyond the fnited “listribuion to teachers amd erlucutors permite by MeCGran Hill joy their indliviclval comese preparation. 4f por are a atue using this Manta, vor are noir it weithuna permission No pare of this Mental may de displaved, 39 PROBLEM 6.4 (Continued) Joint: Fur 240kips 1 4 Truss and loading symmetrical about € PROPRIETARY MATERIAL, & 2010 The MeGraw-Hill Companies, Inc. AL rights reserved. No port of this Menna! mas producer or distributerl à emp form or by att mec, without (he prior writtem permission of the poblisher, or ns distribution to Jeachers amd esitieutors peramitted by Merene-Hitifor vom eme using ir vwithont permission de displaved, ed bexone the Limited sir his Meme), r individual come preparation. df pon aro a star mM 311% Eips 18 ips PROBLEM 6.5 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION Eree boy: Truss Tostips Ing tips + X8=0 A,=0 E A, a 8% q EM Lee 4)EM,=0: D(225)- (108 kips)(22.5)-(10.8 Kipo)(57.5)=0 D=384kipsj EE, = A, =168kips) Free body: Joint 4 Aya |InStiça Ep &y a: er a as 234 ki . " AL CAD » — E Fo =357kips Cd Free body: Joint 3: 104 kips Fo MSkips TA Fon =10.80kips €C € OS kips 7 350 12 [e A+ Fep =33.3kips E = Skips T(Checks) PROPRIETARY MATERIAS. & 2010 The McGraw-lDil Companies, Inc. AlÉ rights reserved. No part of this Mental ney be displayed, reproeduced or distributed bm cny form or by emy means, without the prior iwritiea permission ay the publisher, or used beyond the limited distribution te ioaehers atul edmeators permitted by; MeGrew-LHIl far their individual ento se prepatrarion, If you are a stutslem nsiry this Manal, sou are using ie withon permissian a PROBLEM 6.6 (Continued) Fe + 2025 N-675 N= 0) 22580N Cd 22580 N Pe 2 (2250 N)-1800N=0 (Checks) 378 TARY MATERIAL. & 2016 The MeGraw-Hil] Companies, Inc. AIL cights reserved. No part of this Manmal may he displayed, inded im camp farm or hy any meets, awithout the prior iwritiem permisstom af the publisher, or used beyond! the linrited Hit for Mhcir inclivicdual comse preparation. 4 xou are e studesmi using this Manmal, PROPRI, reproduced or di alistralticos to tecichers: ante ediscators permnitted by Me using él vital pes niesion vou are 4a PROBLEM 6.7 Using the method of joints, determine the force in cach member of the truss Free body: Joint 4: A a RN o “o spo! é E Fo fu E BUEN Free budy: Joint Es Tac U,50AN +[ EM, =0: 13.50EN-DD E, = Ego CL BAN Fop <+15.90 kN Ep =IS90KN 1 4 ilus + 18 — shown, State whether cach member is in tension or compression. 25m SOLUTION Free body: Truss +lz8,=0: 8,=0 x “dem, =07 45) M(SARNKASM)=0 ; D=-B4kN D=84KkN + + Br =0: B, -8SAkKN-84kN-84EN =0 í B,=4252kN B-252kN— Fa o Eu 84 BN E =ISSOkN CA Ec =I350KN 7 4 —Fye 84EN -28(15.90 kN)=0 53 Fm = T680KN Fr I680KN C 4 PROPRIETARY MATERIAL, 3 2010 The MeGraw-Hilk Companies, Inc. AI ights resucvel. No part of this Manual may be displaved, vegrulnend or distribeod im cny foram or by amp amenns, iwilhont he prior serisrem pernnission cf the publisher, ar sed beranel tho limited distribution to teachers and educedors permilfeel he Me Graw-lNiH for thétr individlacd comrse preparaticos. If ya are a student using this Mamual, te care sin E ic prermissiom 745 PROBLEM 6.8 —— ft eusd 1 Using the method of joints, determine the force in each meniber A of the truss shown, State whether cach member is in tension or TA compression. SOLUTION SRA Pê medo af 4 ER 673% | Je 4 & z E É | ap= Palin “Va e g o) - 3 Bopeliie mn Ee é Reaction: 2 =0: Do=0 DM =0 DCUO-(693 ST) =0 D, = [651b | +Íz 0: 165]b-693b 1 E=0 E=S28b | k - + EF =0: 4 q = Joint) E mi EF, =0: 5 Poe =0 (1) ftio p s/ £ 4 ais 4 dos,=0 Le, 42 EA 6sib-o 2) ( £as 13 5 DL De Sb Solving (1) and (2), simultaneousiy: Esp = —260 Eb Fa =260b 4 Foo = 125 Ih Foo 125 TA PROPRIETARY MATERIAL 6: 200 The MeGrow-Hitt Companies, Inc. All vighis resceved. No pere af this Mame may he dispiaved, reproducad or distriluded in amv form or by emp ancans, wilhout the prior serittem permission of the publisher, or used beyond the Tamitedl distrilnattom to teachers and educarors permitted by MeGraw- Hi E for cheir individiral conrse preporation. If you are nu sindent arsing titis Menmual, xom are using ivithome permissicm, 747 i PROBLEM 6.8 (Continued) 5 + EE-0 Dh+ 13 Br HER =0: Er vis men he “Be = x a = Ez rest E pr reoto Lg root e Force polygon is a parallclogram (see Fig. 6.11 p. 209) - 5 4 . Joint Á: A ER=0: 1560) 1 (400 Ibo F Fm 420 1h «sy, 12 3 (260 1h) = "(400 Ib) = 0 13 5 0=0 (Checks) PROPRIETARY MATERIAL. 