Question

The simply supported beam is supported by pin support A and roller support C. It is subjected to a uniform distributed load w, and a couple moment M. If wand Min the image are positive real numbers, select the correct shear force and bending moment diagram: w M B -5 m 5 m .X

Answer 1

w BAM AC Sm sm q Rc RA E Fr=0 RA + Rc = 5W Taking moment about a RAXIO -WX5 x 7.5 + M = 0 RA -M + 37.5W 2:. RC = (-M+ 37.5w )+ Sw Shear Force Now, Im Region AB (SE) RA wa WX = -M+37.5W at x=0, (SE?A= -M + 37.5w at 2:5 · (SF) BE - M+ 37.5 w a5w -M+82.5w In Region BC (SF)x-* - Rc - M+ 37.5W-5w - Mt 32.5w at azon (SF) ca at x 50 (SF) B = -M + 32.5 w M + 32.5 w Bending Moment In Region AB Rar- w x2 (BMD *-* - 2 at x=0 A=0 CBM) 2 at t= 5m, (BM) B = SRA - 25 W - SRA - 125 w =5[-M+ 37.5w) -12'5W -5M +175W

(BM) x * In Region BC Rea [ 5w-(-M+37'50)]? at x=0, [BM]c=0 at a=sm, (BM) [5w w +M - 37.5W]*5 - 162.sw +5M Now, from calculation of Shear force and Bending moments SED BMD (th ti (-