Question

1. Use the moment distribution method to determine the moments at A, B and C. then draw the moment diagram. El is constant. Assume the supports at A and C are fixed. 25 k 25k --8ft ---8- 24

Answer 1

Stiffness Distribution KBA = 4€ – UE L = ET KBC = 4ĘS – YEL – ET factors DF BA= KBA – E116 = 0,50 KBA+KBC E1I6+E116 DFBC = 1- DEBA = 0.50 moment 5 A fixed end MEBA ah MFAB = P ab? Pa2b Here MFAR= -25X8X162 – 25X16882=-133:33 Kony 242 242 MEBA= 25x8x162 .+ 25x16682 - 133.33 Koro-ft 242 २५2 MEB(= MF CB=0

Member AF Joint AB Member DF BABC 0.5 0.5 133.33 - 66166 - 66166 FEM -13333 Balance cabby over -33.33 -33.33 Net moment -166.66 6666 -66-66 -33.33 tt

FBB 166.66k-ft / 25k 8' J 82 GAB 166:666-16 1252 33:33K7 GTA 256 6666kft BTC STB By By 241 1 1.cy EMB= 0 -16666 -25X16-2548 +24 X Ay =0 Ay= 29.167k By, = 50-291167 = 20.833K Emc=0 66:66+33:33 - 24x By2=0 By2 = 41167k diagram moment m=29.167 X16 - 25X8-166-66=100 k-H - 33.33k.ft 0 | | 8'' 8181 - , 166.66k-ft 6666k-ft

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