§3.4 Shearing Force and Bending Moment Diagrams 47 When evaluating B.M.'s it is assumed that a u.d.I.can be replaced by a concentrated load of equal value acting at the middle of its spread.When taking moments about C,therefore,the portion of the u.d.1.between A and C has an effect equivalent to that of a concentrated load of 25 x 2=50 kN acting the centre of AC,i.e.1m from C. B.M.atC=(RA×2)-(50×1)=300-50=250kNm Similarly,for moments at D the u.d.I.on AD can be replaced by a concentrated load of 25 x 4 100kN at the centre of AD,i.e.at C. B.M.atD=(RA×4)-(100×2)=600-200=400kNm Similarly, B.M.atE=(RA×6-(25×6)3=900-450=450kNm The B.M.diagram will be symmetrical about the beam centre line;therefore the values of B.M.at F and G will be the same as those at D and C respectively.The final diagram is therefore as shown in Fig.3.8 and is parabolic. Point (a)of the summary is clearly illustrated here,since the B.M.is a maximum when the S.F.is zero.Again,the reason for this will be shown later. 3.4.S.F.and B.M.diagrams for combined concentrated and uniformly distributed loads Consider the beam shown in Fig.3.9 loaded with a combination of concentrated loads and u.d.I.s. Taking moments about E (R4×8)+(40×2)=(10×2×7)+(20×6)+(20×3)+(10×1)+(20×3×1.5) 8R4+80=420 R=42.5kN(=S.F.at A) Now R4+RE=(10×2)+20+20+10+(20×3)+40=170 RE=127.5kN Working from the left-hand support it is now possible to construct the S.F.diagram,as indicated previously,by following the direction arrows of the loads.In the case of the u.d.I.'s the S.F.diagram will decrease gradually by the amount of the total load until the end of the u.d.I.or the next concentrated load is reached.Where there is no u.d.I.the S.F.diagram remains horizontal between load points. In order to plot the B.M.diagram the following values must be determined: B.M.at A 0 B.M.atB=(42.5×2)-(10×2×1)=85-20 65kNm B.M.atC=(42.5×5)-(10×2×4)-(20×3)=212.5-80-60 =72.5kNm B.M.atD=(42.5×7)-(10×2×6)-(20×5)-(20×2) -(20×2×1)=297.5-120-100-40-40=297.5-300=-2.5kNm B.M.at E (-40 x 2)working from r.h.s. =-80kNm B.M.at F =0$3.4 Shearing Force and Bending Moment Diagrams 47 When evaluating B.M.’s it is assumed that a u.d.1. can be replaced by a concentrated load of equal value acting at the middle of its spread. When taking moments about C, therefore, the portion of the u.d.1. between A and C has an effect equivalent to that of a concentrated load of 25 x 2 = 50 kN acting the centre of AC, i.e. 1 m from C. B.M. at C = (RA x 2)- (50 x 1) = 300-50 = 250kNm Similarly, for moments at D the u.d.1. on AD can be replaced by a concentrated load of 25 x 4 = 100 kN at the centre of AD, i.e. at C. B.M. at D = (R A x 4) - ( 100 x 2) = 600 - 200 = 400 kN m B.M. at E = (RA x 6)- (25 x 6)3 = 900-450 = 450kNm The B.M. diagram will be symmetrical about the beam centre line; therefore the values of B.M. at F and G will be the same as those at D and C respectively. The final diagram is therefore as shown in Fig. 3.8 and is parabolic. Point (a) of the summary is clearly illustrated here, since the B.M. is a maximum when the S.F. is zero. Again, the reason for this will be shown later. Similarly, 3.4. S.F. and B.M. diagrams for combined concentrated and uniformly distributed loads Consider the beam shown in Fig. 3.9 loaded with a combination of concentrated loads and u.d.1.s. Taking moments about E (RA x 8) + (40 x 2) = (10 x 2 x 7) + (20 x 6) + (20 x 3) + (10 x 1) + (20 x 3 x 1.5) 8RA + 80 = 420 R A = 42.5 kN ( = S.F. at A) Now RA+RE = (10 x 2)+20+20 + 10+ (20 x 3)+40 = 170 RE = 127.5 kN Working from the left-hand support it is now possible to construct the S.F. diagram, as indicated previously, by following the direction arrows of the loads. In the case of the u.d.l.’s the S.F. diagram will decrease gradually by the amount of the total load until the end of the u.d.1. or the next concentrated load is reached. Where there is no u.d.1. the S.F. diagram remains horizontal between load points. In order to plot the B.M. diagram the following values must be determined: B.M. at A =o B.M. at B = B.M. at C = B.M. at D = B.M. at E = B.M. at F =o (42.5 x 2) - (10 x 2 x 1) = 85 - 20 (42.5 x 5) - (10 x 2 x 4) - (20 x 3) = 212.5 - 80 - 60 (42.5 x 7) - (10 x 2 x 6) - (20 x 5) - (20 x 2) (- 40 x 2) working from r.h.s. = 65kNm = 72.5 kNm - (20 x 2 x 1) = 297.5- 120- 100 -40-40 = 297.5 - 300 = -2.5 kNm = -80kNm