2010 The MeGraw-Hill Companies. | reprodaced ar eostrilinterl a cm feria or dos cy mecuas, weithor the prio distribution te teachers and educators permitted fr McGraw Bill for tiscie inais dem are sim dt veithonta permission. 748 AM rights reserved. No part of ihis Mental meg: be displ inte permission af He preblisino dual conse preparation. ff pow are ca studems using this Manual, Em e832b CA FM 74 Fes MO 14 =125b 74 Fe Fast € 4 am seserl beyond Mhe limited PROBLEM 6.9 (Continued) 3 sun 105=0 pSTSGEN TA Be = 10.5 4N +Eqursoy-44 625=0 Ea 44,625 EN há 30,625 kN Freg body: Joint F: DE isa zero-force member Eee a AF E Jo u25 km dE Ea Truss and loading syrunetrical about € PROPRIBTARY MATERIAL & 2010 The MeGiraw-Hill Companies, inc. Alt righis reserved. Av part uf this Manual may bo displayed, reproduced ar distributed io mp foras or by amy mens, withanet the prior written permission of the publisher, or axe beyond the Tinuted distribution to teachers amd edncators permiticd by MeUraw-Hilt Jor their indivichual cmuse prepematicm. If van are a studena Hist chis Memmal, sem areorsirg it withend permission. dSm sm PROBLEM 6.10 fu Determine the force in each member of the fan roof — e truss shown, State whether cach member is in tensiou 2m or compressivn. SOLUTION Free body: Truss Tee body: fruss bm Sm 15m 25% L5m 18, Eis jet (o Z7,=0: A,=0 Y Etom symmetry of truss and loading: Es A Erge body: Joint A: F LAB = 9.849 ps =I231kN CA Fa NZEN TA 2kN :5 2. uh si +, 28, = (12.31 KN + Tg + bic) = O o 9.849 o Free body: Joint 8: um “a E or Em + Fac =I231EN q) +20, = ass IBILANA+ Ep é Fu) -24N=0 or Ego Ee = 1386 EN (2) Add (1) and (2): 2Fy ==19.70EN Fay = 985EN CA Subtract (2) From (1): 2F yo s=4.924 EN Fu E2AGKN CA PROPRIETARY MATERIAL. & 2010 The MoGraw-Hil Companies, Inc. AIL tigus reserved. No pari of this Mammal may be displayed, reprodluced or distributed im ame form or By emp means, avititont ihe prior written permission of he publisher, or used beyond the fimited distribution to teachers and ectucaiors permite bp MeGiraw-Hi for their individital course preparation. If you are a student using this Mamuci, sos are using dr withoue permission 751 gue E i E ETA 2 EURO i i a PROBLEM 6.11 Determine the force in cach member of lhe Howe roof truss shown. State whether cael compression: member is in tension or SOLUTION Freg body: Truss sr Because of the symmetry of the truss and loading: Free body: Joim +.4, E =1500lb (4 Fr OM Td s Fx, fee, E O Es TA hMobib C Ec PROPRIETARY MATERIA, repreducer or lisiriburee! im amy Jerim or by cy arc distribution to teachers und ecicinves perimitted bo) era asi de vvithaut permission Hull Companies, vs, vettlene her felGrenw Hifi tor their cota ceu ty DB RC À Hab GR F. + sh HA, é Es] no H =0 e 900 Ib inc. ALL rights es Net grart Of his Meomal mas be displaved, tur wridtem permission ef the publisher. or used beyond the fimrited indivictua! conse preparation. If pon er sr atende sen his Memual, TSA PROBLEM 6.11 (Continued) Free body: Joint 8: + XFr=0: Arm sm +2 500 Ib)=0 5 5 5 or Foo + =1500 b (1) tzr, =): 2 F, 2 Fo + â (1500 b) -- 600 1b = O [db = lo ola + do or Fo — Ea = 500 1b e) Add Egs. (1) and (2): 2a, =: 2000 Hb Fry =1000lb € 4 Subtract (2) from (1): 2Ey =10001b Fog =5001b C 4 Free Body: Joinl O: 4 + EF, q 1000 [6955 for = 60n tb] EN Fou =—1000]b Ee =1000b CA E. JÊRE Ho 3 For qr Hi 2H, 07 (1000 [b) ->(-1000 1b) — 600 Ib— 75, =0 oca |b he Er 5 5 Pe Egg = 4600 lb Fog = 6001 TA Because of the symmetry of the truss and loading, wo deduce that Fer =5001b Cd Fr = 1200 TA FT 0 4 Feu =1500]b C 4 Eu =I200lb 1 4 PROPRIETARY MATERIAL. & 2010 Ehe MelGraw-Hill Companies, Inc. All rights reserved, No peré of this Manual pay be eispluped, chistributed im any form or by any means, oeittcut the prior written permission of the publisher, or used beyond the limited distribution fo ienchers cond educalors permiicd by MeGrem-Hill for deir indiviclnal emerse preparalion. Jrorr are a sitiden nsing this Menna, vou are using it ivithout permission. 754