CE270HW2_S08sol

CE270HW2_S08sol - CE 270 Introductory Structural Mechanics...

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Unformatted text preview: CE 270 Introductory Structural Mechanics Spring 2008 HW #2 — Trusses Due date: Friday 1/25/08 Problem 1: Members AB and BC can each support a maximum compressive force of 800 1b, and members AD, DC, and BD can support a maximum tensile force of 2000 lb. If a=6 ft, determine the greatest load P the truss can support. Use the method of joints. B LL5H 1.59. $2FX=O :. AX=O +TZF =17: Ay-l'Cy—‘P +j2lla‘0'; C7029?) “PLGF‘I’5 I. CY: ~E' Tmfis FAB‘ T FAB = - 1612; FA? F/Z 4r? :0: FAB StnqSO+FAD SW1 “+0 + P/Z- o :51,ng FAP)SV3L{E° + FAD Sl'n H" + P/z 0.128 FAp = P12 F89 FAv : 0-1001 P FAB = ~1312Lae‘if’) pig ‘ ‘ ’ mew—0324? F” FDIC, +TZFY ‘0 = FED - ZFAD 5‘93 l‘t°- P l’ 0 = Fa» « 2mm P’BSMHD’F‘ “ 5' F819 = LaaP FAD“ FPO ( symmw) Fkfi : F c :- ~‘0.01L{P _ Ooolq P i . B compresstm- P - It . FAD: Fm a 0. LM P 5. 2000119 Conwg PS- Z‘Bolfllb 1:39 a may? 5 zooolb lPs ‘85! “0| {2 s:- LSOLt lb- Problem 2: Snow on a roof supported by the truss shown can be approximated as a distributed load of 60 lb/fi. Treat this distributed load as you would the self-weight of the members. Determine the forces in members BD, BE, and DE. State whether these forces are tension or compression. loH w (avg-b; we) KR = 300% +0 M659 éam‘l’ each 801A b°°”’ i, 2 Fx = 0 = Rm. :. Rm =0 H‘ZFY =0= ~300w- BlbOolb)-Soolb 309" calla “EVA + FYH A RX” 0: EVA-Ham "Z‘loOlb. T m e H UZMH :0 = ~K\/A (324304- Boolbtazfi) y+ MA Y “900% ( aqfi+>+ booleHH boo M8430. ‘3‘” Exp» : lzoo no ‘i‘ Rx“ = [L00 lb f 5 0H - - 8 Easy}; 3 2 ME :9 : (a cowtwfih 0,001.9an 'lZoolbUO-F'l) avo'b m — Fed—“gum - Fan (13x 8% lb ‘1’ .E 0‘ ’élboo’ 5M» Fap i- FBD= "[000 - ’I‘ 3 3 - lZoolb he +1‘ ZFY=0= Fan “‘63 " FEE (75)” $001k “00% 4— [200119. 3 0 = L’l000)(%7’FeELf5)+ $00k- p Lboolb V F8: ="500 Up. F Mb FDF 1* 5Fx=0 " "FBD‘P FD’F :. F39; FDF Bo +1~ 21:7“): -L,oolls 4:95 *flooolb)(25-)LL). FPE '2 (young. Problem 3: For the given loading, determine the zero—force members. Justify your answers using appropriate joint free-body diagrams. 400 N zazo Fame mmeaas me. 3L) BK)ACK,HF, FL,IE) 53 Problem 4: Determine the forces in members AC and BE of the truss shown. State whether these forces are tension or compression. (Hint: use section a-a) K“ 5 i5“ 15"” 15“ SM) SM 40‘ C F FY :«2‘ . . Km . m- ' ' - 1 2m . . v AA :5- B ,‘ E H L3amJ—3m‘l‘3ln * a x :. FM: [5 4w? +12FY =0 = K“ — 3st) I 325070: KxA—que u RXA=-Exs var/two: (“SKN)L3m>-'Slm(bm)—SkN6m3+stl‘tm) KXB = --P MA = —22.s I‘M 4— 11, in =0 = FAC‘I' Fe: + 225m «zzspx O " FAL= w +32 Mp =0 = e mum) + name») Aswan) +£22.5ch )L2m)+ g, (Zm). (9 = C’FAcX‘i m3 4' Hg LN‘m -FAL FALS- H-25 KN‘ I’d : "ths FAG: 3T Fa: = “.25 NBC; can; alSo “VJZME =0 = “FM, (Hm) "' [SFN(3m) +225 #No-i'm) Ll FA; = ‘45- FAC = [1-25 k—N ,T, Problem 5: Determine the forces in members GF, GD, and CD of the truss and state if the members are in tension or compression. Use method of sections. 0: — Zloolb (%)L3£+) + Fbc. NH) + Fpo (fiflgxc'f’) 10120 lb'P-I' = 5¢8€5Fbc Fee: Llasno. l + T ZF‘] =0 = FPO "‘ F951 fizbo WOT?) 0 s qflslbbt'fz) "* F9669?) ‘ “0”” wa' .‘L _. .15.. «F $25K“) = ‘FDC.(Z%F;)"FD&1(5) Z‘oolb(l3> Fe. 0 : aqq‘slbclfivl-Z) “("200lb)(%) —- lOOibfiFFa Ffie‘ = Pap: 2/00le C F09: qu'filbfl'. Problem 6: The tower truss is subjected to the loads shown. Determine the forces in members BC, BF, and F G, and state if the members are in tension or compression. The left side, ABCD, stands vertical. The support at H is a roller. 3kN (1 RN 6 kN =0 0 = (—am)t8m)~ Lowtqm) ~21<~N (2-54”) " Fpa (Q'Dm) D. 4m~2~9m= L'Sm. Fm (4.02,)t‘tm) "5“ : 0.5m Unmyfn 5‘5 lcN-m = «am Fm 3 X “t W“ W FM” + tzrfl =o= «zw e F5; —- radii-3* Fm (5‘39) 4 0 = ’ZFN ~ Fee” FBF (3g) -‘ ("'53 “xi—5‘5) FBL: a D'XFBF ...
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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.

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CE270HW2_S08sol - CE 270 Introductory Structural Mechanics...

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