Question

Problem 4. (65 points). For the beam shown, calculate the Reactions at A and B, and then draw the Shear and Bending-Moment diagrams. Consider the loading P= 29 kN/m. P. D | В 40 KN 40 KN ton 0.6 m 1.8 m 0.6 m

Answer 1

29KN/m A C B 1 HORN 40KN a RA RB let RA and RB be the reactions developed at A and B respectively. EMAIO (Summation of moments about > RBX3 -40%(0.6+1.8) -39x3.8x10.6+1.8 २ - 40 X 0.6 =

Х x 29KN/m A C To B HOKN HOKN a RA RB Х

S at A =RA - 66.1 KN SF just to the left ofc - 26.1 +40 = 66.1 KN A and C SF between = 66.1 KN BM calculation BM at B=0 [As it is a simply Supported and BM between D and B = 66.12 (where' n'vis measured from B BM at D = 66.1% 0.6 = 39.66 kNm BM between cand D -40 (x-0.6) 29X(22-0.6)x(2-0.0) = 66.1 x 2. = -14.52² +43.5% + 18.78 BM at C =-14.5x(2.4 +43.5X2.4+18.78 22=2.4) = 39.66 KN m

BM between A and C= 66.1x2 - 40(x-0.6) -29x1.844,8 +(2-2.6) -40 (2-2.4 =-66.12 +198.3 BM at A= - 66.1 x 3 +198-3 29 KN C B AA 140KN HOKNE 66.1 + 26.1 + SED 26.1 66.1 51.405 BMD 39.66 + 39.66 + + +
SF Diagram is constant between A&C and D&B and linearly varying between C&D.

BM Diagram is linearly varying between A&C and D&B and parabolically varying between C&D.