Question

6ITUATION 3.C 2m C am Ek 27 51 MOMENT DISTRIBUTION METHOD Cmini mum 5 cycles) DE TERMINE THE REACTIONS AND MOMENTS, AND PRAW SHEAR AND MOMENT DIAGRAM SOLU TION

Determine the reactions and moments using Moment distribution method and draw V, M diagram

Answer 1

Fixed end moments M_FAD = -41.3 - 13.92 = -55.22 KNm M_FDA = 23.6 + 64.05 = 87.65 kNm M_F_DE = -123-67.72 = 190.72 kNm M-FED = 123.+ 45.14 = 168.14 kNm sigma M_A = 0 implies R_D1 times 5.5 - 38.5 + 121.09 + 51 times 2 + 78.62 times 1.5^2/2 + 9.38/2 times 1.5 times (4 + 2/3 times 1.5) = 0 R_D1 = 56.04 kN sigma F_y = O implies R_A + R_D1 = 51 + 88 + 78.62/2 times 1.5 R_A = 119.92 kN R_A = 119.92 kN sigma M-B = O implies -R_E times 6 - 121.1 + 203.04 + 41 times 6^2/2 + 37.62/2 times 6 times 2/3 times 6 = 0 R_E = 211.9 kN Sigma F_y = 0 implies R_D2 + R_E = 78.62 + 41/2 times 6 R_D2 = 146.92 kN Analysis results M-AD = -38.5 kNm M-DE = -121.1 kN R-A = 119.92 kN M_DA = 121.09 kNm M_ED = 203.04 kN \r\n