Question

For the beam shown draw the shear and bending moment diagram. You need to label all transition points and clearly show the shape of the curves to receive full credit. 15 m 05 m P40 KN W = 20KN/m dob D

Answer 1

40KN + 10.75 E 0.75 : 20 kN/m Bm to.smc am. D 3 m Кр RB - - - 1 curve 1 KN - - - im - (+) - 1 20KN - 25 I KN - - - ME = 10 kN-ml 27 2 cuve - / 11:25N- + G I BMD) curve Reactions - {fy=0 = RA+RB+ Ro = 40+ (2002-5) = 90 KN {Mc=0 (20x221) - 2 Ro=0 Ro = 20 kN / RAT Rg= 20 kN (EM) left = 0 a) 2RA - 40(1-25) + 0.5 kg- (20x0.5X0.75) 20 solving RA=15KN R = 55 KN Stall BM Equations- Portion at s sign convention (tve SF - Vy = 15 KN BM = Mp = 15x MA=15(0)=0 Mp = 15(275) = 11-25 kNm

Portion 'EB Vy = 15-40 = -25 kN My = 11.25 +150-40x ME= 11:25 KN MB = -7.5 KN portion BC v = 15-40 + 55-20x V2 = 30 kN VC= 20KN Mx = -7.5 +15 x 40 x 20 x ² MC=0 portion cd ( from right - to test) Vx = -20+20x ID = -20 kN Mx = 20 x 20 x² Mo=0 ME = 10kN-m Deffected shape Deflected shape i t-sag * Hog at point 'G' and 'c BM =0 saga so curve changes its curvature.
i have explained each and every aspect. Thank you