Question

oints) The beam shown in the figure below is in supported at A and simple su Q3)(100 points) The b a) Use singularity function ularity functions to find the shear and bending moment equations. singularity functions and find the equations of shear and bending spin supported at A and simple supported at b) Evaluate the singularis segment of the beam. Shear and bending moment as a function of x) for each 20 kN/m -4 m 2m- eace to draw an appropriated tree body diagram and to solve the equations of static equilibrium.

Answer 1

[Q.3. Page 1 420KN/m 4m 2m Frron e static equilibrium, is Ay+By = (20+4) = 80k ü) IMA =O ( +) =>-Byx 4 + 2004+ 20 => By = 2004X2 = 40 kn Ay = (90-40) = 40 KN For shecor Force :- AB section :- Let, at a distance x_x seen in there at AB COLX 54 ) Tip S. F. at 7-4 = +40 – 20xx = (40-20%) (4) BC section :- let at a distance xx seen is there at Be (4¢X 46) Sofaat % -% = +40 -(204) + 40 = 80-80=0 For Bending moment :- AB Section &- let at a dist. x-x sea is there at AB COLX & 4) (+) I B.M. at x4 = (+ 40xx)-(20xxx) = (40x- 10x2) BC section - Let at a dist. xx sea is there at BC scanned with (12x = 6 El canBo M. at toto = (40xx)– 20x4x2+x-4)+ 40x(x-4) x

2.3. Page 21 continued... al for Bending moment ? - BC section - Bim. at ay = 40X – 80 042) Than=6) +4062-4) = 40x480x+160 +407-160 A 16 For shear force :- AB section - S.F. at 1-40 = 40-20x (OLX = 4) sofo at A. (n=0) => 40-2000 = 40 UN sif at B (989) 5) 40-2004 = 40-80 =6240 kor. when, sif. changes sign. B.m. is mas there. I'. Suf. 0 40-20150 -) a = 40 = 2m @ n=2m) sofoco & Bom, u mash BC secbon ! Safn=0. for Bending Momenline A B section. - Bom, at sy loca24) => > 40x_10x² Bimo at A @ X=0 y 4000-1000 = 0. B.Me at B @NE 4 -40x4-10047 = 160-160=0, @ n=2m =) B.Me may = 4002– 1092-80-40 = 40KN.m BC section : - BMa = 0. LAW Scanned with CamScanner