hw9_soln - _—_———7 11—8 Determine the moments at...

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Unformatted text preview: ________—_._———7 11—8. Determine the moments at B. C, and D. then -T‘J'W 11w moment diagram for ABDE. Assume A is :.nncd, D is a roller, and C is fixed. El is constant. 2( ml gym“ = T a 56.2.5bfx r vztml _ j .me = E2 - "-4 “1 50k. 424;; at 42-9! K4! 20.5; r-j_r_ "-1 7 mama. = 24.0 k-fi (Wu # (Ffiwc. = 0 lanai (an llafi’i f 5 - r IZ-l k. . $3 at. I. x503; 11,, EMIGN mummy 4 1786:. “is”; 56: .r u” = 35(E)(e, - 0} + 56.25 Map“; I‘ u” = (fi)¢29,+ a” — 0} w 2.: u” = 0.3333519; + 016611-289 — 24 (2) M” I j ) 8+0 tk fl" = 25(E)(239 + 93 — 0) + 24 ‘ M“ MD“ 1 Mac up, 3 03333518,, + 0.166658,+ 24 (3) . I. u“ = 2£{fi)(28. + o - 0} + o u“ : 0.333356, (4) 501m; Eqs. 1.1: J 1' —66.79 153.35 ua=25[fi)(2(m+e,—0)+o a,=_.b_' an: H HE, 9'- “.156739. M” s 42.9 k-fi Ans Wm u", :40; H: Ans W; + MH: '9’ Min: 0 (5) M“ = “Din-fl All: M“ = 412“: Ans _ = (7} “D- 5“ a Mazqukn Am 11—18. Determine the moments at each joint and 1200mm support of the ballcrcdrcolunm frame. The joints and ’ supports are fixed connecled. E] is constant. - 16ft i MN 225(E){28N + a; — m + (mm J 25:: MawfierBfiJrfl L7~12u_ 712f1—— 12ft J 251 u m was +9) +0 IA 20 8 25: 12(12): u“ = 334265 4- er) '— I2 25129 + 9 L202)2 = --—- + M" 11 i 5 “ 12 251‘ Ma, = $036 +0) + o 251 M m —- 0 + 9 ) 4- {3 “C 20‘ ‘ qulibrium Mk Mu “ Mn: a 0 ' M“ .5! M” 21-? “ M "ii—{239 + “IE-(26, + BC} '- 14.4 = 0 3* 14.4 0.533399 + 0.16519c = m— (1) £1 1"". + ,MC” 2 0 2H 2 m' m HA "— 23 = 0 Rue. as; + + m( c) Whig! 053338. + 9.16679, 3 m (2} 5? Solving Eqs, i-E: 39.27 ' = ’57 “u 3 3.93 k-fi Aug 3: h. I} 7.83 k-fl Am Ml": 3’ ABS 2* N 7‘85 k-Et Ans “7.85 k-fl Ans z? MM = 4.93 m a» 11—22. Determine the moments at A, B, C, and I) then draw the moment diagram. The members are fixed connected at the supports and joints. El is constant. ~t§fi31 =~22.5k-fl {mum = (FEM;Jim = 22.5 k f: WINS): (FEM),,, = = «75.0 k-fl (FEM:m :15“ k-f: (Puma, = (FIEch = a m Wu: ‘2 War a“ Mn Mr» I M... r. 25(3126‘ + 8; — 31y) + (mu M“ “a “I I H” = 2£(r5)l2(0) + 8, — 3V“) + 0 us: M; 4h] M” = “36675103 — 0.59m; - 22.5 (I) J .r "‘4. A Mn: ‘0 ' I M“ : unifies + e — 3w”) + 22: M8,. = usaswe, — Ojflm,+12.5 (2) “debit 1' gm. = 24:3)(29, + 96 — 3(0)) 45.0 MEL- : 0265739. + 9.1333565—753 (3} I 7 M” = uhgyfif + 6. — 3(0)) +753 Mu: M55 x 0.25.5139, + 0.1333mc+15.0 (4) “'7” 3”“ I Ma) 2 2£(E}i295 +0 » 3w”) 4- 0 Soh‘ing these figuration? 2 Mo, = (133335196 - 05.9w” {5} I I W 159.83 a —113.33 uM=2£{fi)(1lm+9c»3wup+o “ " £1 ' C" H 7 5m»; MM = ujué'iEIBC m 0.557%, (6) Wu 3 E, Eqmflbrium u” D2417 k-f! 2—24.: H: Ans = ‘ M“ + M“ } (7) u“ x 47.48 k-r‘: m 41.5 H: Ans Mm + M” = u (8) M“ =v-47.