Question

Q4: Using slope-deflection method, determine the reactions of the supports for the frame shown in Figure (4). Then draw shear and bending moment diagrams for the frame. E is constant 5W/ 10 ft B 10K 5 ft -20 tra- 21 Econstant

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Answer 1

Mode = MDE Solution in Criven frame :- 5 kift C than BE L Tur E 21 ΣΙ I loft * lok E = constant. 5 ft 2oft 5ft STEPD:- Fix End Moment in Mojca le 2 5A20 12 202 -166.67 Koft 12 Made 2 fde = + wall & 5x20 = st166.67 Koff 12 5*5*=-62.5 K-ff And - 10 Mada fda = Wab? l² Mlad 10x10452 152 = -11,11 k-ft Alonlotas 152 =+22:22 Krft wath

Maca 166.67 to 2ELO & OMELO, no STEP2: Slope deflection equation in >f=0 Relative displacement of any span enda Mij Meis 2E1 (20; + Og t Reis Mas = -166.67 +2€129) (20+0s) 20 -166.67 70-4EI OC +0.2ELOS MDC = 166.67 & ZE(2L) (200+ Oc) Mor -11 * 2 (200+ MOA =-11:11 + 0.267E1D, + 0.133 ELOA M AD = 22.22.21 SE (200+ os) AD = 22:22 +0.133EL OG +0.267 EOA STEP 3: Joint equilibrium equation in Eme=0 Mcs 0:40 00:2 EL DO = 166.67 EMD=0 & M. Moc t MOE 166.67 +0.2518ct 0-4 ELO, -62.5 -11:11+0.267EROS +0.133 ETOA = 0 0.2E1 Oct 0.667 ELOJ +0.133 ETOA =-93.06 EMA=0 MAD=0 0.133E10 to 267 EIDA =-22-22

Solving equation ® ® and ©. we get A El Oc=581.75 ELOD -330.16 EIOA = 81.24 STEPY".- find End moment i > from equation o McD = -166-67 +10.4*. 581.35)-(0.24330-16) Ok-ft And Mcs 330.16 > From equation Moc = 165.67 +0.24581.95) – (ovvero Moc =150.95 kaft Moe =-62.5 kritt Ang > from equation ③ (0.267*330.16) + 10:13381.24) -88.45 kaft Ang MDA MDA Check Moc & MDE AMDA o ork > from equation a MAO - 22:22 + (-0.133*330.15) + (0-26781.243 MAD AD Fokoff Any

o kaft 150.95 625 Skifte 5ft 150.95 160 2 100 150.95 20 182-55 88.45 longa 15 88:45 15 loft are e".. lok Sft to the long aniss of Member either STEP 5 :- Support Reaction - 5klat 22 Front zoft lask V = 82.556 Ve=42145k 9.23K (N Hence Reactions at supports Ve=4245k (1) He=9.23K() VA= 82.55k(*) Any th=0.77K HA -0.77k(y v=82.55k STEP 6.- Shears force. Diagram and Bending Moments siagram > Shear force in it is algebric Sum of all the transverse load (perpendicular Moving left to right or right to left (tp, (1-, +) > Maximum Bending moment (Beam Moments in Maman for CD Mmas for DA= Wally 10 1045 15 88451151 10*10 15 } Oktt A t we 57202 =+ 250 kaft 33:33krift

U245 25 t + (9.23 observer Position For DE q 57.55 7 observer Position for CD 0.77 9.23 B observer Position for А. ni Shear force Diagram in And 250 + 150.95 62.5 OE 88.45 ☆ observer R Position for respective Member EE.EE A - Bending moment Diagram. And