Question

Question3: Using slope deflection method, calculate the moment at joints B and C, and draw the shear and moment diagram for each member of the frame. Assume the supports at A and D are pins. El is constant. X1 =101 X2=218 (X1+x2) kN/m B VC 12 m D A 5 m 10 m 5 m

Answer 1

0 319 PN/m 2, the = 101+218 B = 319KN/m 13 112m Fixed End moments- -> Но ME = ME BAMF = MFOCO HA AB 1 RA RD MFBC - we2 12 -319x102 12 sm som 5m MFBC = -26 58.33 KN-m MFB = 2658.33 kN-m now by slope deflection equation. MEBA- MEA BEI FAB + BEI ( 081 (OB- A) MBA 2 3EL 13 (OB-o) 0 - 0 + MBA BEI OB 13 MBCE ME BC + 21 (208 PĘ? (208 +0,- +0,- 30) MBL - 2 658.33 + 25 २EI 10 (20.83 +0 -0) MBC = -2658:33 + २EL98 + ELOC 5 5

McB MEB + 21 I (20 +OB-34) REIOC ELOR MCB = 2658.33 + + 5 MED 3EIO 19 B MBC MBC MCB PreA Moc at joint B' MBA+MBC = 0 EI OC 2 EI OB ह + = 0 5 3 ELOB - 2658.33+ 13 0- 83 + 2 = 2658.33 ЕГ atjojut & MCB + MCD 0.208 +0.630c = - 2658.33 by ① & ② 6182.16 EL EI OR - 6182.16 EI

now, MBA = 1426.65 EN-m MBC = -1421.89 IN-m MCB = 1421.89 kNm MED = 1426 65 kN-m Due to symmetry RA Rp 319x10 15958N 2 EMacao 1426.65 = H x 12+ Rox5 1426.65 = Hox12 + 1596x5 Ho- – 545.69 KN & Fx=0 HA+Hy=0 HA=545.69 KN 1595kN 112 13 SED tve sm HA RA (+) 1595N Nol the 601 S KACoro-Masino 13 812 To4 747 ley => 1595x5 -545? Se = 109.747

WIR BMD 8 = 3987.5kN bo-m 1421.89 1421, 89 In-n 1426.65 1426.65 kN