HW12 - 9.1 through 9.4 For the loadin 3 shown, determine...

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Unformatted text preview: 9.1 through 9.4 For the loadin 3 shown, determine (a) the equation of the elastic 1 curve for the cantilever beam A B, (b) the deflection at the free end, (c) the slope at thcfieecnd. >5 7!. - mo zwc “Swafidx = ‘5 T“ _L‘JO)CIX -—- ~%xz + wax + c. [x=OJV=O] o=c3+0+ci .'-CI=O Mo 2. MC?!) = = 5(_TX + woxhdx wt, 3 L4,, + = - + —‘r&. C. 3L y‘ “*- [x=o,M=o] c3=c3+o+c1 .'.CZ=O 6‘2 — — _fl 3 £2 1 EIEZE - M — .5 .1 X m” 4 u 3 Eidx- ~Ti—Lx + 6"): +C3 3 3 a _ woL w,L b“: 1E1=01 0’: 12 + (a +C3 3 . awoL -- C3 ‘ :2 “a ‘1' £10. 4 _ MOE EIY _ ‘60Lx + 2+“ 12 x + (1" 4. ~_ _.L'-JDL4 wolf _ wal— EK=LJY=01 O- ZT+T 1.2 +C4 . 7waL.‘ " C4" :20 ' “J0 S 4 4- __ 5 a.) ELAsTw. mama: ~1* “IZOEIL(21 SL-r. + IOLx 7L ) a 4. 7w {.4 __ 7w°L 7.. = = + 5 _ a b) w A 12951 7" IZDEI. 1 3 a d‘t _. . _ _ Wal— waL. 4 Jié’x‘o' chant 17.2: 9A .251 q; _-__——‘—\—__ PrOblem 9.7 For the beam and loading shown, dclenfiine (a ) the e - for portion BC of the beam, (2:) the deflection a1 quahon ol' '- midspanl (6} “I: “q. Using ABC as a. ‘Fr‘ee'. food-Y) Civic: 0: (%X%) ‘- RBL + LwLxggh RB = ng F01" For‘l'lkon Only) (0 4 25 < DEF/5:9: ‘g—L 1;“) -g“wi>< + W13“ mm"?- wLxs— J-wx" —§15wtzx‘ 4 0,): + C‘ o—o-o-o+O+C,_ C2=0 o=(f5—fi-fi;flwt“+C.L+o ClumLowL‘ 17%}(fih3 fif—fgfix‘ I:L3><§ an fi=é(fi-sz—gxs~J-L‘§<~figfi) L‘ L" _L 21.7- a}: “03 yew—'35:! 3w fii‘rflm #1:) "29L(-1)',3I;L(931 a “Mil. __I_ _..L __I_. -_ IBWL" ‘ "5f So 3H 9o 2% H20 53::J _ IEWL .Yn‘moEIJ" 3 .. . 5' __ 3L _ _J_.3 __ h W'L (c) fiQx 0 EMS. EI(0- o 0 ML) 1703 e- ”“3 <15: B [2061: 9.25 through 9.28 Determine the reaction atthe rul moment diagram for the beam and loading shown. 18f Supmn w - Problem 9.27 REALT\or-J‘$ ARE 51px“;an IHDEEKM‘HNYE M3 +¢ZF1=o: ak+aa-:= Rh Re, +5214! =01 "RgL+ TX?)+ 3:0 MB=R*L_..%I (rifle-'3: gz—Eflx V=Rk"t‘—:3¥- M=RAx—§J—:-13 51$: ka— 335:? 51% = than? ~ $331“ + c. E-L‘r = 167%,? $1545 + cm. + (1 M=P~Ax—“1:L(-% Eli—:1; = ka _waL(_tl{-‘K ‘fiQ 51““ = 41Rsz — Mal—(TEP‘Z - fiLK5+ C3 E'L-T = éRAxs — w,L(Ele~3* fiaL-iz) + 53% + c4 [x=o,1=o]: ozoflo+o+c1 .'