Question

Solve the following using Moment Distribution Method

200 KN 200 KN 21 kN/m 20kw/m A "B اے D 1at EI ET uit 4m 2.5m a.sm 3 m

Answer 1

200KN 200KN 20 kN/m A 21км / B D 4m | gm M 2.5m 3 m. Step 1 : 12 fixed end moment: MTAB = - W1² - WI - 21x8 2 - 2188 -133 KN. m. 8 12 8. MBA = +W1²+ wil 21482 +2188 +133 KN.m. 8 8 MBC = -200x5 - 125 KN.m. 8 200x5 125KN.m. 8 8 McD - 20x3x3 90 KN.m. 2 Step2 : Distribution factor - WL 8 MCB +WL a Sum D. F. Joint member 0.45 BA stiffness 4 EI 8 361 5 1.1 EI B 0.55 Вс 4 EI 1 CB 4 EI 5 C CD 0 0

D 0 -133 + 133 step3. Moment Distribution table: --- А B 0.45 0.55 1 1-125 +125-90 3.6 1-4.4 - 35 -1.8 |-17.5 +7.875+9.625 + 202 +3.94 141.1 - 0.495 -0.605 -2-2 - 0.2475 - 130.61 +136.78 -136.78 +90 1-90 Final moment 136.78 130.61 90 Mab Ra 4m MBA "Roi 200KN Reactions: 21kN/m AB: 4m 5 Fy-o; RA+RB) = 200+ 2178 RA+ RBI = 368 - (1) RAX8 - MAB - 200x4- 21x8 x 4 +MBA = 0 RAX8 - 130.61 + 800+ 672- 136.78 RA = 183.23 KN. EMB=0; RB = 184.77 KN.

200 For BC! MBC TRA DMCB keer Efyo; RB z t Rei 200 -(1) EMC2O; RB2 x 5 - 200x215- MBC - MCB=0 RB₂ X5 2 2008 2.5 + 136.78-900 RB₂ = 109.36 KN. 200-109.36 2 900 644 KN. gt1 For CD :₃ Efyz RC₂ 20x3 = 0 RC22 60 Now, RAZ 183.23 KN. RB² RB1 +RB2 = 184.77+ 109.36 294.13 KN. RC= RGI + RC2 90.644 + 60 RC= 150.649 KN