Question

Problem #2. Using the moment distribution method, determine the internal end moments, support reactions and plot the shear and moment diagrams for the beam shown. Label all points on the shear and moment diagrams. El is constant 9/ 20 k 9 kr BO D 771m 101 6 61 101

Answer 1

qkift A B 20K qkift C D 6ft t -loft x latt x loft. 1. fixed end moment : MAB = 12 9x102 = -75 k-ft; MBA = + 9x10² _ 75 k-ft MBC - - 1 - 20x12 = -30 k-ft; MCB = +20x 12 = +30k-ft- 9x102 = -75 k-ft.; MDc = + 9x102 = +75k ft. . 12 McD = 12 2. Distribution factor sum D.F Joint member stiffness 4E1 0.55 BA 10 0.733EL 0.45 B 4E1/12 Вс 0.53 4E2/12 CB 0.633 EL C 3 E1/10 0.47

3. Moment distribution table: -75 +75 -75 - 12.375 -3.967 B 0.55 10.45 +75 1-30 - 24075 - 20.25 1+14.425 -7.934 -6.491 + 12.62 -6.941 -5.68 0.86 -0.473 -0.387 +0.76 - 0.42 -0.342 +34.48 1-34.48 0.53 10.47 +301-75 23.85 45/+21.15 -10.125-37.5 +25.241 +22.383 -3.25 +1.72 +1.53 -2.84 + 1.51 +1.33 - 0.1935 +0.103 14001 +00 -66.02 66.02 -3.4705 -0.24 - 95.03 Bending moment 112-5-kft diagram: -95.05k-ft 112.5144 60kft 66.0 24/7 34.486.ft A B. 5. MAB 9 MBA Reactions: Efy=0; RA+RB, = 10x9 - ) EMB=0; Rax 10 - 9X10X5-95-13 +34.48=0 7 Ra= 510.55 = 5/855 kips: (1) Rby - 38. 945. kips (1) TRA loft TRBI 10

Вс 20 K. 2 Fy=0; RB₂ + Rc, = 20 (1) MBLE 12 ft Rol EMC = 0; RB2 RB₂ X 12 - MBE 20x6+ MacB=0 RB₂ X 12 = 34.48 +120 - 66.02 7.37 K. - 7.37 12.63K. RB - RB, + RB2 = 38.945+ 7.37 46.315k. RB₂ Rei 20- CD:- 2 MDC MCD IRC2 10 ft TRO EMD=0; & Fy=0; Regt RD=9x10 - (1) - 66.02 + 2 c 2 X 10 - 9x10²=0 516.02 Rc2 2 51-602ko RD = 2 Ra= Voc 90-51.602 38.398 K RC= Reit RC2 = 12.63+ 51.602 2 64.232k. Shear fora Diagram :- Na= 51.055k. Ra - WXL 51.055-9x10 = -38.945 left Ra-WL + Rb - 38.945 + 46.315 = 7.37 (Right) Ra- WXL + Rb- 20 7.37- 20 = - 12.63 k. (just left) - 12.63 + Rc = -12.63 +64.232 = 51.602 ( just right, Vo = 51.602K. Vc

Snear force - diagram: 51.055k 51.602K +7.87 +7.87K B С A - 12.63k - 12-63k - 38.945k 51.602K