Hw1 Solution - Prob 1.4-8 A weight W is supported by a...

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Unformatted text preview: Prob. 1.4-8. A weight W is supported by a cable attached to beam BE at E, as shown in Fig. PIA-8. The beam is supported by a fixed. frictionless pin at B and by an inclined cable from A to D. (:1) Determine the tension in cable AD, and determine the reaction force at pin 3. (b) Determine the internal resultants (axial force. shear force. and bending moment) on the cross section at point C. P1 .4-8 Solwion ‘- ta)+cmion 0¥- cable so ovt' (tad-10H 'n' : m: beam . PP EGAN—Bum LEFXSO: BfiF'fis'TAD so 4 sz'l'f'rho +4555“ 5W '2" TAP -w=0 (mime/had resulmwls pap: laws—ham beam semen EarU'liibViW I LZPX eozay +Fceo Fc= “ex—W 3 *4‘Zfi$=o= -VL r, O VL5—W i- 12(ZMlc‘ 0‘ BM?) ‘Mc = O Prob. 1.4-1.1. Beam AD in Fig. 131.441 has a frictionless-pin suppon al A and a roller support at D. The beam has a linearly varying distributed load. of maximum intensity w. {farce per unit length). over two-thirds of its length. (a) m H C Determine the reactions at a and D, and (b) determine if the internal resultants [axial force. shear force, and bending U3—+—u3 U3 moment) on the cross section at B. [14.11 and P1 .442 52%: 52M 3 kfil (4} etecf/Oaf fi/fifiwa/fl- ‘57 i I W Eel-(‘A'Jrr'mm o jc’fim 59.. 4-)éfffn4/ Kerk/{1026' frr'fl‘gn'um of Kram fflimdu/ $2790 éffltfio 6:0 so .244 ’4? r {z #Vé ‘0 van-flu Prob. 1.4-1. Each member in the pin-jointed planar trusa in Fig. P1.4-1 is 6 ft long. The truss is attached to a firm base by a frictionless pin at A, and it rests on a roller support at B. For the loading shown, (a) determine the reactions at A and B, and (b) determine. the force in each of the three members, labeled (1) through (3). 33/6425}??? .' (a) flfécfr’oru' 4/1? atria/6:: Jjélkbr r” fl; r 761/ 444: r 5} = +4.70 éw r (/35) fixr'fi/ féfr-I r'a mamfi'r—r (U '6’). :3; ‘42??? war 6=fi+¢£4§v ff: fiféé’éWz’rl 6: 445. £02: [a] +T P1 .4—1 and P1 .4-2 aX/r'én'am .' 51-»; F559 9/ m/m firm, +( {2'qu r 0 — 344w ( 6ffJKrr5 66")— ééaér/é/J’F H3 MW 4; -' iffy £911.! $- 275 c 0 a; n’éo'u =0 19:: = '5? +7“ ‘56 -"0 ,% fl 6, 4+; 1- fi; = o I?) = A 70 / MW Ezmfl' 031»: 5 him Fifi Jr" 27%;"? M?” 1/154 Kara; +f(Z/W/y :52 gKf/HAH m .t is; (4,?) =0 1?:-% #454?! Zr? :0 é/ffln 40") — :54ng ~03 - O 6" /. 764 1:}; Few. Fm) afflm/fi) €018 60") 1t 5, =0 —: 4:; “fig 455;, flE/iw we: *fiz’ww‘) «2 5- 2. m 4!» Prob. 1.4-3. For the pin-hinted truss in Fig. PIA—3, (a) deter- mine the reactions at the pin supports at C and E, and (b) detennine the axial force in each of the following members: F3 (in member AD), F. (in member CD), and F3 (in mem- ber 3D}. fi/Qéloq " W W Phi-3 and P2.2-13 f fiI'A-Jr/‘erm a (3/3}? flu.” .' F{(Z7’?)C =0 ZéN{i?~m}fjifi/{EW} ff} (who 4? “flabby 19273; =0 [ME-o mama; 4‘27; to warm/=0 fgmfi'in'um a)“, {la-all :5 .' L25 :9 first/6 =0 5- 75:5 Hfgeo J r 513%»; =0 5;- rf/gjfizmm Lj jfwéfl t;— Amo M! g 5. F" fgu.‘//Jn’un cg {EH31 [/95 3 {— +756 ‘0 ‘ZW'§6 =0 5 = *a-éwp-zmw 6 5 ' EEflXIA-Jfi'f-Mm 9F {ha/5! x—g Frag} w awflgw €=—£m{fl If. C} C- Egg/A‘o/H'wa gF {lei-l! ’1 K _ r x; c aim «a? 74 ,5 = may (a fig: 2J0 Mm 25; = 3.00 44/ (t) Prob. 1.4-4. The pin-jointed truss in Fig. PIA-4 supports a ioad of P = 2 kips at joint C. Determine the axial forces F3 through F5 in the truss members. fa/a 519/2} flfll‘P/M/flc firK/a/ fé/Lgf 5 IL. I D a» 0] 5: A #7 M; m 6.- Z,00 41/“! {7! g: flé/é 49% {5} g,- /,/H ér/«x [75 5, .— /./41;> £03; [7! {91.44 and MEIER éZMr/r'gfl-UM cfz 514/724? Zémrrf T $27.. so a :0 24% flat/I 'DJM/z‘] =0 Jr, 97 ’ fl /¢’Z? 4?” 0; +FCZ—M/g 2 o a] my; 1%,; {who 5'; T 0-5577 Apr fk/flXr/‘am a hiya/fl; 5 +727;ch a} # 3:6 =0 2? —M‘Zfié AW 925: to 5 $90 6 ‘75 F. = #42,; MA. 55’%/@) Lip/Aim}... 0%j/‘01‘I1 / (’.' 277; :0 5:5; ,- / /4Zf/n//;ar E650 5’ - 54/“ figuflén‘m 0; {WM 0 : wig-.0 IFKfl/Z)*0y:0 53 ~ 170; = -/.é/62 Mu Prob. 1.440. A unflame distributed vertical load 19.] = 3.2 erafm acts over the entire 25 m length of tho Horizontal mambo: BC of frame ABC, as shown in Fig. PIA—20. The flame is supp-med by a frictionless pin at C and rests on a rollcr support at A. (3} Datermine the mactfions at A and C, and (1:) determine the internal resultants (axial force, shear force, and bending moment) at arbitrary cross sections of members AB and BC, using the rsspective notations shown in the two cutaways. fi‘éfi’é‘éfl.‘ 5?} flrécfifw; Q/fimc/C’. r KEN“ rum/m [a A'Jnd/fe’ffl/f‘éw [If fin we = 3.2. thn 2.3 m PTA-20, PEA-40, and Pad-24 Ezuxfl'én’um df [JP/amt *((gm)c=u (43%“ fiéflénf/Mrn} =0 flydflu Hit“; =0 {1}Jr -6’£N+ (.3 =0 Q, - mfg/u” 0 {f} :0 fl; vim I"r are, 5,: HINT ffur'A'é/‘I'MM car/f man/r9 .' Militia 6'; 3i/gflcfl}:0 gbzmgu “73'va %(£1féM)-V,«z0 Mama: (K r Wm Me ME = 321’ r in; Mm; 3M: 1,4 JEAN/2,91%?) 1‘ mm :5, 2195.2,“ (km )5 :0 ML “:2 rz.:-xl)’{£)—rrJ’/z.f—X=J =0 “ ZN" fzfin *Kéxal {’W'M ...
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