Question

2. Find the nodal moments in the following frame. Note: Perfect loading and geometry symmetry, therefore there is no sidesway. Use hand manipulations to reduce the four equations in the four unknown nodal rotations to two or three and use a calculator to solve the 2x2 or 3x3 system similar to the example done in class. (25 points) I 5A 5A 10 K 10 ft EI = Constant

Answer 1

LLLL el solution :- tok 5ft *5ft 7 Giren:- €1 - constant | loft Apply slope deflection method step fixed end moment (mp):- MEAB =MEBA =0 [since no external loading] MFCD - MEDC = 0 MFBc -- - 10 x 10 = -12.5 k-ff cha ) MFCB = MECB = PL = PL = log10 = 10x10 = 1205 kaft MEAB MERA K a=soft 10 k Step2 slope de flection equation : MAB = MFAB + 2€1 (204+ 0B -34). 10 - No side sway thus A=0 Also for Au member usert => MAB = 261 (204 +0B) .. MBA = MABA TREI (20B +O A - 34) 10 MBA = 265 (208+0n) similarly, for spa. Bc. and CD As. Mac = - 12+5 + 3 (29 4 + 9) McB = 12.5 t. 2EL (20c + OB) 10

Meg = 0 + ?EJ (20c +00) Moc = 261 (20) + 0c) Step (3 equilibrium equation to fin equation to find out nodal displacements :- since support A and I are linge support Support thus - MAR MDC =0 1o 10 > MAB=0 . MORE = 0 ? 261 (204+0p) = 0 Also > 261 (20.+0c) = 0 => 20A = -08 » 20p - - - 7 OD = -0 > OA = -06/2 Al joint ® MBA + M B C =0 - -5 + 2 ( 2 a + b) + 16 (.- 9g) = 0 à ? [ 3-5 % +0. D = 11.5 @ 3.508 + c = 6205 -- Beq At Joint ② McB + McD=0 = = 12.5 + 2 (206 +08) + 286 (206 - 0) = 0 3.50 0c + Op =-6215 - @ea ea 1 and ② we get as OB = 25 Oc=-25 EI EL thus OAS Y ,OD = 12