Question

1. Analyze the frame shown in the figure using both Slope Deflection and Moment Distribution Methods
2. Draw the V, and M diagrams
   20 kN/m SEI BEI CEI 2E1 2E1 3 m JE - 3 m - - 3 m - - 3 m - - a) Analyze the frame shown in the figure using both Slope Deflection and Moment Distribution Methods b) Draw the V, and M diagrams.
   

Answer 1

20kN/m El B El. El te ... 261 2 EL 2 EN 261 3m 3m 3 m " 20KN/m 20kn/m LNm Fixed and rements (FEM) chandanks MtaB = - Web-20X3-15\m MABA = +15 kom. anan MABC = -.Wxl = - 15 kom. Mfrases = 15 kom. tan tin Miepe - 15km Motoc= 15kerm. C . MA BA = TER="fCF = MHF (= 0 kNm. at point A and since only one member meeting the joint hence Distribution factor (D.F) for A& Dis 1. and being fixed support. At E and F roof at D.F. =0. A=1 E=0" D.F. Table joint I Members. ) I stiffness (K) 1 D. F( 4 361/3 0.2 BC 461/ 0.267 B 214 E1/3 0,533 4E1 13 0.267 CD 301/3 881/3 " D=1 F=0 / BA CB o o 0.533 C

Joints. ко от TBI C. ITO IEF 3 BA BE BCсвер со | DC. ТЕСТ FC 0-2 д-S33 4-21ə o249 (-533 0-2 о | 15 o-sis o-tri- D. FEMS Release I. 115 АХ). СОН Balance -t:4 -3t3| -|- gt 1.87 3.Эзі 1:| 1.86 СО И 4-73 |-0-135 (3) - 918 -2vgi s • 478 • 182 | 1:24] - Balance 1 Find To Ігочны-1 , не е з-за о -|- фик НА - Мен = 20. 13 Lлт "ВЕ - - 4. 27 LN ) МВС = - 16. 18ч т МЕ3 - - - Се со т Мрі - Нео) - - 70. ЧІ 3 у м. Mers 4-229 KNM MeB = 16.184 kom. Mire = 1:865 karm.

By slope deflection Method Гутц цо sinking of MAB= MHAB + 2 E1 (20A + OB - 34) support in observed MAB= -15+ 4E10A + 2EL03 hence so for all case] Нst - PoE I LEAF fixed OF Tsupport kene DE = OF=0 MBA = MABA + REI (20B + OA - MBA = ist 4 ELOR + 2EL CA | Hee - voz + Bes) ( n + 9+ A) ... - MBET BEI @B MEB = Mg8 + 2(2)(28€+ @6-35%) MEB= 4E1QB MBC = -15+ 4E1 @+ REF. Be Mer= 15+ 4E1 OC + REI OR Meda -15 + 485 Ct + 2 EL 3 . - 15 A 2 B Moca G 15a UEI OD 3 e 2EI OC CF = ot & B1 Oc 3 MFC= 4 EI OL

AAD being hinge support MAB= Moc to.. НА - ). 461,8A + 2ELOR = 15 2 FI OA + El Q3= 22.5 - © MDC=0 2F1 BD it Elol = - 22.5 Considering joint equillibriun. at B. MBA + MBCF MBE FO.. 1+ 4E1 OR + 2€LGA 1-15 + YEL® + 2F) de + 8 E103 -0. 2 E 1 &13 + EI OA + QE1013 + El QC Ef 4E1OB20 8 El 013+ E104 + El OL soi- j cint equillibrium ante Имп HHu) HMce - е. 1 + YEJ De + 2€1053 - Y64451 + zelen .+ 8E104 20 .8 El OC + El OB + El Do = 0 ->6. M . . 3

③. From El A = 11:05 – 0-5 El 013 tence egno can be writtenas. 8E1 013 7.11.75 -0.5 ET OR telo o. 7.5 EIOR & Elor=-11-25 - Fromd EloD = -11.45-0-5 Fl OL. Hena egu 6 cambe written as. DEL QC + F103 -11.25-0-5E1 De so 7.5 E 1 Ol +El O3 = 11.95 – - by doing (7.5x)]- @ weget. . 56.25 Elo +7.5 E OB = 84.375 - Elol - 7.5 El O13 = tll. 35 55.15 Elde = 95.625. Elle= 17301 from ean FIOB = -1.7301 From egn ☺ ElA = 12:115 From egh TE ! Op = -12.115

MBA= 16+ 4 (-1.9301) + (12.115) MABA= 20:774 Kom I MAB=O= MDU MBC= -15+ų (-1.9367)+(1.7307) MBC = -16.153 kam I'M CB = +16.153 ream MCD = -20.774 kom MBE = $(-1.9307) = -4.615 kom MCF = 4.615 kam MEB= 4 (-1.7307) MEB= -2,30 krm] | MFC=2-30 kNm] F.B.D . 20KN/m 20.774. KNm 0.77 korm e kom 20.774 KNM Per copil 0 km 50 cc.per mon = 36.92KN 23.08 6.44 CTB4.615-kam 4.615 KN 2.305 KN KN 2.305 O kNm. tahs 2.305 KN 12-30 kNm 66.92 kn 66.92 KN

S.F-D 30 KN 36.92 KN 23.08 UN 36.32 23.08 KN 2.30 5 KN 2.305 KN. BMD. 22.5 krm 22-50Nm 20.774 KNM Titre list , 16.163 22.5kWm 20.774 kNm 4.615 V KNM. I frel 2.30 KNR 2. 30 kNm