Question

13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13-P13.25 using the method of con- sistent deformations.

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Answer 1

solution 18. 2kIFt ti 10' 301 RB det Ro be reduntant: so, primary smucture without reduntant; 18K. akift Atoom de. 30 B Here, deflection at B 3 SB 18x30j 3(361) t. 18.x(30)2 X10 + 2x (3014 2(361) 8 (361) Sg=148500 El. Now, Primony smucture with reduntant AX B c 301 lor Ro Here, deflection at B; AB Rox (30)3 3 (352) 3000 RB EL

but deflection at B=0. . SB = DB 148500 = 3000 RB El El 2 RB = 49.5 kip Now, shear force between CB; Ve 18k Bending moment; Mn=18x at a, mn=0 at B; mons-18x10 -180 K-fe Now, Shear Force berween BA; (n=0 to 30') Vn= 18+22-RB = 18+22-ug.5 Vy= -31.5 +2%. at B; Vyeo -3)'S K. at A.; Vya -37.5+2x30 =28.5K, For Zero, Shear; Vn=0 =) 29= 31.5 > na 15;75'. now, Bending moment, Mix Roxa - 18% (u+(0) - 2xnxx Mu= 49.57-18%-180-3?: Mic = 39-5-1?-180.

at B; n=0, Mans -180 k-ft. of A; 1=30', Mus 3115x30 - 302-180 .2-135 k-fo: at n=15.75' ; Mn = 31.5x15 75 15 75² -180 68-0G2017: Now for zero for zero bending moment; Muro = -1? + 31:581-180. n= 7.5' J and n=24' or 6' from A. So shear force diagram; 28-5 18 18 15.75 А. B -31.5 Bending moment diagram, 68.0625: 6' 75! -135 -180