Question

Q2: Application of Force Method on Frame Determine reactions at the fixed end support A and roller support C of the frame as shown in Fig 02 8KN/m AR-1250 (10) mmn -9m 3m 20 kN Inc-625 (10) mn 3m

Answer 1

Given frame. let the Support reaction at ~ 8 KN/mAh HAT am i 3m ® OT 20KN c be the redundant fonce. IAB = 1250 x 106 mmy IBC = 695 x 100 mmy By equilibrium eq? <fg=0 - Ro = 3(4) = 13 VN - ¿fn-o → HA = 20 kN. EMA=O MA = 20 (3) + 72 (9/2) MA = 384 KN-m. 3m MBC (in from c) = -20 (n-3) MP8 ( h ) B) = - 8 (x)(x-2) - 60 MAB (m from B) = -4(X2) - 60. for displacemut at c, apply wit load at c in downward direction.. mbc (a fome) = 0 MAB (n from B) = -1(n) = -X. By virtual work methood - M. mi du [-4(42) - 60] [-u ] da * Aca ΕΙ EI AB 1+ 4(x3) + 60 (x)] dx Ac = + 30 (x?)! Ac z (x EI AB ETAB - Ac = 8991 EIAB displacemeh in downward direction.

Now apply redundant force to the hame. Mu = + Rc (u) JA qm ! I T EIAR (d²4/242) = Rc (1) 16m EIAB (o) = Rc (x²/2) + EIAB (Y) = Re (x²%) + C (0) + C2 applying bound ary condition at x=9, y=0 = EIlo) = Re (92%) + 4 (9) + Cq.. = 1 c + c = - | R 1.5 Rc. at n=9, o=0 » Ello) = Rc (9²/2) +6 7 G = - 40.5 Re.. So, q = 243 Re.. Now deflection at c = deflection at B n=o yoyo EIAB (Y) = 0+0 + 243 Rc. 243 Rc EIAB displacemuh in upward direction Compatability Condition Ac=o. 8991 ETAB 243 Rc ELAB Re = 37 KN. Redundech hoc force Now by taking whole beam & by equillibrium egre {fu=0 [HA = 20 kN/ (→ 4=0 = R + R = 81 ) = 13 = | Ro = 35 KN

EMA=0 → MA = (8x9x4.5) + (2003) - (3789) MA = 51 KNM.. anticlockwise direction

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