Question

4. Calculate the moments at A, B, C and Din Figure 4 by using Slope Deflection Method, then draw the shear force and moment diagram for the frame. Assume A is pinned, D is a roller and C is fixed. El is constant. 80 KN 20 kN/m -15 m 12 m- 12 m Figure 4 (30 marks) (CLO3: PLO2: C4)

Answer 1

Date slope deflechan Method: 20 kalm 80 KN A B 15 m 14.3 12m-tomt fix endt Oc=0 с are will find Fix and Mements: FEMAB 2 we -202152: خودم - 20x 18² - 375 knim 12 (FEMBA t we? 12 +375 kom FEMBc - FEMEB=0 FEM BOR -we? 2 - 20 X 12 12 240 kNm 12 FEMDB = +240 KNim Lasite egation for Momento? MAB 2 FEMAB+ 2616204 +03) 15 LA BA= FEMBAH 261 (208 + 04) ef1 (228.7020) 15 Mcs ot 12 0 McB a ot 2£1 C 202 + B) CS Scanned with CamScanner

Page No. Date MBD = -240 + 261 (208+ @D) 2 T2 DB = 240 + 261 Tz (200+08) cool MDES + 80X85 640 kw.m MDBE -640 kNim and MAB=0 Chiogel MBATM BC + M BD=0 crisid suppost) (3 by there cauchen we can finde O B, 040 MAB? -- 315 + 2€7 (204403) .: 2€1 (204)+ 016) = 375 15 0.267 FJOA t 0.133 Ezog 375 2f24208104) + 287(208) + (240) + 242(288tes) 20 0-133 EIGA + 0.67 E10B + y fluB +0267 €1652-235 .: 0; 133 6104_t 0.932 EIQB+ 0167 6100 = -135 y -6402 290 + 2€? (206+0.13) 12 C-880 : 0.167 Eldr + 0.333 EI OD CS Scanned with CamScanner

Page No. Date ८ (B) FIRA 1328.36 $100 152.84 21Q0 2+19.29 - : MAB20, MBA - 592.812 NIL Mgea 50.95 kNim Mege - 25:48 PM ७पर MSD- -12.268 Knim MDB - -C240.Edim B. 45 'S92.872 घ82.2G8 642 640.9 25.48 A S42.872 B B 12836 5 D u 7 B 39.53 1283.18 6.37 10693 592-872 TS 39.53 6.90/- D76.43 H 46.43 →IC le 56.3+ A B B d V क |150 T 150 । 120 120 Totat Reachen 637 B A B - 44 226.93 10.47 18953 13.0+ 6. CS Scanned with CamScanner-

Page No. Date a SAD - 226.93 + 80 110.47 BI А 16.37 13.07 189.53 .6.37 Р BMDI 642.268 5625 592.871 360 LED B E 50.951 640 25.48 Seanned-with- CamScanner