CE270HW3_S08sol - CE 270 Spring 2008 HW #3 — Cable...

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Unformatted text preview: CE 270 Spring 2008 HW #3 — Cable Structures Due date: Friday 2/1/08 Problem 1: For the cable ABCDE and the loading shown below, the value kg = 7 m. Given this information, determine (a) the reactions at A and E, (b) the distances kg and 11,) , and (c) the maximum tension in the cable. EVA 5m 5m 5m r "I" 3kN 00 i." 2Fx=o '1 R‘A ‘1' RXE o'o KxA= “RXE. +1‘ Eta/=0: KyA + RYE -' 50921410) = RyA+leg~°llcN +3 EMA“; = ~3mL5m) as“) (10m) —* BMLIem)+ R15 (20m) + Rxelipm)‘ 0 c 20 K1; + UKXE - 010 lam-m. TSC. +52Mccv: RYE(|W)~RXg(|m)~3kM(‘—DM) ‘m E to RYE -—- Rxe -‘ 15 mm = 0 C ’ [2:21 10KY5‘23¥E‘30tH-M=D 3"” D 3““ subtract 7/05: ~l~ Maw: “4094” = 0 {72'4" “62345 + bokd-M‘JO RXE‘; '75 {CM 0a 20 [ATE + Lo (15%!) "to KN‘M :. Km: :15“, 4.. RYE 5 1 Ky». 4— 2-25 lLN -—- qLN’O I. lax/A = (0.16 Ian) 1‘ commas —> CE’L’LO MW 5 PrDblen/u \, mm. b.7SkN b) 1.9 ‘ UMB=D= ’LSLM (hB)-*lo-7SKN(SM> KN A = QIEM‘ 4.525%; 0 = 2.7,9\c~(5m3 ~7-Slm (hog? k a To» W4 . km; : “5m .1 149:: km; 4- Ian --= {orrH— luSm=75m 7" ngm hp 2: c7 Maxlmwn Ramon. +yPiCallut m“ Supper-1’s WoK 0m Les 3 GAB-=+M4(.%S—rng‘)= "m00 b "W‘s em = 1—M-l No.7" gm 9.19 w TEN 9A5 11;. Z Fx=0 = ~75 m + TAB cos 42.12O TAE TAE = 12:“; r 104 KN cos L’erD" mammum Jrensm = [DJ {:M (I?) AB) Ibmmcw.» $945 1‘, 682:??? éoLOhiw» ‘Pmb 2 . :1) P253) 3+ a. 6 fib\"~—-‘*" m H6“ "949‘ l 5M1 : _. I 4| — W a, 5,949. ' J7 [50" 51 WW mfg-=27" / wlwfiwum 42—" =0 ’ @37: 1%? 7}: [X Eogmfla T); :po/fz-(zo‘s 26,51);- ~\80 2N, “F: 108 KM 73 =2o/'2 (5m .256?) .1 ‘ro ZN mg” Problem 3: Each cable of the side spans of the Golden Gate Bridge supports a load w = 10.2 kips/fi alo from each tension in Not-c: ng the horizontal. Knowing that for the side spans the maximum vertical distance h cable to the chord AB is 30 ft and occurs at midspan, determine (a) the maximum each cable, and (b) the slope at B. TB 98 w=uo.2*"/F+3(H 00H) H.220 Kips- +j ZMA = o = ~13 603195'CLHUH) + T3 SMQB ([100 H) «(than K)(§$0Ft> H 1350195 *LfiQUTBCCéBB ': (élflio .- .- hc = mg +30% = 276% 2 +32Mc=o = Tgsm‘EB (550£+>~Tsc«>$93[278¥+3 —- 5WD" (2169+) 5,5T5 512393 —« 2.12Tscos 9:; ‘-‘ fSHZ'Ti‘E ll T6 5 ' 98 " 5.5UTgcos 93 :- 30,8 '59 H T 51?: 93 “ Mu Tacos 93 = (olle 0427860393 =- 50,895 T3w593= 51,425 :» T55m95= 237 ore '4 POW” A 3 Wefore cam/wt“ be. used meow ( ' Xa, We , Xe,i§g,,erc..) MA use. all awhbflm cosmth as, abate, ls m batty»- approach“ Hawaii; $3 , Cézf0 40/qu . ‘ 4" 97w —;. '2 #5172. ' x: '73:“. :1 2143?):«9' / 7Q 4;??45 1647/2, A #455??? #5 Summer 35 :15539’9’74/ - 9455250 g ~.—. WWI? {’2‘ my ave/5.1mm, =2. (21711.57): 4555, mm A :2 435(,/g)»43n.m a» 3.75% 5mm ' 5 Problem 5: A counterweight D of mass 60 kg is attached to a cable, which passes over a small pulley atA and is attached to a support at B. Knowing that L = 12 m and h = 3 m, determine (a) the length of the cable from A to B, and (b) the mass per unit length of cable. Neglect the mass of the cable from A to D. (m kjl‘i-S 13,.) = 555 N DC‘lCrrm/‘Ie a. (my a+ 8) fix): c [Cosk(%)~ I] 3m: C[w5h(25"1)»1:1 gnu: CoshUQérL) 0, 1+ J2. a) 5(x)=C5r‘nh(-%) T; i; cf-El-h-‘léfs—l WM“: 1 1:2 1:05 2 (545'? 5'“ “ (fin—g7") 55 1-375 [.201 Uf , _ (J, [,8 [.843 (9.5 mm: 1.457 b) T MA = 555M (since odoLc. 0]an— om Pullafl , .5.— Apfiggw Tu» magma) W 593“ 3 H Cosh é-LfSM 5:? H = 401.4 N c=iL (A) (9.45m = UfDloLl’M' bu Problem 6: Two cables of the same gage are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the resultant of the forces exerted by the cables at B is to be zero. Determine the required sag h of cable AB. ‘EXB: sme ‘er‘ CnbLeBC FLO la lo 2.190 loot? 1' '85; $00 1:04 [toal 500 L024. [.0291 (000 1.0 2.0 1.020 U96 C5 600 For Cable Pvt; Yéx): Ci:th (%)-*l] h = (6mfiicwsh("fi12309)"l] 1*qu SW34“ Cable AB. ...
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This note was uploaded on 04/27/2008 for the course CE 270 taught by Professor Ramirez during the Spring '08 term at Purdue University-West Lafayette.

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CE270HW3_S08sol - CE 270 Spring 2008 HW #3 — Cable...

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