-18 k-E: : 41.5 k-ft Ans VA + VD — :3 = 0 MG : mm k-t‘t = 66.0 k-ft Ans "(Mu + M“ "‘90) _ (M60 + Mac) ‘ 15 = a Mm a was»: k-ftm —5en k—f: Am 12 12 “X = 41,21 k-fi = «412 Hz Ans ______———— 11—23. The side of the frame is subjected to the hydrostatic loading shown. Beta-mine the moments at each joint and support, E] is constant. ~61”: ih'iz (Fianna = ‘0 2 ~97: krfi mrlSil mam“ = m = 6-18 H: (mm = (FEM'IL-g w 0 (Fang, = u 'hn ’ War 1 A!“ = 25639.5 ¢ 9,? u m + (534),, 1 J M” : whammy + as — 3v“) - 972 M“, 7. U.i11lE!8, — 0333333!“ - 972 (l) 1 M“ = ZEtm](283 + u — 3%...) + 648 18 M“ = 0.2222516, — 0.333319%, + 648 (2) t M“ = 25(F)(385 + ac — 3(0)) + 9 M“. : 0.265739, + 0133389.: (3) 2’ MC: = 1£(-g)(286 + 6, m 3(0)) + D M” 3 (3.1667366 + 0.5333518, (4) I. Auw3 E18, ~v)+iFEM), I Hm = 3E(£—§)(8€ — 1p”) + 0 Solving 545, | n 3 : Mm = 0.156751% - G.£666P3%a (5} [254,594 _ - 1222.839 8‘. " B as = H Ethbdum _ 4359.174 u“ 4r MM- : u (63 Van - H Mug 4- MZ-o : 0 {'1} _ Mu: x “323(303) k'fl Am i .. __ a X; + UL1 m 7,[fiUjUfi :- 0 Mrs; ‘- 49: k-fi A“ Mn:- = 4‘)? krft A “MM 7 M” “54003) - (Mm) ‘ m z o as 13 18 M“ = 497 it.“ A“ u“ + M.. + Me» = "3149 (3) “m a "49‘! k-a Ans *lleld. Determine the moment at each joint of the gable frame.Tlm roof load is transmitted to each 01' the purlins over simply supported sections of the reef decking, Assume the supports at A and E are pins and the joints are fixed connected. E1 is constant. ~2i65t15t (Fawn mew“: ——;——n = aux-n (FEWH :(F‘Emm = 20 k-fl I My = 35(E)(9N “ V) + (Fma' 35! “IA 5 T3493) I My z 15(I)t26y + 8; — 3m + (FEMQ , gas, + of) — 20 S? "x 3 Ma é-Iafic + 95) + 20 t L 1.: nm 4. Q u v r O at”) “ 3H “m: = 1‘24 9) Equilibrium MIA + use = 0 Mcn‘HKrD‘o MM+MD£=0 m SH 251' ?as*filles+arl*20=0 0.51579, + 0.13339C 2 Elk“; 1 .5 2E! 3129C + a.) + 20+ 734296 + an) ~ 20 a u sect-9,449,350 25 351 $080 + 9:) + 20+ as” a 0 20 (madden + 0.13333: 4 “H. 5f 1200 nun [_ 12 [t W.» Max. Mr; I Mac Solving mesa nqmtions ; BC 3 I) 38.7! “”‘%*7r Thus. Mu = 9-53 k-fi Ans M“; = ‘968 k-fi Ans Mg: = 25.2 kit A.“ May = - 25.2 k-fi Ans Mac 968 am A”, M95 '-'~‘ "9.68 k-ft Ans 11—25. Solve Prob. 1144 assuming the supports at A and E are lixed. —2!(m m mam“ maxim: ——9-— = an“: mam." = [FEB-11m- : 20 k-I‘: MN = 1£(£)(28. + 8, « 3w 4- (FEMlh. 251 “‘8 : T2433) IE! MEA z xvi-[235) MM : T5429, + of) — 20 2E! ' M” : TS—{zef + 9,) + 2:) "kn 2H “'6 MCD = + Bo} "’ .0 “(5 Na: MM = ——(:9D + BC) + 29 2E:r M MM = 724330} M Mn: 25! MED 3 “5(99) Equilibrium M“ + M56 = 9 Solvmgthmccqualims: Mas + MED z 0 9r 2 9 Mac + MOE : 33 33 9 z —9 x __'__ DI'. _ B 0 E7 M = 5.56k-f1 Ans m AB Eggs“ + Fmaa + ac; — 20 = o ' M" =mu: Am 10 0-639 + 0J3339c = E u“ x 41.1“: Ans 23 '25? M“ = RAE-ft Ans Feet + 8,) + 20+ TS—(ZGE + 85,) - 20 m 0 I.ch m -24.¢ k-fi Ans (153339,: + 0.13339, 11 0.133390 = 0 Mac = IL] Ha Ans 7 2H —-28 +8 +20+w28 =0 5‘ D ‘} l2( 9} M9,: = 41.: k-fi has 20 A (166,, + 03336: a *E Mm — wfifié k-fi Am; ...
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