.c2=o [Kr-LIg—O]: gag} woLffiL‘fi—lfi) “:3: o _- C3: - ZRAL + $144. EH, v=01= zRAL *wol—Gzl—a $6) +( é-Rklfq'wfixm + afiwofi:o RA: fiwoL 1‘ RB= “3L —%uoL RB- :20 w,L+ MB: fi—‘oufi— 3;; Ma: -g‘Su‘J. = —0.0354rw:,13 4 OVER. Oéx-CE: V=RA‘%RZ=‘I%%L*%KZ v-_-o m- x =~AM= oficollaL 74 “w 3 M: [—52:ng if: 1W: “(F03 = o 2 | Me: M(x= % = 002400.313 M": Ma: cam-3.3L) = 0.031‘IwcL Problem 9.29 9.29 and 9.30 Determine the reaction at the roller support and the deflection at m- C- Int W M“. iéx< L f) M - Pnc- Mac-:1: 1i'(%-E3=‘ Robs—1W5? R '- ol“ L 1 as w a ELK-iii: pr—éwQ-aj IWDM Eli : % Axh—g—wcx-w +c3 2 a? __ E‘EJ _ JéTRhXS'ifiWG-iyfi 63% +6? ‘ V [15:0ero] o+o+£12=’ 32:0 _ &\ A. '1' z Iw-aafiw ware. — gimme eye. Eva,wa dawwcuwmar—micaum a <23: 0 DH; i=0] %.RAL‘ — #:st + Cs= o 0.3- —(&R.L‘—%gwt’3 BrzLJj=oj JéQ‘LS 31—Hqu (fifia—qlslejL+ O Rn-%wr'_1\ ~43 CI : C3“.- — % WL3~T4I§WL$) '—" 'i‘EWLS -2 _ _. 3 a A+ X' 1 w éRfi +(‘7563W1L3 g. + O = ‘Glriq ML? _ :3 M." ‘Yfi'em E ‘1’ d. Problem 9.45 9.45 For the timber beam and loadin 3 shown, dererm' the deflection at the midpoint C. Use "1e (" E: L6X 106 pg )lhcslom' 2 Rip: 3.5 in. 351’} Ih’ft 3.5 ft 135 Ft 1.?5 ft Uni+sl Forces in 161953 QEMJ3 = o: jaw-3%: .5 H. — 7 Qt +(2:(5.251+(1.22§)0.7s) : C, 2 L225 F?n = Lsoezs' Laps w(x) I 0-350 (x —3-.5>' Q“ 1:21: fl 7 — W '-' -O.35<X-3.5>o a? = V: 1.054925 - I(x-I,7S)° — 0.356e—34- Eigfifi = M : r,soc;25x — 2<><-I.75>'~ C:=.I75<><-3-5>z >' k-f :1 _ a 1 ! tffi = 0.903115 Xz— J<X-L7S‘) — o_05333<x_3_$>3 +Cf kr {H HI I T 0.30l0q2X3_ é<X-L7h—) -— O_oi5¢5'83 <x_3-5>‘r + Cl x4 CL k1, h. I [x:o;J"—U] CL: C) ; 5”": 3w] (0.301042%?)3— §(s.2s)3- O,ol~+5€3(3.5)" + elm + O r U CI = — 7.54??? k.-P_ 42+?- %: E7 LGXIO‘. F5" = Lgxtos LEI. I= fi(B.S)C5.s)’ = 142.52.; m 5-1:: (F.6rro‘)Cu8.§2é)-‘ 771cc“? 1694;“ __ 53%;? 10.13%; (0-) Siege a'f A. 3:0) J _ EIJI- O-O—O ~7.5477? klpgfi J 94! r " ZSHTH : —H.oo K104 rum! Qtir-Looaqo mgr]; 4 53%”: A an} De—Leerlfout 03* C. 0’1 X? 3.5 Eli : (0.30:042)(3_sf - in.st ~ 0 (7.5477065) + o = — I529? M‘s-1P1! y: -‘ — fl = —2€_37H.{0‘5 'F'!‘ 3:. = 0.340inJ/d Problem 9.57 9.57 and 9.58 For the beam and loading shown, determin A, (b) the deflection at midpoint C. Ha] I}: wa‘: : w<flb-%>° .' ‘ .-.- dv‘ ___. _£N(y‘) -: _w<%__£:>b L—m—i—m—v-f j; M 7 I . r]? ? V ? 12A '- w<x—-_‘%> w 2 I_ M; C! 1 MP; M T Mg‘l’ Pfix—§w<¥_%>l E3231: : MA0¢+§P,12—ijw(y—§‘>3+CI R. E; i.- é—MAXI-LéQfiXg “fiwa- if)?“ C‘lac. I. [x:03£=91 O+O~Q+C,=O cl”: [x:ogy=o] O+O-O+O+Czr0 CZ EszJ$=oj MAL-LéQHL: -fiw(%\3=o C.) 1 1 if \ [9c:f_3 j=01 +é‘Q‘L ‘fiwfié‘j = O (i; Soil/Jung E7vw'pr‘0ns (F) Aux-JG” simuj'g‘ane‘odst‘pj) 3M. LL 3__W_L (a) R" H W Mfi fit: I?» " 31 2" Mn' E .51 1 EIJ : “g—gjf—WL x14 Iii—wine“ — fiw<x=§j Efimshg curve. : W 5 1'; _3__ 3 __ “' h" gri'fi‘t‘x *JM‘L” ‘I:F<x :7E Clo} De‘FJED'FI'Ow of Ma'apeofhf C. at x: U 3‘ _L_3 Ya 2 (7%“)(5) +(r7a W23 " 0% :‘ wL” _ NU .