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Doc1statics Reviewer

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50 views16 pages

Doc1statics Reviewer

xgzfzg

Uploaded by

Katrina Celeste Revelo
Copyright
© © All Rights Reserved
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Part |. Equilibrium of a Rigid Body. Solve the problem as indicated. 41. Determine the reactions at the supports. SOLUTION Equations of Equitibeiam, ‘equation of equilibrium about point B by referring 1 the FED of the beam shown, inFiea GEM, = 0; 8005)25) — 44) = 0 Ny ~ 383333 N = 333kN Ams Liege rento wie fae stool ein sgt nyc sxnne asm (? 5, = 2008 = aan me stsp=o sss woa(!) 2-0 1333 = 133 Ams. @ 2. Ifthe roller at A and the pin at B can support a load up to 4 KN and 8 KN, respectively, determine the maximum intensity of the distributed load w, measured in kN/m, so that failure of the supports does not ‘occur. 4m SOLUTION Equations of Equilibrium. N, can be determined directly by writing the moment ‘equation of equilibrium about poiatB by referring wo the FBD ofthe beam shown inFiea eX My = 0; w(4)(2) ~ Nasi 4° (3 sin. F) ~ Nicos M0 Boos + 4 Na = 123766 Using this result to write the free equation of equilibrium along + and y axes, SX = 0) 125%6wsin— B= 0B 6188 +138, B, + 1.2376 60830" — wid) B, = 29282w Thus, Fy = VBE +B} = VES WF + 292K wy = 2.99290 eis required that Fg < SKN; 299290 <8 w <2673KN/m And Na <4kN; 12H 6w <4 WC 3QKN/m Thus the maximum intensity ofthe distributed load is 2.673 N/m = 2.67 KN/m Ans. wt) @ 3. The bent rod is supported at A, B, and C by smooth journal bearings, Determine the components of teaction at the bearings ifthe rod is subjected tothe force F = 800 N. The bearings are in proper alignment ‘and exert only force reactions on the rod. SOLUTION — "|| x wom equation of Equibriam. The xy and e components of force are F, = 80005 60 05.0" = 641. E 800 cas sin 30° = 200 N = S00sin 6" = 692. ‘Rofertng tothe FRD of the beat red shown in Fi, EG BD) RTO) o AMO BG) + CQ) e BM.= 0) -G(075)~ C2) ~ B11) ~ Mo1Q) = ° BR=O A+C.+3641=0 o Bh=0 +B+6,= ° TR=0 A+B, onw2=0 o Solving Eas (1) 106) 6, = SN B= 10718 N= 107N 8 = ON An C,=S339N=SRON A= ADDN A, = SON Ans. 1, and A, ate directed inthe senses opposite to ‘The negative signs indicate that C, hoe shown in FAD. Ame ¢,= ON = WN B,= «oN C= 56N A. = 400N 1. Thesmooth pipe rests against the opening atthe points of contact A, 8, and C. Determine th ‘needed to support the force of 300 N. Neglect the pipe’s thickness inthe calculation. (Answer: A~173N, c=e16N) SOLUTION Equations of Equiitrium. N, can be determined direcly by writing the force equation of esqlirium along the x axis by referring tothe FAD of the pipe shown in Fi. SER = 0, NycoeM~ 300sin3P = 0 Ny = 17321N = 173N Ans Using this resto write the moment equations of equilibrium about points B and C. GHEMy = 0; 300 0830°(1) ~ 17821 cor 307026) ~ 17321 sin 37(015) ~ Ne(OS) = 0 Ne= 41363 = 416N Aas. GhEMe = 0; 300.60830°(05) ~ 17321 c08 37(026) ~ 17321 sin 37(045) ~ N4(05) Np = 6820 = 692N Ne’ 020m 05m osm en Jou Ne @) reactions at these points 12. The member's supported by a square rod which fits Loosely through the smooth square hole ofthe attached collar at ‘Aand by a roller at 8. Determine the components of reaction at these supports when the member is subjected tothe loading shown (Ansiver: Ax=300N, Ay=500N, Nb=400N, Mie TkN-m, My=200N-m, Mz=1.SkN-m) oN ‘SOLUTION Farce And Potion Vectors. Ths coordinates of point and Care wT gos BOAO) and etd 2m aby = + nj ~ sau Ma = (Mah + (ad ~ (Me ay Nm tyo = 28 Equation af gl forveequnton oem (300 — A+ (500 A + (Ne tn) = 0 Equating ij and k components, 300 ~ A, Ans A, = S00N Ams. Ans. Ny = 400N ‘The moment equation of equilibrium gives EM, = 0; Ma + tan Nu + tac x F=0 pep deg & (Mit (di (dar i2 0 of+]3 0 -2]=0 0 0 