Question

determine the moments at all supports using moment distribution method. EI is constant.Perform unit distribution 5 only

3. Determine the moments at all supports using Moment Distribution Method. El is constant. Perform until Distribution 5 only. All lengths are expressed in meters. The 20 KN acts 3 meters from the fixed support. Set w = 2 kN/m. (20 points) 20 KN w 오 오 오 4 5 5 t

Answer 1

LOUN Q 2 kN/m K 3m um B 5m 5m E soy joint Member Ik ) (Ek) 10+) 112 BA EI B Вс 2ti EI 112 CB с FI 1.8t1 CD 0.84 519 419 0.861 1.661 1/2 0.847 112 fem BA =) 80x9x1/16 = 11:25 KN m , fmo -> - 20X 3x1 3x1116 = -3.75kNm feo=) . -U 166 KN m, 2x25 B 1/2 tembe 4.166 kNm D A 1519 419 11212 11:25 E -3.75 -8.8125. -4.166 4.66 -51625 -5.625 2.31 1.85 -2.08 -2.08 1:04 -0.287 * 1:15 0.675 -0.575 -2.8125 -1.04 2.14 4.71 1.42 -071 -0.7) -0.355 107 - 00635 -0.535 -02871 -0.355 0.356 0.285 0.85 -0.425 -0.425 -0.2125 0.178 -0.267 - 0212 -.044 0.142 -0.089 -0.089 012660.22 -0.071 -0.071 0.0355

A 1/2 12 519 4/9 1/2 1/2 -0.044 -0.089 -0.089 0.266 0.212 -0.07)-0.07) -0.0355 0.133 -0.044 -0.0355 0.106 -0.033 4 -0.066 -0.066 044 0.035 -0.053 -0.053 4-0.0265 022 -0.033-0.0265 0.0175 -0.0055 -0.01 -0.01) 0.033 0.0265 -0.0087 -0.0087 → 0.004375 - 7.2 KNM 4.350 1-4.350 KNM KNM 1.710 -1.710 KN M KNM -3.35 -3.35 KNM -1.6 75kNm KNm MA MBA MBC MCB MED MDC MDE Ans MED MA =) - 712 KNm, MCD = -1-71 KN-M MBA ) +4.35 KN-M MDC =) 3.3 5 KNM MBC – 1.35 kNm MDE = - 3.35 KNM Ans MED = -1.675 KNM MCB =) 1.710 KNM support exn's