46 Mechanics of Materials §3.3 S.F.)is then the reaction of 10 kN upwards,i.e.positive,and the bending moment reaction ×zero distance=zero. The following characteristics of the two diagrams are now evident and will be explained later in this chapter: (a)between B and C the S.F.is zero and the B.M.remains constant; (b)between A and B the S.F.is positive and the slope of the B.M.diagram is positive;vice versa between E and F; (c)the difference in B.M.between A and B =20kN m area of S.F.diagram between A and B. 3.3.S.F.and B.M.diagrams for uniformly distributed loads Consider now the simply supported beam shown in Fig.3.8 carrying a u.d.1.w=25 kN/m across the complete span. 25 kN/m A D E G eocoxcccoooxxcoooceeexxoeeeeccccco -12 m- 150 S.F.diagram (kN) 50 B.M.diagram (kN m) 450 400 400 250 250 Fig.3.8. Here again it is necessary to evaluate the reactions,but in this case the problem is simplified by the symmetry of the beam.Each reaction will therefore take half the applied load, i.e. RA=RB=- 2=150kN 25×12 The S.F.at A,using the usual sign convention,is therefore +150kN. Consider now the beam divided into six equal parts 2m long.The S.F.at any other point C is,therefore, 150-load downwards between A and C =150-(25×2)=+100kN The whole diagram may be constructed in this way,or much more quickly by noticing that the S.F.at A is +150kN and that between A and B the S.F.decreases uniformly,producing the required sloping straight line,shown in Fig.3.7.Alternatively,the S.F.at A is +150kN and between A and B this decreases gradually by the amount of the applied load (i.e.by 25×12=300kN)to-150 kN at B
46 Mechanics of Materials 53.3 S.F.) is then the reaction of 10 kN upwards, Le. positive, and the bending moment = reaction x zero distance = zero. The following characteristics of the two diagrams are now evident and will be explained later in this chapter: (a) between B and C the S.F. is zero and the B.M. remains constant; (b) between A and B the S.F. is positive and the slope of the B.M. diagram is positive; vice (c) the difference in B.M. between A and B = 20 kN m = area of S.F. diagram between A versa between E and F; and B. 3.3. S.F. and B.M. diagrams for uniformly distributed loads Consider now the simply supported beam shown in Fig. 3.8 carrying a u.d.1. w = 25 kN/m across the complete span. 25 kN/rn A C D E FG 0 RA R.2 I” 150 I50 0.M. dmgrorn (kN rn) 450 Fig. 3.8. Here again it is necessary to evaluate the reactions, but in this case the problem is simplified by the symmetry of the beam. Each reaction will therefore take half the applied load, i.e. 25 x 12 RA= Rs= ~ - 150 kN 2 The S.F. at A, using the usual sign convention, is therefore + 150kN. is, therefore, Consider now the beam divided into six equal parts 2 m long. The S.F. at any other point C 150 - load downwards between A and C = 150 - (25 x 2) = + 100 kN The whole diagram may be constructed in this way, or much more quickly by noticing that the S.F. at A is + 150 kN and that between A and B the S.F. decreases uniformly, producing the required sloping straight line, shown in Fig. 3.7. Alternatively, the S.F. at A is + 150 kN and between A and B this decreases gradually by the amount of the applied load (Le. by 25 x 12 = 300kN) to - 150kN at B
§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