Shearing Force and Bending Moment Diagrams 57 and since R+Rg=5+7+2+(4×5)=34 R4=34-19=15kN The S.F.diagram may now be constructed as described in $3.4 and is shown in Fig.3.17. Calculation of bending moments B.M.at A and C=0 B.M.at B =-2×1=-2kNm B.M.at D =-(2×2)+(19×1)-(4×1×)=+13kNm B.M.at E =+(15×1)-(4×1×)=+13kNm The maximum B.M.will be given by the point (or points)at which dM/dx (i.e.the shear force)is zero.By inspection of the S.F.diagram this occurs midway between D and E,i.e.at 1.5m from E. BM.=25×1-6x15)-（4x25×2) =+17.5kNm There will also be local maxima at the other points where the S.F.diagram crosses its zero axis,i.e.at point B. Owing to the presence of the concentrated loads(reactions)at these positions,however, these will appear as discontinuities in the diagram;there will not be a smooth contour change. The value of the B.M.s at these points should be checked since the position of maximum stress in the beam depends upon the numerical maximum value of the B.M.;this does not necessarily occur at the mathematical maximum obtained above. The B.M.diagram is therefore as shown in Fig.3.17.Alternatively,the B.M.at any point between D and E at a distance of x from A will be given by 4x2 Ma=15x-5k-)-2=10x+5-2x2 dM The maximum B.M.position is then given where- 0. dM d=10-4x=0 ..x=2.5m 1.e. 1.5m from E,as found previously. (b)Since the B.M.diagram only crosses the zero axis once there is only one point of contraflexure,i.e.between B and D.Then,B.M.at distance y from C will be given by Mw=-2y+19(y-1)-4(y-1)(y-1) =-2y+19y-19-2y2+4y-2=0 The point of contraflexure occurs where B.M.=0,i.e.where Myy=0, 0=-2y2+21y-21Shearing Force and Bending Moment Diagrams 57 and since R,+R,= 5+7+2+(4~5)= 34 RA=34-19=15kN The S.F. diagram may now be constructed as described in 43.4 and is shown in Fig. 3.17. Calculation of bending moments B.M. at A and C = 0 B.M. at B B.M. at D B.M. at E = -2x 1 = -2kNm = -(2~2)+(19~1)-(4xlxi)= +13kNm = +(15~1)-(4xlx~)= +13kNm The maximum B.M. will be given by the point (or points) at which dM/dx (Le. the shear force) is zero. By inspection of the S.F. diagram this occurs midway between D and E, i.e. at 1.5 m from E. B.M. at this point = (2.5 x 15) - (5 x 1.5) - 4 x 2.5 x - ( 25) = + 17.5 kNm There will also be local maxima at the other points where the S.F. diagram crosses its zero axis, i.e. at point B. Owing to the presence of the concentrated loads (reactions) at these positions, however, these will appear as discontinuities in the diagram; there will not be a smooth contour change. The value of the B.M.s at these points should be checked since the position of maximum stress in the beam depends upon the numerical maximum value of the B.M.; this does not necessarily occur at the mathematical maximum obtained above. The B.M. diagram is therefore as shown in Fig. 3.17. Alternatively, the B.M. at any point between D and E at a distance of x from A will be given by 42 2 M,,= 15~-5(~-1)--= 1Ox+5-2x2 dM dx The maximum B.M. position is then given where - = 0. x = 2.5m i.e. 1.5m from E, as found previously. (b) Since the B.M. diagram only crosses the zero axis once there is only one point of contraflexure, i.e. between B and D. Then, B.M. at distance y from C will be given by My, = - 2y + 19(y - 1) - 4(y - 1)i (y - 1) = -2~~+19y-19-2~~+4~-2 =O The point of contraflexure occurs where B.M. = 0, i.e. where My, = 0, .. 0 = -2yz+21y-21