§3.7 Shearing Force and Bending Moment Diagrams 51 Since this will tend to lift the beam at A,R acts downwards. Moments about B: R4×L=Fd Fd RAL Fd and for vertical equilibrium Ra=RA=L The S.F.diagram can now be drawn as the horizontal loads have no effect on the vertical shear. The B.M.at any section between A and C is M=-Rx=二Fdx -Fd Thus the value of the B.M.increases linearly fromero at Ato at C. Similarly,the B.M.at any section between C and B is Fd M=-R4x+Fd=Rx=元X Fd, i.e.the value of the B.M.again increases linearly from zero at B to L -b at C.The B.M. diagram is therefore as shown in Fig.3.12. Type(b):moment applied with vertical loads Consider the beam AB shown in Fig.3.13;taking moments about B: RL=F(d+b) F(d+b) R4= L F(a-d) Similarly, RB=- L The S.F.diagram can therefore be drawn as in Fig.3.13 and it will be observed that in this case F does affect the diagram. For the B.M.diagram an equivalent system is used.The offset load F is replaced by a moment and a force acting at C,as shown in Fig.3.13.Thus B.M.between A and C=Rx =F(d+b) L F(d+b) i.e.increasing linearly from zero to La at C.$3.7 Shearing Force and Bending Moment Diagrams 51 Since this will tend to lift the beam at A, R, acts downwards. Moments about B: and for vertical equilibrium The S.F. diagram can now shear. R, x L = Fd Fd R, = - L R =R =- Fd L3 ,L be drawn as the horizontal loads have no effect on the vertical The B.M. at any section between A and C is - Fd L Thus the value of the B.M. increases linearly from zero at A to - a at C. Similarly, the B.M. at any section between C and B is Fd L M = -R,x+Fd= R&=-x' Fd L i.e. the value of the B.M. again increases linearly from zero at B to - b at C. The B.M. diagram is therefore as shown in Fig. 3.12. Type (b): moment applied with vertical loads Consider the beam AB shown in Fig. 3.13; taking moments about B: R,L = F(d+b) F(d+b) L .. R, = ____ Similarly, F(a-d) L R,=--- The S.F. diagram can therefore be drawn as in Fig. 3.13 and it will be observed that in this For the B.M. diagram an equivalent system is used. The offset load F is replaced by a case F does affect the diagram. moment and a force acting at C, as shown in Fig. 3.13. Thus B.M. between A and C = R,x X F(d+b) L =--- i.e. increasing linearly from zero to ~ a at C. F(d+b) L