- ma EI “’5 ‘ l” Problem 9.65 Pb vc=- ,8- a- = zLusv g; c semi: (L bk] GEIL 1.! —(L1_§.)(Z-5L] = __§._P_L3 486 E: 9‘: L 9—9) = ~P___<~a>(L:—cm 4 PL: GET—L QEIL = ' 5—1“ (a) 'DEFLELT _ __ 1 PL; 8 PL? 3 Hfififii- vs Harare-L — m‘e—x = 155% 5 PE ( z 2 fit: T67.— E‘ Wk "a b) 5L .. __5...ti ._ iPL 7' Problem 9.80 9.79 and 9.80 For the uniform beam shown. determine (a) the reaction 51A, (5} the reaction at B. RR Conslyfer Q, 4.5 Y‘aJuwaln-rf‘i narrate. Denim!) sys'bew. I mmJIE. 3 Lymph; I. (Cue 1 a? AFF9'J:*D3 if")? 3P5: Logging H. POF+EaiN CB . (Ca-5e Jr AppemJIMD) _ _ ng/z)‘ “fl _ wCL/2)3_ WLS (XE); 85: ' 12851 J (9332' E’EI ' 92 El . - . . _ L_- z. _ 7w“ 'PorJrlam r5- 5‘1'V‘mjLL-t—- (jaL‘ (f8)1_ 2(93}L 33L, EI- Sulpev‘Pos:+I-0M fill/19? CDnSMI-d'i'. ya : (jfll +6931 '= 0 (an 35: —.38‘1’ E1: ' RA ' :23 T ELL 2 M Using: Bud-foe Leann «.5 a. 'Fr‘ec LINDA-.1}: a L j +TZFJ20: Qfi+Re—%=o 1; Q3 L . _ _ '{NL 1:5? wL R! L _"—_+ Raw £9: IR‘B RB {23 T +32ngo: MB—P.L-%L'-T’;:o _ _ yfifi qu‘ MB" “L ‘8 ‘ _ |28 __‘ Problem 9.82 9.81 and 9.82 For the uniform beam shown, deten-ni me the reaction at each of the three supports. CoHSiDEfi RC #5:: REbUr-JDAMT AND REPLACE LOADIAC-r fififiTEH 8‘1 IAN'D'II. A CD: B . Rafi FR i: :3, Lemma 1. (use 4 0F APPENDM V3.3 Yr, = ‘4951 C W Lemme ‘11. (use 1 oF A'FFENDW- D.) M L 3 g M E ® Mo Vt“ = -6_.=_*1'OE[("€) " 6(1)] 7' IGEI A C a) 7'; if 509EKP051TIOM AND Cor-JSTRAINT. l LIZ 1 LIZ ' Yc=Y£+7§ =0 _R(_E MOE _ _ 3M5 4 A c 5M0 +351 + 1651 ' O Rt‘ L. ‘L Mo L +:T__‘—_P +5EM5=03 _?‘AL +(3L. )(i‘ ‘ Mo = 0 RA 3M RB ._ Mo L" RA‘ at. + i _ Mo * 3M0 * RB_ 5Mo+ “a . | Problem 994 9.94 A lG-mm—dtameter rod has been bent into the shape shown. Determine the deflection ofend C after the ZOO-N force is applied. Use E = 200 GPa and G = 80 GPa. LET Zoo rd = F. Coer\bER. TOR-5w»! 0F Rob AB. E_(_P_L.3_LgP_I—Z 435::5'36 ‘36 I- ..-P_13' ‘fr."l—¢e‘ IG- oousmeit BEHDlNG- 0F AB. (casex OF N'WEthix b) .._ __P_L3 7': ' Ya " 3E1 acne-wen Benome 0F SC. (0.52 1 OF meant:st '03) n. _. _P_f: 7c ' 3E1 5UPEKPO$\TtoH'- _ .+ .. _ _P_L?_:LE_P_G _ _P_P(E+; 12— 7c ‘fc. *‘k — :6 351 3&1 ' EI 36 3 DATA '- G = 8000‘} Pa. 3" E =2oo(|o“)Po. I= 1,13 = 3.2no(:o“‘)m"’ ET. = (943.40 N-«F IG = SH-JZ NW? 3 {100)(o.25 (643.40 A) _ _ 7c: ' 643.44: 514.17. + 3 — '- °1.'50°13(|o 3)m jc= £1.3imm ¢ .1 ...
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HW12 - 9.1 through 9.4 For the loadin 3 shown, determine...

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