$3.2 Shearing Force and Bending Moment Diagrams 45 0x3 20x Rx7-210 40 kN m 40kN m Rx5=5020x,30x5 x Fig.3.6.Total B.M.at X-X. B.M.at A =0 B.M.atB=+(10×2) =+20kNm B.M.atC=+(10×4)-(10×2) =+20kNm B.M.atD=+(10×6)+(20×2)-(10×4)=+60kNm B.M.atE=+(30×2) =+60kNm B.M.at F =0 All the above values have been calculated from the moments of the forces to the left of each section considered except for E where forces to the right of the section are taken. IO kN 20kN 30 kN 10kN 30 kN 20 kN S.F diagram (kN) 20 10 -30 60 B.M.diagram (kN m) 20 Fig.3.7. It may be observed at this stage that the S.F.diagram can be obtained very quickly when working from the left-hand side,since after plotting the S.F.value at the support all subsequent steps are in the direction of and equal in magnitude to the applied loads,e.g. 10kN up at A,down 10kN at B,up 20kN at C,etc.,with horizontal lines joining the steps to show that the S.F.remains constant between points of application of concentrated loads. The S.F.and B.M.values at the left-hand support are determined by considering a section an infinitely small distance to the right of the support.The only load to the left (and hence the$3.2 Shearing Force and Bending Moment Diagrams 45 R,x5=50 20x1, 30x5 I Ix 'X Fig. 3.6. Total B.M. at X-X. B.M. at A =o B.M. at B = + (10 x 2) = +20kNm B.M.atC= +(lOx4)-(1Ox2) = +20kNm B.M. at D = +(lox 6)+ (20 x 2)- (10 x 4) = +60kNm B.M. at E = + (30 x 2) = +60kNm B.M. at F =o All the above values have been calculated from the moments of the forces to the left of each section considered except for E where forces to the right of the section are taken. 10 Fig. 3.1. It may be observed at this stage that the S.F. diagram can be obtained very quickly when working from the left-hand side, since after plotting the S.F. value at the support all subsequent steps are in the direction of and equal in magnitude to the applied loads, e.g. 10 kN up at A, down 10 kN at B, up 20 kN at C, etc., with horizontal lines joining the steps to show that the S.F. remains constant between points of application of concentrated loads. The S.F. and B.M. values at the left-hand support are determined by considering a section an infinitely small distance to the right of the support. The only load to the left (and hence the