Question

Moment Distribution Method

Determine the moments and the reactions for the continuous beam and frame using the moment
distribution method.

Thank you.

-60 kN/m B с 0 4m A D 3 m

Answer 1

Solution i Criven that - 6o kn/m B C 3m 121) 1 um A > Non-sway Analysis in STEP 1 - fix End Moment in Malab = Mama MAcd = M E O KN.m fde (because on AB and CD) M Mane We2 ião foto E-45 karam 12 Maes = + Wah -- 60*3 +45 KNM

STEP Distribution factor :- {of EK ky: Joint Member K ER Difak ER BA 0.6 als B 51 G BC Whe 0.4 0.57 CB will c 7 I 12 CD 14 0:43 STEP3 :- Non sway moment (Due to applied loads Distribution in C Joint A Dof B 10.6/0.41 10.57 0.43 o 1-45 +450 o +27 718 -12.82 -25.65-19:35 + 9 -9.67 F.E.M Balance C. D.M 413.5 Balance Coim 43.84 Balance C.O.M to76 Balance 77.69 #5:13 1-2.56 -513-3187 +2.56 -1.93 -1.46 -1.10 71.53 +1:02 1-0.73 +0.51 -0.55 118.1 +0.44 +0.29 -0.29 -0.22 +36-66-36.66 +24.54 -24.54 Final End Moment -12:15

To find Sway force in 36.66 24.54 54.76 <->3669 чм um 34.76 36.69 18.10 12:15 + EH at Columm Head .54.76 -5476 36.69 4 5-4.52 KN = 4.52Kn Ca ( Reaction) Hence the Sway force at Column Head = 14.52 kN (A) » Sway Analysis in Due to sway Force 4.52KN (+) frame Will Sway towards Right. let us assume frame will sway by an amount of 'S. wate- San = Sex > Relative displacement of Beam 'BC =s-s=0 and

s B rat dah د ت ܢ ܢܢܢ 0 B ks to с GEIS um Med - Mlde -0.375 E18 S " GE(21)8 42 D. um 6 €121)8 GEIS 42 42 66121) fas Maba -0.75 ELS 42 =o (Rektive displacement is ren reros GEL -0.375EIS 42 5 Masal Mead Moment Ratio = M fab -0.75CIS 0.375518 Meed Mdab=2 Mafed let us assume Macd=-5ko-m Hence Maas -lo korom

STEPY", - sway Moment Distribution: B 10.6 0.41 С. 10.570.43] -100 -5 0-5 +2.85 +2:15 +1.07 Joint A Dif F.EMI-io Balance Coom A3 Balance Com -0.42 Balance final -7.42 End moment + 6 ty 41.92 -0.85 +0.57 -0.57 -1.14 -0.86 -0.28 0:43 +0.34 to 23 toolb tois -4,36 -4.50 +4.51 +3:59 -3:59 To find sway force in 4.59 3.59 11,93 r.gg 2.95 um Alloy 3 k 2.95 7.42 4.36 EH at Column Head 11.93 4 4.99kr (+) 7.95 -4.97 RN caction) 4. Hence SWAY Force

Correction factor (C. f) = 4,52(+) Ito.gog 4.971) STEPS - Final End moment (from Principle of superpon -tong. Joint A 9. Non sway H18 +36-66-36-66 End Moment -12:15 +24,54-24,54 2. Corrected 10.909*(-7.42) Sway Momen =-6.74 rt=cf*som 1+11:36 Moment -4.10/+4,10 -3.96 + 3.26/-3.26 Final End 432.56/-32.56 +27.801-27.80 -16.11 Hence McD = -27.80 Karim AB 5+11.36 karam MBC =-32.56 kr.in MBA =+32:56 Karim McA=+27.80 KN.m STEP 6.- Support Reaction in Moc=r16.1k Ang 32:56 60kN/m 91,58 27.80 88.41 ( -> 32:54 4.76 3 43.9 old, >439 91158 88.41 43.9 43 11.36 1 91:58 16.11 E 88.41

Hence V. 91.58 kN (1) 43 3:9 = 10.975knly प HA Ane 88.42 KN (1) Ho=10.975kN LA