4001 300 suo -400 [1000 — (ta), Ji + [(Ma), — 200]j + [1500 - (,)_ Jk = 0 Equating i,j and k components, 1000 — (My) = 0 (Ma), = 1000N-m = 1.00 kN-m (a), ~ 200=0 (My), = 200N-m 1500 ~ (My), = 0 (My): = 1500.N-m = 150 kN-m FRE sa. “The bulkhead AD is sbjecte to both water and sil teekfil premures Asuming AD i “planed” 10 the found MA, determine the borzon SOLUTION yeas of ivi Teo wn BC he ine ety GEM, = 0 tOIrS(2A167) ~ 2361388) ~ £6) F = 31398 N = 11N Am ARR<0 | A,+ 31S +2%6—10nS=0 A, = 460K Ams TEER @ A780 4, ~ TASK Ams Part. Stuctural Analysis. Solve the problem as indicated. 1. Determine the force in each member ofthe uss and state f the members are in tension oF ‘compression, (Hint: Use Method of Joints) Skip 1 weave SOLUTION Ay(40) + 15(4) +202) — 140) — 300) = 0A, = 1Sp 1842-£,=0 £,=35kip Bt1s-3-320 6, =45Kip Fy AM9EIp ~ 404kIP ©) ‘Ans. Pay = 4089062190" =O yy = 27ST) Am 45 — Fey sin 2.80" ~ 0 Fup ~ LID ~ 2Lkip(C) Ans. REP. = 0 Pup 35 + LIDeo 218-0 Fen = 195K9 (1) Ans. Join 2184, <1 Py <0 Am EFL 0 Fag 3980 Pye = 395 ip (1) Am. soins > itor, =o Foe=0 Ans. EPA Orie RISO Fue = TIS KPT Am Fyceus 4640" ~ 3052180" =0 Fre = AUB kip = 414 kip(C) Am. SABES =O Bag + Sen 21M + 40sinabaE — 1D =a Fig = 8UTBkip = 808% (C) Ans. Jon SEA G2 Foci 0 Fryg = 2454 215k) Am 21S n 21ND ~ Fy = 0 Fy = SIP ET) am 2. Determine the force in members BC, CH, GH, and CG ofthe truss and state ifthe members are in tension ‘or compression. SKN SOLUTION Sepport Reectiom. Referring to the FID ofthe entre truss shown in Fg, A, can te determined dicey by writing the moment equation of equilibrium about point F CABMy = 0: Sid) +818) +4112) ~ A(16) = 0A, = 825KN 3K, Ano Method of Sections. Referring to the FID of the let portion of the truss section Utwough aa shown in Fig, Rye, Fy ad Rey can be determined diecly by writing ‘the moment equations of equilibrium about points H, Cand O, respectively, G48My =O; Facl3) ~ 82568) =O Fe = 1.ORN (TY Ans came Fale) 9 +c - 1288-0 Fay = SV KN (C) = 11.2 kN (C) Ans, ety = 0: fn(2)u0 + 252) 0 fey =1284N(0) Am Meet di Ui te lh Fan aim ft eis san=0 (5V5)(2)~to(2)=0 tur= (sv8)anco sty 5V9) 2) teo= 0 Fee a0 even Ans Fae = WORNCT) Few = N2EN(C) Fey = 1L2S4N(O) Fea = 1OOKNCT) (Spssan ee nparoncen waar aurgmonae oS) SOLUTION Method of Jolare tn this ease, the support reactions are wot required for determining the member forces Joint A: aL, ral va * ts Fg ~ 6A2AN (T) = 646KN (T) Ans 2 3 Pac(3)- Fail) = 0 Fae= o 4 4 uc($) + Fal $) — oa Frc Fan = 300 ® Solving Bs (1) and (2) iets Fre 0% (©) Am. Fee Fao o ) + Fao{ $2) - sa $2) OVS, va, Fre + Fay = 1397 ® Saving as (1) am) yes ne = Fup = 3009 N (C) = 3708 (C) Am ® [san se Va Fae = AB0KN (7) Ans [Note:The support reactions at supports C and D can be determined by analyzing joints Cand De respectively using the results obtained above. Ame Fp = 646 KN (1) Fre = Fan = 150 N (©) ye = Figg = 3.70 EN (C) Fup = 480 KN CT) 3. Determine he ec inmanbers A, Fon 8, a att te memes ein aso of cameron ose BPP OCS COPING) o_o SOLUTION am Support Reactions. Not requited “Method of Sections: Referring othe FID of the upper portion ofthe tus section through a-a shown in Fig. a, Fyy and Fye can be determined diecly by writing the ‘moment equations of equilibrium about points B and F, respectively AEM = 0; Fy(LS) ~ 82) - 444) = 0 Fay 2394N (1) ~ 213 KN) Am GoEMe =O: Fah) ~ 40) =0 Fac = S33EN(©) = 533 EN (0) Ame Aso, wit he fore eatin of ei slong the we can oan Far area LF = + 3). a sansa 468 mu(2)<0 Fp ano ENco) ron 15m 2m @ “a 4 a m oN Te SOLUTION |——1s0 Tic CrsM.= e106) ama) + B,05)- asm 05) b.wt seamen COD Berm ke : a Ne = 266 6 +12P, = 6 = n+ sta 64802 = 6091 GrEMe= 0-61) ~ 120015) + HAIL 40) n OU) — Me= 0 Me = ~#23138Ib+ = —423kipe Segment Ds ute Ans. topo sf 2 Vo Eom =0 ON) ee e375 Ans.

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