Question

Solve Problem by the conjugate beam method Compute the deflection and the slopes on both sides of the hinge at B in Figure P7.36. El is constant. Figure P7.36: 20 KN 40 KN с A F B D E k 5 m *5m 5m - 5 m 5 m

Answer 1

20KN 40KN с CE sm sm RE IB D VE F "A Hinge A tinge * X sm sm Sm Rc {My = 0 => Rp x5x2) – 40x5 = 0 Rf = 20 kw EM₂=0 = 20X20 + src-40X15-0 =) Rc= = 40 KN { Fy=0 = R+ 40+20 = 99 + 40 >>RAFO MA =-20x25 - 40x10 + 40X20 +20x5 = 0 ) MASO thown The Bending moment diagram of the above structure is below. lookum JookNm G) (+)'(+) BMD) + A B Son c sm sm sm E sm llow, apply the conjugate beum method. Fixed end in AB in free, Hinged end point ind pinned syport. Paned support Ma Hinged. as shown in Fig. The positive BB causes docon wald Loading and are Bm causes upward Locoding, in my en de diagram

100 ET LOO JE B A F sm sm X sm RBL sm sm Conjugate beam the given X The above fig. shows the of the Real beam. Here, Aib free, Bob pinned 'b in pinned and c'sh hinged (To get 10.8 Mib CB = No deflection in RB). consider span BM)&=0 (No Loading Deflection at point 6,108= 0 Shear free left to the pond-8, SFBL = 0 (No loaderg) slope at B' on left side, &L=0 AB, Take BM)=0 (Hinged ), Let us consider Regelin at 8' on Right = RBLE ReLX5-(x xsx r RBL 83.33 EL 83.33 EI 'BR 83.33 EL shear force Right to the point B, slope at s' on Right Side, As shee' is Antirclockwise sheas in Bc'. There here-fore in Real beam is anticlockwisse. OBR
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