Question

* m For the following Beam Use the Moment Distribution method to determine the reaction at the support B (kN) w=9 kN/m P=6.25 w (kN); For example, if w=24, then P=24*6.25 - 150 KN E-200GPa 1=6,000 cm Remember Forces to the right and up are entered as positive, left and down are entered as negative. Counterclockwise moments are entered as positive, clockwise moments are entered as negative w(kN/m) P(kN) P(kN) А B С be 20 m . 5 m n 5 m

Answer 1

Solution :-

Given Beam :- p=56-25kn P= 56.25kn w kN/m = 9 kN/m V V A с B 20m 5m 5m * * 5a). * va WagkN/m. AS P= 6.25 W KN p = 6.25 kg = 56,25 Kn 200 Gla I= 6,000 cm4 Step I : Calculation of fixed And Moments :- * MFAB Wh 12 9x20 12 -300 kNm -0% MEBA we? ।२ 9x20² 12 300 kNm. # MFBC 2PL - 2x 56.25x15 -187-5 kN-m MACB 2PL 2x 56.25 x 15 187.5 kNm g

Step II : Calculation of Distribution factor :- Joint Member Relative Stiftforess (k] 4EI L EI BA 4£I २० 1/2 =0.5 Sum (EK) Distrybution / Df=K factan ΣΚ. £I/5 £I PEI REI/S ells |REI/5 BC ЗЕІ 3EI EI 1) 1/2=0.5 L 15 5 Step II : Moment Distuibution Table :- A B с AB BA Вс. CB Joint Member D'F f.E.M. FEM 0.5 10.5 -300 300 -187.5 187.5 * 187.5 Release? 1/2 - 93.75 Balance Inital Moments Balance - 300 300 - 281.25 .-9.375 -9.375 13 carry overy -4.6875 final Moments (kN-m) -304.6875 2900625 -290.625 MAB S-304.6875 kwang MBC = -290.625 kNm Mega O KNm. MBA 290.625 knim

Step N : Calculation of support Reachions : free Body Deguan of BC: IP E MB Me=-290.625 MCB no с Rc 5m Som * T using equilli berium equations; * Ef=0 Rot Re-pop RB tre = QP {p= 56.25kN -0 EMB=0; >>-290.625 +(Rexis) -(Px10)-(Pxs) 20 +290.625 t ISRC -15P to => ISRC = 15P +290-625 P-56.25 KN 3) 15RC = (15% 58.25) +290.625 -> ISRC = 843-75+290.625 RC = 1 34-375 KN 15 Rca 75.625 KN .* RB + Rc=2p -> RB + 75.625 = x 56.25 RB = 36.875 KN from eq0 +

free Body Diagram of AB:- -304.6875kN MAB wa gkN/m Mea= 290.625 kN A ® RB RA K २०m * using equillibulum equations : # EMA =0; > - 304.6875 +(R8x20) -(gx20x20) » 20RB +290.625 20 = 1814.0625 - RB = 1814.0625 KN २० RB = goo70KN 4 ² Efy =0; RA TRB - (8x20) = 0 » RA+go. 70 1800 RA 289.2968 KN IRA 89-30 KN from equation ® ®; 36.875 +90: 70 Total Reaction at Ba RB = 187.575 KN final Answey

Reaction at B = R_{​​​​​​B} = 127.575 KN (upwards)

shear force and bending moment diagram :-

Given Beam: p=56.25kn P= 56.25kn w kN/m = 9 kN/m Itt A C 20m | RB 5m RA 5m sa JRc B9.30 KN SFD 36.875 KN В C A -19.375 kn -75.625 kW -90.7 KN BMD JA Storaght line - 290.625 km -304-6875 KN-m Parabolic curve

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